problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
Let $P(x) = x^2-1$ be a polynomial, and let $a$ be a positive real number satisfying $$ P(P(P(a))) = 99. $$ The value of $a^2$ can be written as $m+\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ .
*Proposed by **HrishiP*** | <details><summary>Solution</summary>First, we obtain $P(a)=a^2-1$ . Upon plugging in this value into the polynomial again, we obtain $$ P(P(a))=(a^2-1)^2-1=(a^2-1+1)(a^2-1-1)=a^2(a^2-2)=a^4-2a^2. $$ Finally, upon plugging in this value into the polynomial again, we obtain
\begin{align*}
P(P(P(a)))&=(a^4-2a^2)^2-1
... | [
"alternatively you can just note that the answer is constant so we just need to find one value of $a$ that works\n\nwe have $P(P(a))=10$ so that $a^2-1=\\sqrt{11}\\rightarrow \\boxed{012}$ ",
"Wait yes, working backwards here is probably cleaner.",
"i did it a diff way $a^2=x$ so you get $x^4-4x^3+4x^2-1... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1028,
"boxed": true,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 DIME/2673239.json"
} |
In $\triangle ABC$ with $AC>AB$ , let $D$ be the foot of the altitude from $A$ to side $\overline{BC}$ , and let $M$ be the midpoint of side $\overline{AC}$ . Let lines $AB$ and $DM$ intersect at a point $E$ . If $AC=8$ , $AE=5$ , and $EM=6$ , find the square of the area of $\triangle ABC$ .
*Propo... | Interesting problem!
Trivially $AM=MC=DM=4$ when drawing the circumcircle of $\triangle ADC$ , so $D=2$ . Applying LoC on $\triangle AME$ we get $25=36+16-48 \cos(\angle AME) \implies \frac{9}{16} = \cos(\angle AME)$ Applying LoC now on $\triangle AMD$ we get $AD^2 = 32-32 (\tfrac{9}{16}) = 14 \implies AD=\... | [
"MD=4, DE=2, then do some area ratios and heron bash to get the answer",
"oops I fakesolved by using the fact that aime answers are only integers. :blush: ",
"<blockquote>oops I fakesolved by using the fact that aime answers are only integers. :blush:</blockquote>\n\nHuh? Now I'm curious as to how you did it.... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1038,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 DIME/2673241.json"
} |
Let $a_1,a_2,\ldots,a_6$ be a sequence of integers such that for all $1 \le i \le 5$ , $$ a_{i+1}=\frac{a_i}{3} \quad \text{or} \quad a_{i+1}={-}2a_i. $$ Find the number of possible positive values of $a_1+a_2+\cdots+a_6$ less than $1000$ .
*Proposed by **stayhomedomath*** | AHHHH HOW IS EVERYONE SO SMART
For this problem I basically spent an hour bashing, first mapping a triangle that had all possible values of each set of $a_n$ terms for $1 \le n \le 6$ , of course setting the initial value to $x$ . Then, I spend EVEN LONGER finding all possible values of sums with making an EVEN LA... | [
"the sequence 27, -54, -18, 36, 12, 4 and all its multiples gives 7x => 142, but i cant figure out how to explicitly prove",
"Note that ${-}2 \\equiv \\tfrac{1}{3} \\equiv 5 \\pmod{7}$ , so $$ a_1+a_2+\\cdots + a_6 \\equiv a_1(1+5+5^2+\\cdots+5^5) \\equiv 0 \\pmod{7}, $$ which means that $a_1+a_2+\\cdots+a_... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1010,
"boxed": true,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 DIME/2673242.json"
} |
Given a parallelogram $ABCD$ , let $\mathcal{P}$ be a plane such that the distance from vertex $A$ to $\mathcal{P}$ is $49$ , the distance from vertex $B$ to $\mathcal{P}$ is $25$ , and the distance from vertex $C$ to $\mathcal{P}$ is $36$ . Find the sum of all possible distances from vertex $D$ to ... | Since $ABCD$ must be a parallelogram, the plane $\mathcal{P}$ will always be fixed for some configuration of $ABCD$ . Thus, we fix the plane $\mathcal{P}$ and then proceed.
Let the notation $\{\mathcal{F} \}\{\mathcal{B} \} \in \mathcal{P}$ denote the set of all points relative to $\mathcal{P}$ such that $... | [
"let the z-coordinate for $A$ be $z_A$ , and similarly for the other points\n\nthen we have $|z_A|=49,|z_B|=25,|z_C|=36$ .\n\nnotice that because $ABCD$ is a parallelogram, we have $z_A+z_D=z_B+z_C\\rightarrow z_D=z_B+z_C-z_A$ .\n\nthe possible values for $z_B+z_C$ are $-61,-11,11,61$ and the possible va... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1062,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 DIME/2673243.json"
} |
Richard has an infinite row of empty boxes labeled $1, 2, 3, \ldots$ and an infinite supply of balls. Each minute, Richard finds the smallest positive integer $k$ such that box $k$ is empty. Then, Richard puts a ball into box $k$ , and if $k \geq 3$ , he removes one ball from each of boxes $1,2,\ldots,k-2$ . F... | Simple recursion suffices. Let each marble be labeled $1,2,3 \cdots $ such that for every $n$ th minute, the marble with label $a$ is placed into a box such that there exists no marbles with labels $b<a$ . Let the function $f(c)$ denote the number of the marble that was placed into box $c$ first. We get $$ ... | [
"clearly the first time that box $10$ is filled is when we have $$ 1111111110\\rightarrow 0000000011 $$ which means that box $9$ will also be filled, so we just need to smallest positive integer $n$ such that after $n$ minutes box $10$ is filled.\n\nlet $f(n)$ be the smallest positive integer such t... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1056,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 DIME/2673244.json"
} |
Let $a$ and $b$ be real numbers such that $$ \left(8^a+2^{b+7}\right)\left(2^{a+3}+8^{b-2}\right)=4^{a+b+2}. $$ The value of the product $ab$ can be written as $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
*Proposed by **stayhomedomath*** | Everything was right until somehow I got $a=\tfrac{29}{4}$ ....
First off, make everything base $2$ to get $$ \left(2^{3a} + 2^{b+7} \right)\left(2^{a+3} + 2^{3b-6} \right) = 2^{2a+2b+4} $$ Clearly the RHS is a power of $2$ , which means both expressions in the LHS product are also powers of $2$ .
Now, supp... | [
"the LHS is $(2^{3a}+2^{b+7})(2^{a+3}+2^{3b-6})=4^{a+b+2}$ , this motivates us to set $3a=b+7$ and $a+3=3b-6$ giving $(a,b)=(\\frac{15}{4},\\frac{17}{4})$ . this works so the answer is $\\frac{255}{16}\\rightarrow \\boxed{271}$ .",
"wait i don't really understand the motivation for the above solution\n\nes... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1050,
"boxed": false,
"end_of_proof": false,
"n_reply": 16,
"path": "Contest Collections/2022 Contests/2022 DIME/2673245.json"
} |
A positive integer $n$ is called $\textit{un-two}$ if there does not exist an ordered triple of integers $(a,b,c)$ such that exactly two of $$ \dfrac{7a+b}{n},\;\dfrac{7b+c}{n},\;\dfrac{7c+a}{n} $$ are integers. Find the sum of all un-two positive integers.
*Proposed by **stayhomedomath*** | If $n \equiv 0 \pmod{7}$ it's not hard to see that $n$ is not un-two. Otherwise, suppose we have $7a + b \equiv 0 \pmod{n}$ and $7b+c \equiv 0 \pmod{n}$ which means that $49a \equiv c \pmod{n}.$ We know $n$ is un-two if and only if $7c+a \equiv 344a \equiv 0 \pmod{n}$ for all $a$ so we just sum up all f... | [
"<details><summary>Solution</summary>Set $a=1,b=-7,c=49$ to get $$ \\frac{7a+b}{n}=0,\\frac{7b+c}{n}=0,\\frac{7c+a}{n}=\\frac{344}{n}. $$ Then clearly we must have $n\\mid 344$ . <insert explanation that i got during the test but forgot> it is sufficient so we have $(15)(44)=660$ .</details>"
] | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1020,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 DIME/2673246.json"
} |
A sequence of polynomials is defined by the recursion $P_1(x) = x+1$ and $$ P_{n}(x) = \frac{(P_{n-1}(x)+1)^5 - (P_{n-1}(-x)+1)^5}{2} $$ for all $n \geq 2$ . Find the remainder when $P_{2022}(1)$ is divided by $1000$ .
*Proposed by **treemath*** | <details><summary>Author's/Official Solution</summary>The recursion is applying the second roots of unity filter to $(P_n(x)+1)^5$ that only returns terms with odd degree. For the first step, $P_2(x)$ is the odd-degree terms of $(x+2)^5$ , which is $x^5+40x^3+80x$ . For all subsequent steps, we again apply the se... | [
"the only thing i was able to get on the test was $P_{n-1}(x)=-P_{n-1}(-x)$ for all $n\\geq 2$ giving $$ P_{n+1}(x)=\\frac{(P_n(x)+1)^5+(P_n(x)-1)^5}{2} $$ for all $n\\geq 2$ ",
"Try showing that $$ P_{n+1}(x)=(P_n(x))^5+10(P_n(x))^3+5(P_n(x)) $$ for all $n \\geq 2$ .\n\nNow, use CRT with mod $125$... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1070,
"boxed": true,
"end_of_proof": false,
"n_reply": 27,
"path": "Contest Collections/2022 Contests/2022 DIME/2673247.json"
} |
A spinner has five sectors numbered ${-}1.25$ , ${-}1$ , $0$ , $1$ , and $1.25$ , each of which are equally likely to be spun. Ryan starts by writing the number $1$ on a blank piece of paper. Each minute, Ryan spins the spinner randomly and overwrites the number currently on the paper with the number multiplied ... | Cool problem!
<details><summary>Sol</summary>Let $a_k$ denote the $k^{\text{th}}$ term in our sequence of numbers, where the $0^{\text{th}}$ term is $1$ .
Now let's figure out the probability that $a_n$ is the last positive term in the sequence for a given $n$ . Following this positive we can have:
A $0... | [
"PROBABLY WRONG! but here goes\n\n<details><summary>eek</summary>Let $f(n)=\\text{E}[\\text{max}(\\text{largest number Ryan writes if he starts with the number }n,1)]$ . Then $$ f(1)=\\frac{f(-1.25)+f(-1)+f(0)+f(1)+f(1.25)}{5} $$ or $$ f(1)=\\frac{f(-1.25)+f(-1)+1+f(1.25)}{4}. $$ Notice that $f(1.25)=1.25... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1120,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 DIME/2673248.json"
} |
Let $\triangle ABC$ be acute with $\angle BAC = 45^{\circ}$ . Let $\overline{AD}$ be an altitude of $\triangle ABC$ , let $E$ be the midpoint of $\overline{BC}$ , and let $F$ be the midpoint of $\overline{AD}$ . Let $O$ be the center of the circumcircle of $\triangle ABC$ , let $K$ be the intersection ... | <details><summary>Solution?</summary>Reflect $O$ over side $\overline{BC}$ to a point $P$ . Due to reflections, we have that $E$ is the midpoint of $\overline{OP}$ , and we are given that $E$ is also the midpoint of $\overline{BC}$ , so $BOCP$ is a parallelogram. We also have that $\angle BAC = 45^{\circ}... | [
"it suffices to show that B, O, L, C concyclic and i half guessed this during the test but basically BLC is right by 6-8-10 (very cringe) and then BOC is 90 cuz BAC = 45 so B O L C concyclic. now it's simple by pythag and ptolemy => 48",
"official sol?",
"Guessed $50$ :wallbash: ",
"<blockquote>official so... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1076,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 DIME/2673249.json"
} |
For positive integers $n$ , let $f(n)$ denote the number of integers $1 \leq a \leq 130$ for which there exists some integer $b$ such that $a^b-n$ is divisible by $131$ , and let $g(n)$ denote the sum of all such $a$ . Find the remainder when $$ \sum_{n = 1}^{130} [f(n) \cdot g(n)] $$ is divided by $13... | <details><summary>Author's Solution</summary>Let all sums in this solution be taken modulo $131$ , and let $\text{ord}_{131}(x)$ be the order of $x$ modulo $131$ .
We claim that there exists an integer $s$ such that $r^{s} \equiv x \pmod{131}$ if and only if $\text{ord}_{131}(x) \mid \text{ord}_{131}(r)$ . ... | [
"any solutions to this?",
"i think its smt with primitive roots",
"sol pls?",
"neither do I @above\n\nyou guys rlly should've made a DJMO that included this problem",
"yeah thats a rly hard order problem\n\n",
"<blockquote>neither do I @above\n\nyou guys rlly should've made a DJMO that included this probl... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1226,
"boxed": false,
"end_of_proof": false,
"n_reply": 24,
"path": "Contest Collections/2022 Contests/2022 DIME/2673252.json"
} |
Let $f:\mathbb{N}^*\rightarrow \mathbb{N}^*$ be a function such that $\frac{x^3+3x^2f(y)}{x+f(y)}+\frac{y^3+3y^2f(x)}{y+f(x)}=\frac{(x+y)^3}{f(x+y)},~(\forall)x,y\in\mathbb{N}^*.$ $a)$ Prove that $f(1)=1.$ $b)$ Find function $f.$ | Giả sử tồn tại hàm số thỏa mãn $$ \frac{x^3+3x^2f(y)}{x+f(y)}+\frac{y^3+3y^2f(x)}{y+f(x)}=\frac{(x+y)^3}{f(x+y)},\forall x,y\in \mathbb{Z^+}.(1) $$ Từ $(1)$ cho $x=y=1$ ta được $$ \frac{2(1+3f(1))}{1+f(1)}=\frac{8}{f(2)} $$ Hay ta được $$ f(2)=\frac{4(1+f(1))}{1+3f(1)}. $$ Suy ra $(1+f(1))|(4+4f(1))$ hay... | [
"Let $a=f(1)$ . $$ (x,y)=(1,1) \\rightarrow \\dfrac{2(1+3a)}{1+a}=\\dfrac{8}{f(2)} \\rightarrow 1+3a\\; |\\; 4a+4 \\rightarrow f(1)=a=1. $$ Using induction with the relation $(x,1)$ we deduce that $f(x)=x$ ."
] | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 48,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810677.json"
} |
$a)$ Prove that $2x^3-3x^2+1\geq 0,~(\forall)x\geq0.$ $b)$ Let $x,y,z\geq 0$ such that $\frac{2}{1+x^3}+\frac{2}{1+y^3}+\frac{2}{1+z^3}=3.$ Prove that $\frac{1-x}{1-x+x^2}+\frac{1-y}{1-y+y^2}+\frac{1-z}{1-z+z^2}\geq 0.$ | $a)$ The expression factorizes as $(x-1)^2(2x+1)$ which is clearly positive for $x\geq 0$ . Moreover, equality occurs for $x=1$ . $b)$ Using the result from point $a)$ , we get $1-x^3\leq \frac{3}{2}(1-x^2)$ , thus $$ 3=\sum_{cyc}\frac{2}{1+x^3}\implies 0=\sum_{cyc}\frac{1-x^3}{1+x^3}\leq \frac{3}{2}\sum_{cy... | [
"Let $x,y,z\\geq 0$ such that $\\frac{2}{1+x^3}+\\frac{2}{1+y^3}+\\frac{2}{1+z^3}=3.$ Prove that $$ \\frac{2-x}{2-x+x^2}+\\frac{2-y}{2-y+y^2}+\\frac{2-z}{2-z+z^2}\\leq \\frac{3}{2} $$ \n"
] | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
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"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810679.json"
} |
$a)$ Solve over the positive integers $3^x=x+2.$ $b)$ Find pairs $(x,y)\in\mathbb{N}\times\mathbb{N}$ such that $(x+3^y)$ and $(y+3^x)$ are consecutive. | Part a) $x=1$ only solution. For $x>1, LHS>RHS$ Part b)
If $(x+3^y)$ and $(y+3^x)$ are consecutive $\Longrightarrow x+3^y+1= y+3^x$ Case $x=y$ $\Longrightarrow x+3^x+1= x+3^x \Longrightarrow 1= 0$ impossible
Case $y>x$ Let $y=x+n\Longrightarrow x+3^n3^x+1=x+n+3^x\Longrightarrow 3^x=\frac{n-1}{3^n-1}<1$ ... | [] | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810681.json"
} |
Let $e$ be the identity of monoid $(M,\cdot)$ and $a\in M$ an invertible element. Prove that
[list=a]
[*]The set $M_a:=\{x\in M:ax^2a=e\}$ is nonempty;
[*]If $b\in M_a$ is invertible, then $b^{-1}\in M_a$ if and only if $a^4=e$ ;
[*]If $(M_a,\cdot)$ is a monoid, then $x^2=e$ for all $x\in M_a.$ [/list... | Notice that the set can be redefined as $M_a=\{x\in M : x^2=a^{-2}\}$ . $a)$ Observe that $a^{-1}\in M_a$ . $b)$ Since $b\in M_a$ , we have $b^2=a^{-2}$ , thus $$ b^{-1}\in M_a\iff b^{-2}=a^{-2}\iff e=b^2b^{-2}=a^{-2}a^{-2}=a^{-4}\iff a^4=e. $$ $c)$ Let $f$ be the identity of $M_a$ . $f\in M_a$ , thus $... | [] | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810712.json"
} |
Let $(G,\cdot)$ be a group and $H\neq G$ be a subgroup so that $x^2=y^2$ for all $x,y\in G\setminus H.$ Show that $(H,\cdot)$ is an Abelian group. | Let $x \in G \setminus H.$ For any $a \in H,$ we have $ax \notin H$ and so $(ax)^2=x^2,$ which gives $axa=x$ hence $$ a=xa^{-1}x^{-1}, \ \ \ \ \ \forall a \in H. $$ So if $a,b \in H,$ then $$ xb^{-1}a^{-1}x^{-1}=x(ab)^{-1}x^{-1}=ab=(xa^{-1}x^{-1})(xb^{-1}x^{-1})=xa^{-1}b^{-1}x^{-1}, $$ which gives $b... | [] | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810713.json"
} |
Find all values of $n\in\mathbb{N}^*$ for which \[I_n:=\int_0^\pi\cos(x)\cdot\cos(2x)\cdot\ldots\cdot\cos(nx) \ dx=0.\] | [
"[url=https://artofproblemsolving.com/community/c7h1646205p10391941] here [ \\url ]\n\nhttps://artofproblemsolving.com/community/c7h1646205p10391941"
] | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810715.json"
} | |
Let $I\subseteq \mathbb{R}$ be an open interval and $f:I\to\mathbb{R}$ a strictly monotonous function. Prove that for all $c\in I$ there exist $a,b\in I$ such that $c\in (a,b)$ and \[\int_a^bf(x) \ dx=f(c)\cdot (b-a).\] | Let $I=(k,l)$ and suppose $f$ is strictly decreasing (otherwise, look at the function $-f$ ). Consider some real number (which we will define later) $y_0\in (c,l)$ and let $g:(k,c)\rightarrow \mathbb{R}$ with $$ g(x)=\int_x^{y_0}f(t)dt - f(c)(y_0-x). $$ Observe that $g(c)<0$ since from the monotonicity o... | [
"Assume $f$ is strictly increasing (otherwise work with $-f$ ). Since $I$ is open, there exists $\\delta>0$ such that $(c-\\delta, c+\\delta) \\subseteq I$ . Fix such a $\\delta$ and consider the function (for $x>c$ )\n\\[\ng(x) = \\int_{c-\\delta}^{x} f(t)\\mathrm{d}t.\n\\]\nSince $f$ is strictly incr... | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810716.json"
} |
Let $f,g:\mathbb{R}\to\mathbb{R}$ be functions which satisfy \[\inf_{x>a}f(x)=g(a)\text{ and }\sup_{x<a}g(x)=f(a),\]for all $a\in\mathbb{R}.$ Given that $f$ has Darboux's Property (intermediate value property), show that functions $f$ and $g$ are continuous and equal to each other.
*Mathematical Gazette* | <blockquote>Let $f,g:\mathbb{R}\to\mathbb{R}$ be functions which satisfy \[\inf_{x>a}f(x)=g(a)\text{ and }\sup_{x<a}g(x)=f(a),\]for all $a\in\mathbb{R}.$ Given that $f$ has Darboux's Property (intermediate value property), show that functions $f$ and $g$ are continuous and equal to each other.
*Mathematical ... | [
"https://ssmr.ro/",
"<blockquote>[ see here ](https://ssmr.ro/) </blockquote>\n\nRomanian maths problems are very nice",
"Just a question, what is the name of this competition, can you attach website or something because these questions are college level yet they are for grade 11, so I want to know. Also i sear... | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810717.json"
} |
Let $A,B\in\mathcal{M}_3(\mathbb{R})$ de matrices such that $A^2+B^2=O_3.$ Prove that $\det(aA+bB)=0$ for any real numbers $a$ and $b.$ | Interesting problem $A^2=-B^2$ take the det then $det(A)^2=-det(B)^2$ ==> det(A)=det(B)=0 (1)
If a=0 or b=0 with (1) ==> det(aA+bB)=0
If a=/=0 and b=/=0 we must prove det(A+xB)=0 for any x in R $Q(x)=det(A+xB)=det(B)x^3 + px^2 + qx +det(A)= px^2 + qx$ where p,q reals numbers $Q(i)Q(-i)=|Q(i)|^2 = (-p+iq)(-p-iq... | [
"What does the $O_3$ mean?",
"zero matrix $ 3\\times 3$ ",
"<blockquote> $A^2=-B^2$ take the det then $det(A)^2=-det(B)^2$ ==> det(A)=det(B)=0 (1)\nIf a=0 or b=0 with (1) ==> det(aA+bB)=0\n\nIf a=/=0 and b=/=0 we must prove det(A+xB)=0 for any x in R $Q(x)=det(A+xB)=det(B)x^3 + px^2 + qx +det(A)= px^2 ... | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810718.json"
} |
Let $(x_n)_{n\geq 1}$ be the sequence defined recursively as such: \[x_1=1, \ x_{n+1}=\frac{x_1}{n+1}+\frac{x_2}{n+2}+\cdots+\frac{x_n}{2n} \ \forall n\geq 1.\]Consider the sequence $(y_n)_{n\geq 1}$ such that $y_n=(x_1^2+x_2^2+\cdots x_n^2)/n$ for all $n\geq 1.$ Prove that
[list=a]
[*] $x_{n+1}^2<y_n/2$ and ... | Here is yet another way of doing part b) without the help of the auxiliary sequence.
Firstly, we will prove that $(x_n)$ is decreasing. Notice that $$ x_{n+1}-x_n = \left(\sum_{k=1}^n \frac{x_k}{n+k}\right) - x_n = \left(\sum_{k=1}^{n-1} \frac{x_k}{n+k}\right) + \frac{x_n}{2n} - x_n = \sum_{k=1}^{n-1} x_k\left(\fr... | [
"<blockquote>\n[list=a]\n[*] $x_{n+1}^2<y_n/2$ \n[/list]</blockquote>\n\nThis fails even for $n=2$ . Check your definition of $y_n$ ...",
"I am very sorry, I made a typo. It's supposed to say $y_n=(x_1^2+x_2^2+\\cdots+x_n^2)/n.$ Fixing it.",
"You can do b without using the auxiliary sequence $(y_n)_n$ .\n... | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 26,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810720.json"
} |
Let $A\in\mathcal{M}_n(\mathbb{C})$ where $n\geq 2.$ Prove that if $m=|\{\text{rank}(A^k)-\text{rank}(A^{k+1})":k\in\mathbb{N}^*\}|$ then $n+1\geq m(m+1)/2.$ | Let $m = \{ d_1 , d_2, \dots, d_m\}$ we have $$ \frac{m(m+1)}{2}=1+2+\cdots + m \leq d_1 +d_2 +\cdots d_m $$ $$ \leq \text{rank}(A^{i_1})-\text{rank}(A^{i_1+1})+\text{rank}(A^{i_2})-\text{rank}(A^{i_2+1})+\cdots +\text{rank}(A^{i_m})-\text{rank}(A^{i_m+1}) $$ $$ \leq \text{rank}(A^{i_1})-\text{rank}(A^{i_m+1... | [] | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810731.json"
} |
We call a set of $6$ points in the plane *splittable* if we if can denote its elements by $A,B,C,D,E$ and $F$ in such a way that $\triangle ABC$ and $\triangle DEF$ have the same centroid.
[list=a]
[*]Construct a splittable set.
[*]Show that any set of $7$ points has a subset of $6$ points which is *not* ... | a). Let $ABC$ and $DEF$ are equilateral triangles with the same circumcircle. Then the center is the centroid.
b). Let $O$ be the origin and $\vec{v}_{P} $ the position vector of the point $P$ . Now if $ABC$ and $DEF$ have the same centroid $G$ then $$ \vec{v}_{A}+\vec{v}_{B}+\vec{v}_{C}=3\vec{v}_{G}=... | [] | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 64,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2811494.json"
} |
Determine all $x\in(0,3/4)$ which satisfy \[\log_x(1-x)+\log_2\frac{1-x}{x}=\frac{1}{(\log_2x)^2}.\] | I claim $x=1/2$ is the only possibility.
We have
\begin{align*}
&\frac{\log_2(1-x)}{\log_2 x} + \log_2(1-x) = \log_2 x + \frac{1}{(\log_2 x)^2}
&\iff \log_2(1-x)\left(\frac{1+\log_2 x}{\log_2 x}\right) = \frac{(\log_2 x+1)((\log_2 x)^2-\log_2 x+1)}{(\log_2 x)^2}.
\end{align*}
If $\log_2 x+1=0$ , that is if $x=1/2... | [] | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2811496.json"
} |
Let $z_1,z_2$ and $z_3$ be complex numbers of modulus $1,$ such that $|z_i-z_j|\geq\sqrt{2}$ for all $i\neq j\in\{1,2,3\}.$ Prove that \[|z_1+z_2|+|z_2+z_3|+|z_3+z_2|\leq 3.\]*Mathematical Gazette* | Joy Ma Tara! :bruce: Let's look at the geometric interpretation of the problem:
<details><summary>New problem</summary>Let $ABC$ be a non obtuse triangle such that all the interior angles are atleast $45^o$ each. Let $D,E,F$ be the midpoints of sides $BC,CA,AB$ respectively. If $O$ is the circumcenter prove t... | [
"This is simple geo. Just use Jensen after introducing antipodes of these points.",
"<blockquote>This is simple geo. Just use Jensen after introducing antipodes of these points.</blockquote>\nExactly; but before doing so, prove that the triangle determined by $z_1,z_2$ and $z_3$ is either acute or right, so J... | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2811498.json"
} |
A positive integer $n\geq 4$ is called *interesting* if there exists a complex number $z$ such that $|z|=1$ and \[1+z+z^2+z^{n-1}+z^n=0.\] Find how many interesting numbers are smaller than $2022.$ | <details><summary>storage</summary><span style="color:#f00">**Claim:-**</span>all $n$ of the form $5\rho-1$ , for some $\rho \in \mathbb{Z^{+}}$ works
<span style="color:#600">pf:-</span>since $|z|=1\implies z\cdot \overline{z}=1$
now taking conjugate of given equation we arise at $z^{n}+z^{n-1}+z^{n-2}+z+1... | [
"We observe: $1$ and $-1$ are not solutions of the equation.\nLet be $z_0=\\cos\\alpha+i\\sin\\alpha$ a solution of the equation, where $\\alpha\\in[0,2\\pi)$ .\nUsing $(\\cos\\alpha+i\\sin\\alpha)^k=\\cos (k\\alpha)+i\\sin (k\\alpha),\\;\\forall k\\in\\mathbb{N}$ , the equation becomes: $1+\\cos\\alpha+\\c... | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 1132,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2811500.json"
} |
Find all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following relationship for all $x,y\in\mathbb{R}:$ \[f(f(y-x)-xf(y))+f(x)=y\cdot(1-f(x)).\] | Let $P(x,y)$ denote the given assertion.
Claim: $f$ is surjective.
Proof: We note that \[f(f(y-x)-xf(y))=y(1-f(x))-f(x)\]
Noting that $f\equiv 1$ is not a solution, set $x$ such that $f(x)\ne1$ . Varying $y$ gives the desired result. $\blacksquare$ Claim: $f$ is injective.
Proof: $P(0,x): f(f(x))+f... | [
"REDACTED",
"<blockquote>Denote $P(x,y)$ as usual be the assertion $P(0,x)\\leftrightarrow f(f(x))+f(0)=x(1-f(0)) \\leftrightarrow f(f(x)) = x(1-f(0)) - f(0)$ From here we get that $f(x)$ is bijective</blockquote>Not yet. What if $f(0)=1?$ ",
"Case 1: $f(0)\\neq 1$ $P(0,y)\\implies f(f(y)) = (1-f(0))y - ... | [
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] | {
"answer_score": 1178,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2811501.json"
} |
Charlotte is playing the hit new web number game, Primle. In this game, the objective is to guess a two-digit positive prime integer between $10$ and $99$ , called the *Primle*. For each guess, a digit is highlighted blue if it is in the *Primle*, but not in the correct place. A digit is highlighted orange if it is ... | <details><summary>Solution</summary>Because we are working with a two digit number and the digit $7$ is not the units digit, then it follows that the *Primle* has a ten's digit of $7$ . The prime numbers in the $70's$ are $71, 73,$ and $79$ . Because when Charlotte put the prime $13$ none of the digits were b... | [
"<details><summary>Solution</summary>The blue $7$ tells us that the number is a prime with $7$ in the tens place. The uncolored numbers tell us that the units digit is not $1, 3$ , or $4$ . The only working prime number that satisfies these conditions can easily be checked as $\\fbox{79}$ .</details>",
"<d... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] | {
"answer_score": 1018,
"boxed": true,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790580.json"
} |
How many ways are there to fill in a $2\times 2$ square grid with the numbers $1,2,3,$ and $4$ such that the numbers in any two grid squares that share an edge have an absolute difference of at most $2$ ?
*Proposed by Andrew Wu* | <details><summary>Solution</summary>Note that the $1$ must be diagonally opposite the $4$ . Then, the $2$ and $3$ can be placed in any way.
Therefore, there are $4$ ways to place the $1$ , $1$ way to place the $4$ after that, and $2$ ways to place the $2$ and $3$ after that, for a total of $$ 4\... | [
"<details><summary>Solution</summary>By symmetry, we can place the $1$ in the top-left corner and then multiply our answer by $4$ for symmetry. This means that the $4$ must go diagonally across from the $1$ since the absolute difference is more than $2$ . Obviously, both orderings of the $2$ and $3$ wi... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] | {
"answer_score": 1026,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790583.json"
} |
The **Collaptz function** is defined as $$ C(n) = \begin{cases} 3n - 1 & n\textrm{~odd}, \frac{n}{2} & n\textrm{~even}.\end{cases} $$
We obtain the **Collaptz sequence** of a number by repeatedly applying the Collaptz function to that number. For example, the Collaptz sequence of $13$ begins with $13, 38, 19, ... | <details><summary>Solution</summary>Hoping that the three smallest positive integers that do not collaptzse are reasonably small, we use brute force:**1:** Obviously, the Collaptz sequence of $1$ contains a $1,$ so $1$ collaptzses**2:** The first two terms of the Collaptz sequence of $2$ are $2,1$ so $2$ al... | [
"<details><summary>Solution</summary>We use brute force with some quick observations. Obviously $1$ to $4$ all do not work by testing. $5$ works because its sequence is $5, 14, 7, 20, 10, 5, ...$ . $6$ doesn't work because it leads into the sequence for $3$ . $7$ works because its sequence leads into th... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] | {
"answer_score": 1078,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790584.json"
} |
Kara rolls a six-sided die, and if on that first roll she rolls an $n$ , she rolls the die $n-1$ more times. She then computes that the product of all her rolls, including the first, is $8$ . How many distinct sequences of rolls could Kara have rolled?
*Proposed by Andrew Wu* | <details><summary>Solution</summary>Since the product of all the rolls is $8,$ the first roll must be a factor of $8$ which can be $1,2,$ or $4$ ( $8$ is out of range).
Obviously, $1$ wouldn't work because then she would have $0$ more rolls, so the product couldn't be $8$ .
If she rolls a $2,$ she has... | [
"<details><summary>sol</summary>We know it must start with a factor of $8$ , so it has to be $1,2,4$ . $1$ clearly doesn't work because we roll the die $0$ times. For $2$ , we have 1 roll so that has to be $4$ . If she rolls a $4$ , she has 3 rolls to get a product of $2$ . This has to be $2,1,1$ in som... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] | {
"answer_score": 1036,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790586.json"
} |
Cat and Claire are having a conversation about Cat's favorite number.
Cat says, "My favorite number is a two-digit positive integer with distinct nonzero digits, $\overline{AB}$ , such that $A$ and $B$ are both factors of $\overline{AB}$ ."
Claire says, "I don't know your favorite number yet, but I do k... | <details><summary>Solution</summary>All two-digit positive integers with distinct nonzero digits $\overline{AB}$ , such that $A$ and $B$ are both factors of $\overline{AB}$ can be listed pretty easily: $$ 12,15,24,36,48. $$ (For $A\geq5,$ for $A$ and $B$ to both divide $\overline{AB},$ we must have ... | [
"<details><summary>Solution</summary>We must have $A|10A+B$ , so $A|B$ . From here, we can see that the only possibilities are $12, 15, 24, 36, 48$ . We have $12 + 48 - 36 = 24$ , so our four numbers are $12, 48, 36, 24$ . We have $12 \\cdot 48 = 24^2$ , so the answer is $\\boxed{24}$ .</details>\n",
"<det... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] | {
"answer_score": 1034,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790588.json"
} |
Carissa is crossing a very, very, very wide street, and did not properly check both ways before doing so. (Don't be like Carissa!) She initially begins walking at $2$ feet per second. Suddenly, she hears a car approaching, and begins running, eventually making it safely to the other side, half a minute after she bega... | <details><summary>Sol</summary>Let $a$ be the number of seconds she spent walking. Then $a+an=30$ and $2a+2an^2=260$ or $a+an^2=130$ . From the first equation, we can divide by $n+1$ to get $a=\frac{30}{n+1}$ . We can substitute to get $\frac{30}{n+1}+\frac{30n^2}{n+1}=130$ . Multiplying by $\frac{n+1}{10}$... | [
"<details><summary>Solution</summary>Let's say Carissa walks at $2$ ft/sec for $x$ seconds. She then travels a distance of $2x$ feet, and after that travels at $2n$ ft/sec for $nx$ seconds. She thus travels a total distance of $2x+2n^2x = 260$ , so $x+n^2x = 130$ . We also have $x + nx = 30$ . Dividing... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] | {
"answer_score": 1028,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790589.json"
} |
Given that six-digit positive integer $\overline{ABCDEF}$ has distinct digits $A,$ $B,$ $C,$ $D,$ $E,$ $F$ between $1$ and $8$ , inclusive, and that it is divisible by $99$ , find the maximum possible value of $\overline{ABCDEF}$ .
*Proposed by Andrew Milas* | <details><summary>Sol</summary>It will always be divisible by $9$ because the sum of the digits is $36$ . For $11$ , we need alternate digits to have a sum that differs by a multiple of $11$ . We can try to set the first few digits in order from $8$ and find a way to place the rest of the digits to satisfy the ... | [
"The problem asks for a six-digit positive integer. ",
"<details><summary>Solution</summary>Since we want to maximize $\\overline{ABCDEF}$ , I let $\\overline{ABCD}=9876$ . Since the sum of these numbers is $\\equiv 3 \\pmod 9$ , $E+F$ needs to be $\\equiv 6 \\mod 9$ . Since we have already used $9,8,7,$ ... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
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"answer_score": 1054,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790590.json"
} |
Triangle $ABC$ has sidelengths $AB=1$ , $BC=\sqrt{3}$ , and $AC=2$ . Points $D,E$ , and $F$ are chosen on $AB, BC$ , and $AC$ respectively, such that $\angle EDF = \angle DFA = 90^{\circ}$ . Given that the maximum possible value of $[DEF]^2$ can be expressed as $\frac{a}{b}$ for positive integers $a, b... |
Let $BD=x,ED=2x,AD=(1-x),DF=\frac{\sqrt{3}(1-x)}{2}$ The area will be noted by $\frac{\sqrt{3}}{2}x(1-x)$ , which is maximum value is got when $x=\frac{1}{2}$ , this time the area is $\frac{\sqrt{3}}{8}$ , the desired answer is $\frac{3}{64}$ leads to $\boxed{67}$ | [
"Diagram!\n\n(points not necessarily optimally chosen to satisfy problem statement.)\n\n[asy]\nunitsize(1.5inch);\npair A = dir(180);\npair B = dir(120);\npair C = dir(0);\ndraw(A--B--C--A);\npair D = (3A + 5B)/(3 + 5);\npair F = foot(D, A, C);\npair E = extension(D, rotate(90, D)*F, C, B); \ndraw(D--E--F--cycle, b... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
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"answer_score": 1012,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790592.json"
} |
Suppose that $P(x)$ is a monic quadratic polynomial satisfying $aP(a) = 20P(20) = 22P(22)$ for some integer $a\neq 20, 22$ . Find the minimum possible positive value of $P(0)$ .
*Proposed by Andrew Wu*
(Note: wording changed from original to specify that $a \neq 20, 22$ .) | The conditions of the problem allow us to define
\[ XP(X) = (X-a)(X-20)(X-22) + C. \]
Since $X$ divides the RHS, then $C = 440a$ . So,
\begin{align*}
P(X) = \dfrac{1}{X} \left[(X-a)(X-20)(X-22) + 400a\right] = X^2 - (42+a)X + (440+42a).
\end{align*}
Henceforth, we want to minimize the value of $P(0) = 440+42a$ wh... | [
"Suppose $P(x) = x^2 + dx + e$ , then let $q(x) = xP(x)$ be a monic cubic of the form $x^3 + dx^2 + ex$ , then it is given that $q(20) = q(22) = q(a) = C$ for some $C$ . Translate down by $C$ , then we get that $q(x) - C$ has roots $20, 22, a$ meaning that $q(x) - C = (x-20)(x-22)(x-a)$ . \n\nThis beco... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
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"answer_score": 1010,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790593.json"
} |
How many ways are there to choose distinct positive integers $a, b, c, d$ dividing $15^6$ such that none of $a, b, c,$ or $d$ divide each other? (Order does not matter.)
*Proposed by Miles Yamner and Andrew Wu*
(Note: wording changed from original to clarify) | We have $15^6 = 3^65^6$ . Let $a = 3^{x_1}5^{y_1}$ , $b = 3^{x_2}5^{y_2}$ , $c = 3^{x_3}5^{y_3}$ , $d = 3^{x_4}5^{y_4}$ .
Note that if any two $x_i$ or $y_i$ are equal, then the condition is violated (one number must divide the other). So they are all distinct, and the problem is equivalent to choosing $\{... | [
"<blockquote>We have $15^6 = 3^65^6$ . Let $a = 3^{x_1}5^{y_1}$ , $b = 3^{x_2}5^{y_2}$ , $c = 3^{x_3}5^{y_3}$ , $d = 3^{x_4}5^{y_4}$ .\nIn total, the answer is just $35^2 * 1 = 1225$ .</blockquote>\n\nNot quite right. Keep in mind that ***<u>none of $a, b, c,$ or $d$ divide each other.</u>***\nThe follo... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] | {
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790594.json"
} |
Georgina calls a $992$ -element subset $A$ of the set $S = \{1, 2, 3, \ldots , 1984\}$ a **halfthink set** if
- the sum of the elements in $A$ is equal to half of the sum of the elements in $S$ , and
- exactly one pair of elements in $A$ differs by $1$ .
She notices that for some values of $n$ , wit... | <blockquote>Georgina calls a $992$ -element subset $A$ of the set $S = \{1, 2, 3, \ldots , 1984\}$ a **halfthink set** if
- the sum of the elements in $A$ is equal to half of the sum of the elements in $S$ , and
- exactly one pair of elements in $A$ differs by $1$ .
She notices that for some values o... | [
"I think this works? If yes, very funny problem. (Though highly doubting this does).\n\n<details><summary>Sol</summary>Consider $n=1000$ . Notice that if we have all the evens or odds below $1000$ , thats $499$ numbers each (we don't include 999 because only one pair of elements differ by 1), and all the evens ... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] | {
"answer_score": 118,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790596.json"
} |
Let $ABC$ be a triangle with $AB = 5$ , $BC = 7$ , and $CA = 8$ , and let $D$ be a point on arc $\widehat{BC}$ of its circumcircle $\Omega$ . Suppose that the angle bisectors of $\angle ADB$ and $\angle ADC$ meet $AB$ and $AC$ at $E$ and $F$ , respectively, and that $EF$ and $BC$ meet at $G$ .... | With **jeteagle.**
The main claim is that $T$ is fixed as the midpoint of arc $\widehat{BC}$ or equivalently $AT$ is the angle bisector of $\angle{BAC}$ . To prove this, we will show that $EF, TD, BC$ are concurrent at $G$ where $T$ is the midpoint of arc $\widehat{BC}$ and $D$ is a variable point on ... | [
"I am being forced to learn asymptote for the first time :P so I decided I might as well practice on my own problems haha\n\n<details><summary>diagram (possible spoilers!)</summary>[asy]\nunitsize(1.5inch);\npair B = dir(210);\npair C = dir(330);\npair A = dir(134.5736);\nfilldraw(circumcircle(A,B,C), opacity(0.3)+... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] | {
"answer_score": 1082,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790598.json"
} |
**p1** How many two-digit positive integers with distinct digits satisfy the conditions that
1) neither digit is $0$ , and
2) the units digit is a multiple of the tens digit?**p2** Mirabel has $47$ candies to pass out to a class with $n$ students, where $10\le n < 20$ . After distributing the candy as evenly as p... | <details><summary>A2 (with sol)</summary>Essentially it means we divide $47-k$ candies to $n$ students evenly, or in other words the largest number less than or equal to $47$ which has a number between $10$ and $19$ inclusive as a factor.
We can try each $k$ starting from $0$ . $k=0$ doesn't work becaus... | [
"<details><summary>A1 (with sol)</summary>We can tell that there's not many integers (and it's the first problem) meaning we can directly do casework: **(i)** When the tens digit is 1, the units digit can be any integer from $2$ to $9$ , which is <u>$8$</u> numbers. **(ii)** When the tens digit is 2, the units d... | [
"origin:aops",
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} |
**p4** Define the sequence ${a_n}$ as follows:
1) $a_1 = -1$ , and
2) for all $n \ge 2$ , $a_n = 1 + 2 + . . . + n - (n + 1)$ .
For example, $a_3 = 1+2+3-4 = 2$ . Find the largest possible value of $k$ such that $a_k+a_{k+1} = a_{k+2}$ .**p5** The taxicab distance between two points $(a, b)$ and $(c, d)$ o... | <details><summary>A6 (with sol)</summary>$f(\overline{AB}) = 10A+B-A\cdot B = (A-1)(10-B) + 10$ , so essentially, we want $(A-1)(10-B) = (C-1)(10-D)$ where the two numbers are $\overline{AB}$ and $\overline{CD}$ If we want this to be large intuitively we should make A as large and B as small as possible . So we sta... | [
"<details><summary>A4 (with sol)</summary>For all $n \\le 2$ , $a_n = \\frac{n(n+1)}{2}-(n+1) = \\frac{n^2-n-2}{2} = \\frac{(n+1)(n-2)}{2}$ . \nSo the problem wants the largest possible value of $k$ where $$ \\frac{(n+1)(n-2)}{2} + \\frac{(n+2)(n-1)}{2} = \\frac{(n+3)(n)}{2} $$ Solving this yields $\\boxed{... | [
"origin:aops",
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"answer_score": 1018,
"boxed": true,
"end_of_proof": false,
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} |
**p7** Cindy cuts regular hexagon $ABCDEF$ out of a sheet of paper. She folds $B$ over $AC$ , resulting in a pentagon. Then, she folds $A$ over $CF$ , resulting in a quadrilateral. The area of $ABCDEF$ is $k$ times the area of the resulting folded shape. Find $k$ .**p8** Call a sequence $\{a_n\} = a_1, a_... | <details><summary>A8 (with sol)</summary>Note that $a_5$ represented in terms of $a_1$ and $a_2$ is $2a_1 + 3a_2$ . If there is exactly one, $m$ and $n$ can't be very large, so one quick sol is to just try each number. For example for $m$ we can start from... $8$ (only 2*1 + 3*2), which works. We don't ne... | [
"<details><summary>A7 (with sol)</summary>Very simple problem - if we fold B over AC the pentagon is $ACDEF$ , and if we fold A over CF the quadrilateral is $FEDC$ which is half of $ABCDEF$ , so $k=\\boxed{2}$ .</details>"
] | [
"origin:aops",
"2022 Contests",
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"answer_score": 1030,
"boxed": true,
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} |
**p10**Kathy has two positive real numbers, $a$ and $b$ . She mistakenly writes $$ \log (a + b) = \log (a) + \log( b), $$ but miraculously, she finds that for her combination of $a$ and $b$ , the equality holds. If $a = 2022b$ , then $b = \frac{p}{q}$ , for positive integers $p, q$ where $gcd(p, q) = 1$ .... | <details><summary>answer p10</summary>$\frac{p}{q} = \frac{2023}{2022}$ . Since they are relatively prime, $4045$ .</details> | [] | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
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"answer_score": 4,
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} |
**p13** Let $ABCD$ be a square. Points $E$ and $F$ lie outside of $ABCD$ such that $ABE$ and $CBF$ are equilateral triangles. If $G$ is the centroid of triangle $DEF$ , then find $\angle AGC$ , in degrees.**p14**The silent reading $s(n)$ of a positive integer $n$ is the number obtained by dropping t... | <details><summary>13</summary>the answer should be $120^{\circ}$ Since $BE=BF;DE=DF$ and $\triangle{DEF}$ is equilateral triangle, it is clear that the centroid of the triangle lies on $BD$ $\angle{DAE}=150^{\circ},\angle{DAG}=75^{\circ}$ , it îs clear that $\triangle{AGD}\cong \triangle{CGD}$ , $\angle{AGC}=2... | [] | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
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} |
**p16** Madelyn is being paid $\$ 50 $/hour to find useful *Non-Functional Trios*, where a Non-Functional Trio is defined as an ordered triple of distinct real numbers $ (a, b, c) $, and a Non- Functional Trio is *useful* if $ (a, b) $, $ (b, c) $, and $ (c, a) $ are collinear in the Cartesian plane. Currently, she’s ... | [] | [
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} | |
**p1.** Suppose that $x$ and $y$ are positive real numbers such that $\log_2 x = \log_x y = \log_y 256$ . Find $xy$ .**p2.** Let the roots of $x^2 + 7x + 11$ be $r$ and $s$ . If f(x) is the monic polynomial with roots $rs + r + s$ and $r^2 + s^2$ , what is $f(3)$ ?**p3.** Call a positive three digit inte... | 3.I will assume $a\not=0$ . Rewriting $100a+10b+c=(10a+b)^2-11c$ gives $12c+25=100(a^2-a)+(b-5)^2+20ab$ .
Then $100(a^2-a)\le 108+25=133\implies a^2-a\le 1\implies a=1\implies 12c+25=(b+5)^2\implies 12c=b(b+10)$ .
Then $b$ is even and RHS is divisible by 3, which gives $100,122,168$ .
5. Note that $x=6,y=112$... | [
"P2:\n<details><summary>Click to expand</summary>$r + s = -7, rs = 11$ $f(x) = x^{2} - (rs + r + s)x + (r^{2} + s{2})$ $= x^{2} - (11 - 7)x + (r + s)^{2} - 2rs$ $ = x^{2} - 4x + 49 - 23$ $ = x^{2} - 4x + 27$ $f(3) = 9 - 12 + 27 = 24$</details>",
"P1:\n<details><summary>Click to expand</summary>Changing base... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/3171127.json"
} |
**p1.** Find the smallest positive integer $N$ such that $2N -1$ and $2N +1$ are both composite.**p2.** Compute the number of ordered pairs of integers $(a, b)$ with $1 \le a, b \le 5$ such that $ab - a - b$ is prime.**p3.** Given a semicircle $\Omega$ with diameter $AB$ , point $C$ is chosen on $\Ome... | <details><summary>Solution for p1</summary>Checking $N=1, 2, 3, \ldots$ going up, we see that the answer is $$ N = \boxed{13}. $$ $\square$</details> | [
"<details><summary>p6</summary>real solutions $$ (x=4,y=16) \\Longrightarrow xy=64 $$</details>",
"P4:\n<details><summary>Click to expand</summary>$r + s = -7, rs = 11$ $rs + r + s = 11 - 7 = 4$ , $r^2 + s^2 = (r + s)^2 - 2rs = 49 - 22 = 27$ $f(x) = x^2 - 4x + 27$ $f(3) = 9 - 12 + 27 = 24$</details>",
"P3:... | [
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
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"answer_score": 1108,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/3195043.json"
} |
(a) Find the value of the real number $k$ , for which the polynomial $P(x)=x^3-kx+2$ has the number $2$ as a root. In addition, for the value of $k$ you will find, write this polynomial as the product of two polynomials with integer coefficients.
(b) If the positive real numbers $a,b$ satisfy the equation $$... | <blockquote>(b) If the positive real numbers $a,b$ satisfy the equation $$ 2a+b+\frac{4}{ab}=10, $$ find the maximum possible value of $a$ .</blockquote>
We have $5-a=\frac b2+\frac 2{ab}>0$ and so $a<5$ Squaring, we have $(5-a)^2=\frac{b^2}4+\frac 2a+\frac 4{a^2b^2}$ $=\frac 4a+\left(\frac b2-\frac 2{ab}\... | [
"<blockquote>(a) Find the value of the real number $k$ , for which the polynomial $P(x)=x^3-kx+2$ has the number $2$ as a root. In addition, for the value of $k$ you will find, write this polynomial as the product of two polynomials with integer coefficients.</blockquote> $2^3-2k+2=0$ $\\implies$ $\\boxe... | [
"origin:aops",
"2022 Contests",
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"answer_score": 1026,
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"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790115.json"
} |
Let $ABC$ be an isosceles triangle, and point $D$ in its interior such that $$ D \hat{B} C=30^\circ, D \hat{B}A=50^\circ, D \hat{C}B=55^\circ $$
(a) Prove that $\hat B=\hat C=80^\circ$ .
(b) Find the measure of the angle $D \hat{A} C$ . | For the part (b) we denote by $\alpha$ measure of the $\angle DAC$ . By the Law of Sines in $\triangle ABD$ and $\triangle ACD$ we have that $$ 1 = \frac{BA}{AD}\cdot \frac{DA}{AC} = \frac{\sin (70^\circ - \alpha)}{\sin 50^\circ} \cdot \frac{\sin 25^\circ}{\sin (25^\circ + \alpha)} $$ **Claim.** $\alpha = 5^\... | [
"If I am not wrong, Part a is trivial because \\[\\angle ABC=\\angle DBA+\\angle DBC=80\\] and since it's isoceles we're done.",
"Pretty sure this is the intended solution:\n\nLet $E$ be the point on the extension of $BD$ such that $AB=AE$ . Thus, $AE=AC$ .\n\nThen, $\\angle ABE=\\angle AEB=50$ , and $\\a... | [
"origin:aops",
"2022 Contests",
"2022 Greece Junior Math Olympiad"
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"answer_score": 1124,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790118.json"
} |
On the board we write a series of $n$ numbers, where $n \geq 40$ , and each one of them is equal to either $1$ or $-1$ , such that the following conditions both hold:
(i) The sum of every $40$ consecutive numbers is equal to $0$ .
(ii) The sum of every $42$ consecutive numbers is not equal to $0$ .
We den... | Let out sequence is $a_1, a_2, ... , a_{40}, ...$ and notice that after $a_{40}$ we must have $a_1$ again, then $a_2$ , etc. So, our sequence is repeated by the first $40$ elements. By the second condition we have that $a_i + a_{i+1} \neq 0$ . Now we have $a_1=a_2=...=a_{20}=x$ and $a_{21}=a_{22}=...=a_{40... | [] | [
"origin:aops",
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"boxed": true,
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"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790120.json"
} |
Find all couples of non-zero integers $(x,y)$ such that, $x^2+y^2$ is a common divisor of $x^5+y$ and $y^5+x$ . | Let $\gcd(x,y)=d,x=ad,y=bd$ . Obviously, $\gcd(a,b)=1$ .
Note that $d^2\mid d^2(a^2+b^2)\implies d^2\mid x^2+y^2$ .
But $x^5+y=d(a^5d^4+b)\implies d^2\mid d(a^5d^4+b)\implies d\mid b$ Similarly, $d\mid a$ . Thus, $d\mid \gcd(a,b)=1\implies d=1\implies\gcd(x,y)=1$ . $x^2+y^2\mid (x^5+y)(x^5-y)=(x^2)^5-y^2\implie... | [
"My solution as I solved it in the competition:\nFirst imply $\\gcd(x,y)=1$ as above. Then we have $x^2+y^2 \\mid x^2+y^2+x^2(x^3y-1)\\implies x^2+y^2 \\mid x^3y-1$ since $\\gcd(x^2,x^2+y^2)=1$ Same we get $x^2+y^2 \\mid y^3x-1$ . Combining these two we get $x^2+y^2 \\mid 2$ and we get the solutions $(x,y)... | [
"origin:aops",
"2022 Contests",
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"answer_score": 40,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790121.json"
} |
Let $ABC$ be a triangle such that $AB<AC<BC$ . Let $D,E$ be points on the segment $BC$ such that $BD=BA$ and $CE=CA$ . If $K$ is the circumcenter of triangle $ADE$ , $F$ is the intersection of lines $AD,KC$ and $G$ is the intersection of lines $AE,KB$ , then prove that the circumcircle of triangle ... | Let $I$ be a incenter of $\triangle ABC$ . Denote by $2\alpha, 2\beta, 2\gamma$ angles $\angle A, \angle B, \angle C$ of the $\triangle ABC$ . Let $\odot (ABI)$ intersect side $\overline{BC}$ in $E'$ .**Claim.** $E' \equiv E$ *Proof.* We have that $\angle KE'C = 180^\circ - \angle AE'B = 180^\circ - \ang... | [
"<blockquote>Let $I$ be a incenter of $\\triangle ABC$ . Denote by $2\\alpha, 2\\beta, 2\\gamma$ angles $\\angle A, \\angle B, \\angle C$ of the $\\triangle ABC$ . Let $\\odot (ABI)$ intersect side $\\overline{BC}$ in $E'$ .**Claim.** $E' \\equiv E$ *Proof.* We have that $\\angle KE'C = 180^\\circ - ... | [
"origin:aops",
"2022 Greece National Olympiad",
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"boxed": false,
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"n_reply": 6,
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} |
Let $n>4$ be a positive integer, which is divisible by $4$ . We denote by $A_n$ the sum of the odd positive divisors of $n$ . We also denote $B_n$ the sum of the even positive divisors of $n$ , excluding the number $n$ itself. Find the least possible value of the expression $$ f(n)=B_n-2A_n, $$ for all po... | I claim the answer is $4$ , and is attained iff $n=4p$ , where $p>2$ is a prime or when $n=8$ . Check that indeed when $n=4p$ , then $A_n=1+p$ whereas $B_n=2+4+2p$ , yielding $B_n-2A_n =4$ .
Let $n=2^k\cdot m$ , where $m$ is odd and $k\ge 2$ . Notice that $A_n=\textstyle\sum_{d\mid m}d$ . On the other h... | [] | [
"origin:aops",
"2022 Greece National Olympiad",
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"answer_score": 64,
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"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 Greece National Olympiad/2790101.json"
} |
The positive real numbers $a,b,c,d$ satisfy the equality $$ a+bc+cd+db+\frac{1}{ab^2c^2d^2}=18. $$ Find the maximum possible value of $a$ . | Note that we can write $a$ as $a=18-\left(bc+cd+db+\frac{1}{ab^2c^2d^2}\right)\leq18-\left(\frac{4}{\sqrt[4]{a}}\right)$
since we need to find $a_{\text{max}}$ , minimizing this part $\left(bc+cd+db+\frac{1}{ab^2c^2d^2}\right)$ will work, hence the above part which is true by AM-GM, we need to solve this $$ ... | [
"Using AM-GM on the last four terms, we obtain\n\\[\n18\\ge a+ 4a^{-\\frac14}\\iff t^4 +\\frac4t-18\\le 0 \\qquad\\text{where}\\qquad a=t^4.\n\\]\nFrom here, $t\\le 2$ and therefore $a\\le 16$ .\n\nWe now give an example attaining the equality. Note that if $b=c=d$ and $b^2 = (ab^6)^{-1}\\iff b^8 = \\frac1a$ ... | [
"origin:aops",
"2022 Greece National Olympiad",
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"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Greece National Olympiad/2790106.json"
} |
Let $Q_n$ be the set of all $n$ -tuples $x=(x_1,\ldots,x_n)$ with $x_i \in \{0,1,2 \}$ , $i=1,2,\ldots,n$ . A triple $(x,y,z)$ (where $x=(x_1,x_2,\ldots,x_n)$ , $y=(y_1,y_2,\ldots,y_n)$ , $z=(z_1,z_2,\ldots,z_n)$ ) of distinct elements of $Q_n$ is called a *good* triple, if there exists at least one $i \... | <details><summary>BRUH</summary>We proceed by induction on $n$ , with the base case, $n=1$ being clear.
For $t=0,1,2$ , let $J_t$ be the set of sequences with $x_n=t$ . We know that any three sequences in $J_t\cup J_u$ must have $x_j,y_j,z_j$ distinct, but $j\ne n$ . First observe $(x_1,\cdots,x_{n-1})$ ... | [
"<details><summary>Can we just?</summary>Note that total number of **unordered** triplets $(x,y,z) \\in Q^3 _{n} = \\frac{6! ^n}{6!} = 2^{n-1}3^{n-1}$ , suppose for the sake of contradiction assume that there exist a good subset with $ > 2 \\cdot (\\frac{3}{2})^n$ , then note that $\\binom{|A|}{3} \\le 2^{n... | [
"origin:aops",
"2022 Greece National Olympiad",
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"path": "Contest Collections/2022 Contests/2022 Greece National Olympiad/2790111.json"
} |
Positive integers $a$ , $b$ , and $c$ are all powers of $k$ for some positive integer $k$ . It is known that the equation $ax^2-bx+c=0$ has exactly one real solution $r$ , and this value $r$ is less than $100$ . Compute the maximum possible value of $r$ . | Since we must have $b^2=4ac$ , this implies $a$ , $b$ , and $c$ are powers of $2$ , so substitute $a=2^m$ , $b=-2^n$ , $c=2^\ell$ . Then $r=-\tfrac{b}{2a}$ , hence we want $r$ to be the largest power of $2$ less than $100$ , which just so happens to be $64$ .
Now we find a construction for this. $b^2=... | [
"solution\n\n**Attachments:**\n\n[Hmmt 2022 P1.docx](https://cdn.artofproblemsolving.com/attachments/e/2/be004fd69ac8fcb133fa636153ecc729ddb017.docx)",
" $\\text{Discriminant }D=b^2-4ac=0 \\implies b \\text{ is even } \\implies k=2.$ It follows that $r$ is a power of $2$ . So max the power of $2$ smaller th... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
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"answer_score": 1040,
"boxed": true,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798886.json"
} |
Compute the number of positive integers that divide at least two of the integers in the set $\{1^1,2^2,3^3,4^4,5^5,6^6,7^7,8^8,9^9,10^{10}\}$ . | First, notice that each of the integers from $1$ to $10$ (excluding $1$ , because it’s hard to define?) have at most two distinct prime factors. Therefore, the integers that divide *at least two* of these power things have only one distinct prime factor, since if it had two prime factors, there’s at most one integ... | [
"22 is the answer ",
"Factorising the set will give you $$ {1^{1}, 2^2, 3^{3}, 2^{8}, 5^{5}, 2^{6} \\cdot 3^{6}, 7^7, 2^{24}, 3^{18}, 2^{10} \\cdot 5^{10}} $$ Let $x = 2^{a} \\cdot 3^{b} \\cdot 5^{c}$ , where $a$ , $b$ , and $c$ are nonnegative integers. (The case when $x = 1$ is obvious, so we will add ... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
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"answer_score": 1038,
"boxed": true,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798887.json"
} |
Let $x_1, x_2, . . . , x_{2022}$ be nonzero real numbers. Suppose that $x_k + \frac{1}{x_{k+1}} < 0$ for each $1 \leq k \leq 2022$ , where $x_{2023}=x_1$ . Compute the maximum possible number of integers $1 \leq n \leq 2022$ such that $x_n > 0$ . | We note that if $x_{k}$ is positive, then $x_{k+1}$ must be negative. Therefore, $n$ must be less than 1012, as if we group our numbers by pairs of $(x_{k}, x_{k+1})$ then we would have 1011 groups (or at most 1011 positive integers).
Now, we will prove that $n = 1011$ is impossible through contradiction: Fi... | [
"My answer is 1010.",
"This is fairly easy to do though prickly to actually prove; I'll only give an outline of how to prove $n \\geq 1011$ is impossible.\n\nFirst, observe as we cannot have consecutive $+$ signs, so $n$ is at most 1011. Consider the case when $n=1011$ . Now, assume without loss of general... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1038,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798905.json"
} |
Compute the sum of all 2-digit prime numbers $p$ such that there exists a prime number $q$ for which $100q + p$ is a perfect square. | Note: $p\equiv11,31,41,61,71,19,29,59,79,89\pmod{100}
\Rightarrow 100q+p\equiv k^2\equiv 11,31,41,61,71,19,29,59,79,89\pmod{100}$ Now observe that the values $k^2$ are congruent to are nothing but the last $2$ digits of the perfect squares. So, we can reduce our calculations to just checking the last $2$ digits ... | [
"With mod 4 and mod 10 we obtain that p must be in the form 4k+1 and the last digit must be 1 or 9. Moreover, since a perfect square must have even number as its tens digit we got the only possibilities are 29, 41, 61, and 89. 89 obviously works when q=2 (17²). 61 when q=3 (19²) . 29 when q= 5 (23²). 41 is rejected... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
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"answer_score": 1032,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
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} |
Given a positive integer $k$ , let $||k||$ denote the absolute difference between $k$ and the nearest perfect square. For example, $||13||=3$ since the nearest perfect square to $13$ is $16$ . Compute the smallest positive integer $n$ such that $\frac{||1|| + ||2|| + ...+ ||n||}{n}=100$ . | Let $k = \lfloor \sqrt n \rfloor$ and let $n = k^2+r$ for some positive integers $r, k$ . Notice that $$ ||a^2|| + ||a^2+1|| + \cdots + ||(a+1)^2|| = a(a+1), $$ so we can write $$ 100n = 100k^2+100r = ||1||+||2||+ \cdots + ||n|| \leq \sum_{i=2}^k i(i-1) + \frac 12 r(r+1) \leq \frac 13 (k^3-k) + \frac 12 r(r... | [
"I'm too lazy to make a mathematical solution xD.\nThe answer is 89800, checked by C++: \n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nint f(int n) {\n int lo_sqrt_n = floor(sqrt(n));\n int up_sqrt_n = lo_sqrt_n + 1;\n int dif1 = n - lo_sqrt_n * lo_sqrt_n;\n int dif2 = up_sqrt_n * up_sqrt_n - ... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1038,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798911.json"
} |
Let f be a function from $\{1, 2, . . . , 22\}$ to the positive integers such that $mn | f(m) + f(n)$ for all $m, n \in \{1, 2, . . . , 22\}$ . If $d$ is the number of positive divisors of $f(20)$ , compute the minimum possible value of $d$ . | The answer is $\boxed{2016}$ .
Claim: The function $f(n) = n\cdot \mathrm{lcm}(1,2,\ldots, 22)$ works.
Proof: Let $a = \mathrm{lcm}(1,2,\ldots, 22)$ . We must check that \[mn |a(m+n)\] for all $1\le m,n\le 22$ . Suppose for some fixed $m,n$ , and prime $p$ , we had $\nu_p(mn) >\nu_p(a(m+n))$ . Then, \[\nu_p... | [
"Does $f(n)=n\\cdot \\mathrm{lcm}(1,2,\\ldots,22)$ work?",
"2016 is my answer .",
"For every prime $p$ , we can look at the minimum power of $p$ that must divide $f(20)$ . In general, we can use the following strategy: if $k$ is the number less than or equal to 22 with the maximum $\\nu_p$ , then we ca... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1178,
"boxed": true,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798917.json"
} |
Let $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ , $(x_4, y_4)$ , and $(x_5, y_5)$ be the vertices of a regular pentagon centered at $(0, 0)$ . Compute the product of all positive integers k such that the equality $x_1^k+x_2^k+x_3^k+x_4^k+x_5^k=y_1^k+y_2^k+y_3^k+y_4^k+y_5^k$ must hold for all possible choices of... | Much more difficult than #6, but a super interesting problem.
Throughout this solution, we drop all constants. By scaling, we may assume that the vertices of the pentagon lie on the unit circle. Now, let $(x_1, y_1) = \exp(\theta)$ for some angle $\theta$ . Then, $$ x_1^k + x_2^k + x_3^k + x_4^k + x_5^k = (\exp(\... | [
"Nice problem, my answer is 1×2×3×4×6×8=1152.",
"Here's a fast unrigorous solution: let $\\arg(x_i,y_i)=\\theta_i$ . Observe $\\textstyle\\sum\\cos k\\theta_i=\\sum\\sin k\\theta_i$ for $5\\nmid k$ ; expanding by Chevyshev gives the sum of degrees $k,k-2,\\dots$ , and so we can inductively build $1\\mapsto3... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1042,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798920.json"
} |
Positive integers $a_1, a_2, ... , a_7, b_1, b_2, ... , b_7$ satisfy $2 \leq a_i \leq 166$ and $a_i^{b_i} \cong a_{i+1}^2$ (mod 167) for each $1 \leq i \leq 7$ (where $a_8=a_1$ ). Compute the minimum possible value of $b_1b_2 ... b_7(b_1 + b_2 + ...+ b_7)$ . | Note that $167$ is prime.
We have \[a_i^{2^{i-1}} \equiv a_1^{b_1 b_2 \cdots b_{i-1}}\pmod{167}\]for all $1<i\le 8$ .
This implies $a_1^{128}\equiv a_1^{b_1 b_2 \cdots b_7}\pmod{167}$ , so $a_1^{128 - b_1 b_2 \cdots b_7}\equiv 1\pmod{167}$ .
Case 1: $a_1 = 166$ .
If $a_i = 166$ , then since $-1$ isn't a Q... | [
"The official solutions are here:\n[https://hmmt-archive.s3.amazonaws.com/tournaments/2022/feb/algnt/solutions.pdf](https://hmmt-archive.s3.amazonaws.com/tournaments/2022/feb/algnt/solutions.pdf).",
"Let $p = b_1b_2\\dots b_7$ . Note that\n\n\\[a_1^p = a_1^{b_1b_2\\dots b_7} \\equiv a_1^{2b_2b_3\\dots} \\dots \\... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1060,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798921.json"
} |
Suppose $P(x)$ is a monic polynomial of degree $2023$ such that $P(k) = k^{2023}P(1-\frac{1}{k})$ for every positive integer $1 \leq k \leq 2023$ . Then $P(-1) = \frac{a}{b}$ where $a$ and $b$ are relatively prime integers. Compute the unique integer $0 \leq n < 2027$ such that $bn-a$ is divisible by t... | bruh i'm not gonna lie this is probably the hardest algebra problem i've ever done :skull:
Firstly, we note that $P(k)-k^{2023}P\left(1-\frac{1}{k}\right)$ evidently has roots $1,2,3, \cdots 2023$ . Therefore, we can write $P(k)-k^{2023}P\left(1-\frac{1}{k}\right)=k(x-1)(x-2) \cdots (x-2023)$ for some $k$ . Let ... | [
"Here's a version of the official solution that I think is a bit easier (or at the very least more time-efficient) for a non-olympiad like the HMMT:\n\nWe begin by constructing the following equation: $P(x) - x^{2023}P(1-\\frac{1}{x}) = c(x-1)(x-2)\\ldots(x-2023)$ ( $c$ is some constant)\n\nWe first evaluate the ... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1066,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798928.json"
} |
Compute the smallest positive integer $n$ for which there are at least two odd primes $p$ such that $\sum_{k=1}^{n} (-1)^{v_p(k!)} < 0$ . Note: for a prime $p$ and a positive integer $m$ , $v_p(m)$ is the exponent of the largest power of $p$ that divides $m$ ; for example, $v_3(18) = 2$ . | [
"It was very tough one in the individual round, though I have solved and found the answer 229"
] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798933.json"
} | |
Let $(a_1, a_2, ..., a_8)$ be a permutation of $(1, 2, ... , 8)$ . Find, with proof, the maximum possible number of elements of the set $$ \{a_1, a_1 + a_2, ... , a_1 + a_2 + ... + a_8\} $$ that can be perfect squares. | The answer is $\boxed{5}$ , achievable with $(1,3,5,7,2,4,6,8)$ .
For the bound, begin by noticing that $1+2+\dots+8 = 36$ , so we can have at most $6$ squares among the set. Thus, it suffices to show we cannot have all $6$ of them. Obviously, the squares must appear in increasing order, and also must be conse... | [
"<blockquote>I submitted this and got full credit</blockquote>\n\ni like ur handwriting"
] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1018,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801545.json"
} |
Find, with proof, the maximum positive integer $k$ for which it is possible to color $6k$ cells of $6 \times 6$ grid such that, for any choice of three distinct rows $R_1$ , $R_2$ , $R_3$ and three distinct columns $C_1$ , $C_2$ , $C_3$ , there exists an uncolored cell $c$ and integers $1 \le i, j \le 3... | The answer is $\boxed{4}$ , and a construction can easily be found (too lazy to use asy).
Obviously, $k=6$ isn't possible. Consider $k=5$ . There are $6$ uncolored squares; pick three of them and report their respective rows as $r_1, r_2, r_3$ . Let the columns of the other three uncolored squares be $c_1, c_2... | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1020,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801546.json"
} |
Let triangle $ABC$ be an acute triangle with circumcircle $\Gamma$ . Let $X$ and $Y$ be the midpoints of minor arcs $AB$ and $AC$ of $\Gamma$ , respectively. If line $XY$ is tangent to the incircle of triangle $ABC$ and the radius of $\Gamma$ is $R$ , find, with proof, the value of $XY$ in terms of... | Let the incenter be $I = \overline{CX} \cap \overline{BY}$ . Since $\overline{XY}$ is tangent to the incircle, we have $\triangle IBC \cong \triangle IXY$ . Thus, $BI = XI = XB$ , meaning that $\angle BXI = \angle BXC = \angle BAC = 60^\circ$ . Hence, $\angle OBC = \angle OXY = 120^\circ$ , giving $XY = \boxed... | [
"Rsqrt3=XY"
] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1014,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801547.json"
} |
Let $ABC$ be a triangle with centroid $G$ , and let $E$ and $F$ be points on side $BC$ such that $BE = EF = F C$ . Points $X$ and $Y$ lie on lines $AB$ and $AC$ , respectively, so that $X$ , $Y$ , and $G$ are not collinear. If the line through $E$ parallel to $XG$ and the line through $F$ pa... | Trivial. Equivalent to $S_{GPX} = S_{GPY}$ and the parallel lines rewrite the latter into $S_{GEX} = S_{GFY}$ . But $GE \parallel AB$ (consider $M = CG \cap AB$ and note $BE/EM = CG/GM$ ) and similarly $GF \parallel AC$ , so we reduce to $S_{GEB} = S_{GFC}$ , which is clear. | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801549.json"
} |
Let $P(x) = x^4 + ax^3 + bx^2 + x$ be a polynomial with four distinct roots that lie on a circle in the complex plane. Prove that $ab\ne 9$ . | This is an incredibly deep problem. BTW, I don't think any teams solved this problem at the contest.
One of the roots is $0$ . Let the other roots be $z_1, z_2, z_3$ . By Vieta's, $z_1+z_2+z_3=-a\implies \frac{z_1+z_2+z_3}{3}=-\frac{a}{3}$ . Also by Vieta's, $\frac{\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}}{3}=-\f... | [
"Suppose by way of contradiction that some $a,b$ work. Let $0,p,q,r$ be the roots, which lie on the circle in that order. From Vieta, we have $pqr = -1$ and\n\\[ab = -(p + q + r)(pq + qr + rp) = (p + q + r)\\left(\\frac1p + \\frac1q + \\frac1r\\right) = 9\\text{.}\\]\nNow we deal with the circle condition. By... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801551.json"
} |
Find, with proof, all functions $f : R - \{0\} \to R$ such that $$ f(x)^2 - f(y)f(z) = x(x + y + z)(f(x) + f(y) + f(z)) $$ for all real $x, y, z$ such that $xyz = 1$ . | <details><summary>good game!</summary>Let $P(x,y,z)$ be the assertion $P(1,1,1)\implies f(1)^2-f(1)^2=9f(1)$ , i.e $\boxed{f(1)=0}$ $P(x,\frac{1}{x},1) \implies f(x)^2=x(x+\frac{1}{x}+1)(f(x)+f(\frac{1}{x}))$ $P(\frac{1}{x},x,1)\implies f(\frac{1}{x})^2=\frac{1}{x}(\frac{1}{x}+x+1)f(\frac{1}{x})+f(x))$ which give... | [
"Bump this",
"plug 1,1,1 to get f(1)=0 and then x,1/x,1 and 1/x,x,1 and solve for f(x), f(1/x)",
"Find, with proof, all functions $f : R - \\{0\\} \\to R$ such that $$ f(x)^2 - f(y)f(z) = x(x + y + z)(f(x) + f(y) + f(z)) $$ for all real $x, y, z$ such that $xyz = 1$ The only solutions are $\\boxed{f\\eq... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1146,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801553.json"
} |
Let $P_1P_2...P_n$ be a regular $n$ -gon in the plane and $a_1, . . . , a_n$ be nonnegative integers. It is possible to draw $m$ circles so that for each $1 \le i \le n$ , there are exactly $a_i$ circles that contain $P_i$ on their interior. Find, with proof, the minimum possible value of $m$ in terms of ... | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801554.json"
} | |
Let $\Gamma_1$ and $\Gamma_2$ be two circles externally tangent to each other at $N$ that are both internally tangent to $\Gamma$ at points $U$ and $V$ , respectively. A common external tangent of $\Gamma_1$ and $\Gamma_2$ is tangent to $\Gamma_1$ and $\Gamma_2$ at $P$ and $Q$ , respectively, and... | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801557.json"
} | |
On a board the following six vectors are written: $$ (1, 0, 0), (-1, 0, 0), (0, 1, 0), (0, -1, 0), (0, 0, 1), (0, 0, -1). $$ Given two vectors $v$ and $w$ on the board, a move consists of erasing $v$ and $w$ and replacing them with $\frac{1}{\sqrt2} (v + w)$ and $\frac{1}{\sqrt2} (v - w)$ . After some nu... | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801562.json"
} | |
Suppose $n \ge 3$ is a positive integer. Let $a_1 < a_2 < ... < a_n$ be an increasing sequence of positive real numbers, and let $a_{n+1} = a_1$ . Prove that $$ \sum_{k=1}^{n}\frac{a_k}{a_{k+1}}>\sum_{k=1}^{n}\frac{a_{k+1}}{a_k} $$ | The LaTeX is supposed to be
\[ \sum^{n}_{k=1} \frac{a_k}{a_{k+1}} > \sum^{n}_{k=1} \frac{a_{k+1}}{a_k}. \]
Anyway here's my solution. It's quite different to the official solutions.
<details><summary>Alternate Solution</summary>We'll show it for $n=3$ first. Notice that we want to show
\[ \frac{a_1}{a_2} + \frac{a... | [
"On the other hand, you can also solve this by smoothing $a_i$ between $a_{i-1}$ and $a_{i+1}$ , and using calculus to note that the inequality is closer when $a_i$ is very near $a_{i-1}$ "
] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801569.json"
} |
Let $ABC$ be a triangle with $\angle A = 60^o$ . Line $\ell$ intersects segments $AB$ and $AC$ and splits triangle $ABC$ into an equilateral triangle and a quadrilateral. Let $X$ and $Y$ be on $\ell$ such that lines $BX$ and $CY$ are perpendicular to ℓ. Given that $AB = 20$ and $AC = 22$ , compu... | <details><summary>Solution</summary>Let the intersection of $\ell$ with $\overline{AB}$ and $\overline{AC}$ be $M$ and $N,$ respectively. Then, let the side length of $\triangle{AMN}$ be $x.$ Then, it follows that $AM=AN=x,$ so we get $BM=20-x$ and $CN=22-x.$ Since $\triangle{AMN}$ is equilateral,... | [
"Let line l intersect AB and AC at P and Q. From 30-60-90 triangle properties, XP = (20-x)/2 and QY = (22-x)/2, where x is the length of the equilateral triangle with angle A of 60 degrees. XY = XP + PQ + QY = (20-x)/2 + (22-x)/2 + x = **21**.",
"Perfect for problem one: clean and simple.\n\nLet the side length o... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1140,
"boxed": true,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801591.json"
} |
Let $ABCD$ and $AEF G$ be unit squares such that the area of their intersection is $\frac{20}{21}$ . Given that $\angle BAE < 45^o$ , $\tan \angle BAE$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$ . Compute $100a + b$ . | <details><summary>Solution</summary>Without loss of generality, let $G$ lie inside square $ABCD.$ Also, let the second intersection of the squares be $X \neq A.$ Furthermore, let $\angle{BAX}=\angle{GAX}=\theta.$ We wish to find $\tan(90^\circ-2\theta).$ Clearly, the area of the overlapping region between the... | [
"Again, trivial problem. Note the area of the intersection of the two unit squares is the sum of two triangles, each of which has a base of 1 and another side x, x < 1. Hence the intersection area = (1*x/2)*2 = x = 20/21. Let the two angles of these triangles, centered at point A, be alpha and beta. The tangent of ... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1116,
"boxed": true,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801593.json"
} |
Parallel lines $\ell_1$ , $\ell_2$ , $\ell_3$ , $\ell_4$ are evenly spaced in the plane, in that order. Square $ABCD$ has the property that $A$ lies on $\ell_1$ and $C$ lies on $\ell_4$ . Let $P$ be a uniformly random point in the interior of $ABCD$ and let $Q$ be a uniformly random point on the per... | Here's a surprisingly clean trigonometric solution.
First, check that $B$ and $D$ must lie between $\ell_2$ and $\ell_3$ ; to prove this is the case, simply notice that the border case yields a probability of $\frac 12$ and the more "tilted" square $ABCD$ is, the greater portion of it will be contained betw... | [
"Different clean trig solution.\n\nFix a unit square $ABCD$ and let $l_1$ be incident from $A$ with angle $\\theta.$ By inspection, $\\theta < 45^\\circ$ <details><summary>equivalent to</summary><blockquote>First, check that $B$ and $D$ must lie between $\\ell_2$ and $\\ell_3$ ; to prove this is th... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1050,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801597.json"
} |
Let triangle $ABC$ be such that $AB = AC = 22$ and $BC = 11$ . Point $D$ is chosen in the interior of the triangle such that $AD = 19$ and $\angle ABD + \angle ACD = 90^o$ . The value of $BD^2 + CD^2$ can be expressed as $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Comput... | Let $C' \neq B$ be a point such that $\triangle ABC \cong \triangle ACC'$ . In addition, let $D'$ be a point in $\triangle ACC'$ such that $\angle DAD' \cong \angle BAC$ .
Notice, $BD^2 + CD^2 = CD^2 + CD'^2 = DD'^2$ , but $\angle DCD' = 90^{\circ}$ . As $\triangle DAD' \sim \triangle BAC$ , then
\[ DD' =... | [
"Rotate $\\triangle{ADC}$ and let $AC$ meet $AB$ , after rotation, we call the triangle be $\\triangle{ABD'}$ with $\\angle{ABD'}=\\angle{ACD}, \\angle AD'B=90^{\\circ}$ We can see that $\\angle{D'AB}=\\angle{DAC}, \\triangle{AD'D}\\sim \\triangle{ABC}, D'D=\\frac{19}{2}$ $BD^2+CD^2=BD^2+D'B^2=DD'^2=\\f... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1018,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801601.json"
} |
Let $ABCD$ be a rectangle inscribed in circle $\Gamma$ , and let $P$ be a point on minor arc $AB$ of $\Gamma$ . Suppose that $P A \cdot P B = 2$ , $P C \cdot P D = 18$ , and $P B \cdot P C = 9$ . The area of rectangle $ABCD$ can be expressed as $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relativ... | Note that we have $PA \cdot PB=2, PC \cdot PD = 18, PB \cdot PC = 9$ , which implies that $PA \cdot PD=4$ . Then, by applying the Law of Cosines on $\triangle PAD, \triangle PDC, \triangle PBC,$ and $\triangle PAB$ , and setting the minor arcs subtended by lines $AB$ and $AD$ to be $2 \beta$ and $2\theta$ ,... | [
"Firstly we can observe that $\\frac{PD}{PB}=2; \\frac{PC}{PA}=\\frac{9}{2}$ Then, since rectangle $ABCD$ is inscribed in the circle, so $\\angle{APC}=\\angle{PBD}=90^{\\circ}; AC=BD$ We denote that $PB=a,PD=2a; PA=2m, PC=9m$ , now with the application of PT, $5a^2=85m^2, a=\\sqrt{17}m$ , the diameter of the ... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1084,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801604.json"
} |
Point $P$ is located inside a square $ABCD$ of side length $10$ . Let $O_1$ , $O_2$ , $O_3$ , $O_4$ be the circumcenters of $P AB$ , $P BC$ , $P CD$ , and $P DA$ , respectively. Given that $P A+P B +P C +P D = 23\sqrt2$ and the area of $O_1O_2O_3O_4$ is $50$ , the second largest of the lengths $O_1O... | Just a quick note:
A shortcut to the official sol is scaling $\square{O_1O_2O_3O_4}$ by a factor of $2$ about $P$ to obtain a cyclic quadrilateral with circumcenter $P$ (This reduces the solution length to 4-5 lines) | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 108,
"boxed": false,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801605.json"
} |
Let $E$ be an ellipse with foci $A$ and $B$ . Suppose there exists a parabola $P$ such that $\bullet$ $P$ passes through $A$ and $B$ , $\bullet$ the focus $F$ of $P$ lies on $E$ , $\bullet$ the orthocenter $H$ of $\vartriangle F AB$ lies on the directrix of $P$ .
If the major and minor axes of... | $AH=BH$ and the distance from $A$ to the directrix of parabola is equivalent to $AF$ which is half of the major axis, $AF=25$ .
Let $D$ be the projection of $A$ onto the directrix, $AD=25$ . Due to symmetry, $DH=\frac{AB}{2}=24$ , $AH^2=25^2+24^2=1201$ . As $AH=BH, AH^2+BH^2=2402$ | [
" $AH^{2} + BH^{2}=2402$ "
] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801606.json"
} |
Let $A_1B_1C_1$ , $A_2B_2C_2$ , and $A_3B_3C_3$ be three triangles in the plane. For $1 \le i \le3$ , let $D_i $ , $E_i$ , and $F_i$ be the midpoints of $B_iC_i$ , $A_iC_i$ , and $A_iB_i$ , respectively. Furthermore, for $1 \le i \le 3$ let $G_i$ be the centroid of $A_iB_iC_i$ .
Suppose that the areas ... | Let $a_i, b_i, c_i, d_i, e_i, f_i, g_i$ be the points' vectors. It suffices to maximize $\frac{1}{2} |(g_2 - g_1) \times (g_3 - g_1)|.$ Observe that,
\begin{align*}
\frac{1}{2} |(g_2 - g_1) \times (g_3 - g_1)| &= \frac{1}{2} |(\frac{(a_2 - a_1) + (b_2 -b_1) + (c_2 - c_1)}{3}) \times (\frac{(a_3 - a_1) + (b_3 -b_1) ... | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1016,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801608.json"
} |
Suppose $\omega$ is a circle centered at $O$ with radius $8$ . Let $AC$ and $BD$ be perpendicular chords of $\omega$ . Let $P$ be a point inside quadrilateral $ABCD$ such that the circumcircles of triangles $ABP$ and $CDP$ are tangent, and the circumcircles of triangles $ADP$ and $BCP$ are tangent... | Let $X=AB \cap CD$ and $Y=AD \cap BC$ . By radical axis theorem on $\omega$ , $(ABP)$ , and $(CDP)$ followed by PoP, $XP^2=XO^2-8^2$ . By similar logic, $YP^2=YO^2-8^2$ . By the perpendicularity lemma, $\overline{OP}$ is perpendicular to $\overline{XY}$ . Let $P' = OP \cap XY$ , and note that $P'$ is the... | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1048,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801610.json"
} |
Rectangle $R_0$ has sides of lengths $3$ and $4$ . Rectangles $R_1$ , $R_2$ , and $R_3$ are formed such that: $\bullet$ all four rectangles share a common vertex $P$ , $\bullet$ for each $n = 1, 2, 3$ , one side of $R_n$ is a diagonal of $R_{n-1}$ , $\bullet$ for each $n = 1, 2, 3$ , the opposite side... | <details><summary>Solution</summary>Note that the overlapping region of any two consecutive triangles is a triangle which makes up half of the area of either triangle, so all triangles have the same area. Therefore, the answer is
\begin{align*}
12 + \frac{12}{2} + \frac{12}{2} + \frac{12}{2} &= 12+6+6+6
&= \boxed{30},... | [
"Trivial, can eyeball this problem. Note that R0, R1, R2, and R3 have equal areas, all which equal to 3*4 = 12. The incremental area added by each R_n+1 is simple 12/2 = 6. Hence the total area of the entire figure is 12+6*3 = **30**",
"Notice that the three rectangles have equal areas, which we will call $S$ . ... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1102,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801613.json"
} |
Sets $A, B$ , and $C$ satisfy $|A| = 92$ , $|B| = 35$ , $|C| = 63$ , $|A\cap B| = 16$ , $|A\cap C| = 51$ , $|B\cap C| = 19$ . Compute the number of possible values of $ |A \cap B \cap C|$ . | @above that's incorrect
<details><summary>Solution</summary>Let $|A\cap B\cap C|=x$ . Then $|A\cap B|=16-x$ , so we have $x\leq 16$ . Also, $|C\backslash \{A\cup B\}|=x-7$ , so we have $x\geq 7$ . We can verify that these are the only constraints on $x$ , so the answer is $\boxed{10}$ .</details> | [
"<details><summary>solution</summary>$|A \\cup B \\cup C| = |A| + |B| + |C| - |A \\cap B| - |A \\cap C| - |B \\cap C| + |A \\cap B \\cap C|$ . Plugging in the above constants, $|A \\cap B \\cap C| = |A \\cup B \\cup C| - 104$ . The intersection of the three sets is at least 0 and at most equal to the cardinality o... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1014,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804643.json"
} |
Compute the number of ways to color $3$ cells in a $3\times 3$ grid so that no two colored cells share an edge. | <details><summary>Solution</summary>If one of the cells is the center cell, then the number of possible colorings is $\tbinom{4}{2}=6,$ and if no cells are the center cell, then it is not difficult to see that there are $4$ possible configuration, each of which yields $4$ possible colorings. Hence, the answer is
... | [
"There are three cases we need to consider for a configuration that violates the rule (in other words, a grid where two colored cells share an edge)\n1. We have three consecutive colored cells in the same row or column. \n2. We have two consecutive colored squares in the same row or column, and one colored cell tha... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1108,
"boxed": true,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804645.json"
} |
Michel starts with the string HMMT. An operation consists of either replacing an occurrence of H with HM, replacing an occurrence of MM with MOM, or replacing an occurrence of T with MT. For example, the two strings that can be reached after one operation are HMMMT and HMOMT. Compute the number of distinct strings Mich... | The final string must be $14$ characters long, and of the form $HM----------MT$ , where each $-$ is either an $O$ or an $M$ such that no two $O$ 's are adjacent. The number of $O$ 's is an integer from $0$ to $5$ inclusive, so we simply do casework on the number of $O$ 's to get $\tbinom{6}{5}+\tbinom{... | [
"HMMT is the first string\n\n2nd string: HMMMT, HMOMT\n\n3rd string: HMMMMT, HMOMMT, HMMOMT\n\n4th string: HMMMMMT, HMOMMMT, HMMOMMT, HMMMOMT, HMOMOMT\n\nengineering induction tells us its just a Fibonacci sequence, 2-3-5-8-13-21-34-55-89-144, $\\boxed{144}$ is the answer "
] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1022,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804646.json"
} |
Compute the number of nonempty subsets $S \subseteq\{-10,-9,-8, . . . , 8, 9, 10\}$ that satisfy $$ |S| +\ min(S) \cdot \max (S) = 0. $$ | <details><summary>Solution</summary>Note that either $\min(S)$ or $\max(S)$ has an absolute value of $1$ or $(\min(S), \max(S))$ $\in$ $\{(-3,2),$ $(-2,2),$ $(-2,3)\}.$ The former case yields
\begin{align*}
2\binom{2}{2}+2\binom{3}{2}+\ldots+2\binom{10}{2} &=2 \binom{11}{3}
&= 2\cdot 165
&= 330,
\end{... | [
"<details><summary>Solution</summary>Note that we must have $\\min S <0, \\max S>0.$ Let $-a= \\min S$ and $b=\\max S$ , then we get $|S|\\leq a+b+1$ and so $ab\\leq a+b+1.$ This shows that one of $a,b$ must be equal to $1$ or $(a,b)$ must be one of the pairs $(2,2), (2,3)$ or $(3,2).$ We can now... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1122,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804647.json"
} |
Five cards labeled $1, 3, 5, 7, 9$ are laid in a row in that order, forming the five-digit number $13579$ when read from left to right. A swap consists of picking two distinct cards, and then swapping them. After three swaps, the cards form a new five-digit number n when read from left to right. Compute the expecte... | By linearity of expectation, we can compute the expected value of each digit.
In general, the probability that the digit remains the same after three swaps is $$ \left(\frac 35\right)^3 + {3 \choose 2} \cdot \frac 35 \cdot \frac 25 \cdot \frac 1{10} + \frac 25 \cdot \left(\frac 25 - \frac 1{10}\right) \cdot \frac 1{... | [
"Find the sum of $1+3+5+7+9=25$ . Find the probability that it will be the same digit. Then, we can have $\\frac{6^3+3\\cdot 6\\cdot 4+4\\cdot 3}{10^3}=\\frac{3}{10}$ . Thus, for each one its essentially the same. By LOE, we have $10000\\cdot \\left(\\frac{3}{10}+\\frac{7}{10}\\cdot \\frac{25-1}{4}\\right)+1000\... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1018,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804648.json"
} |
The numbers $1, 2, . . . , 10$ are randomly arranged in a circle. Let $p$ be the probability that for every positive integer $k < 10$ , there exists an integer $k' > k$ such that there is at most one number between $k$ and $k'$ in the circle. If $p$ can be expressed as $\frac{a}{b}$ for relatively prime ... | The idea is to use recursion. Let $a_n$ be the number of *ways* to permute $1, 2, 3, \cdots, n$ cyclically such that the condition holds. Call a number $a$ *close enough* to $b$ if there is at most one number between $a$ and $b$ .
To develop a recursive formula for $a_n$ , we consider two cases. In the fir... | [
"<details><summary>solution</summary>There are at least 3 numbers less than or equal to 10 that are greater than numbers 1-7, so they will never be directly adjacent to all the numbers that are greater than them. The condition is only not satisfied when 9 and 10 are adjacent or 8 is between 9 and 10. Since reflecti... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1096,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804649.json"
} |
Let $S = \{(x, y) \in Z^2 | 0 \le x \le 11, 0\le y \le 9\}$ . Compute the number of sequences $(s_0, s_1, . . . , s_n)$ of elements in $S$ (for any positive integer $n \ge 2$ ) that satisfy the following conditions: $\bullet$ $s_0 = (0, 0)$ and $s_1 = (1, 0)$ , $\bullet$ $s_0, s_1, . . . , s_n$ are distinc... | Observe that consecutive moves are either one forward move or one left move. Define a *run* as a maximal subsequence of $s$ in a line. Then $s$ can be segmented into alternating runs horizontally and vertically.
Given a set of numbers, we can determine the direction of the runs in the sequence due to the non-inter... | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 1026,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804650.json"
} |
Random sequences $a_1, a_2, . . .$ and $b_1, b_2, . . .$ are chosen so that every element in each sequence is chosen independently and uniformly from the set $\{0, 1, 2, 3, . . . , 100\}$ . Compute the expected value of the smallest nonnegative integer $s$ such that there exist positive integers $m$ and $n$ ... | [] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804651.json"
} | |
Consider permutations $(a_0, a_1, . . . , a_{2022})$ of $(0, 1, . . . , 2022)$ such that $\bullet$ $a_{2022} = 625$ , $\bullet$ for each $0 \le i \le 2022$ , $a_i \ge \frac{625i}{2022}$ , $\bullet$ for each $0 \le i \le 2022$ , $\{a_i, . . . , a_{2022}\}$ is a set of consecutive integers (in some order).... | Typo: In the second condition, $a_i \le \frac{625i}{2022}$ should be $a_i \ge \frac{625i}{2022}$ . | [
"<details><summary>ans idk</summary>216695????????????????????????????????????????</details>"
] | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 4,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804652.json"
} |
Let $S$ be a set of size $11$ . A random $12$ -tuple $(s_1, s_2, . . . , s_{12})$ of elements of $S$ is chosen uniformly at random. Moreover, let $\pi : S \to S$ be a permutation of $S$ chosen uniformly at random. The probability that $s_{i+1}\ne \pi (s_i)$ for all $1 \le i \le 12$ (where $s_{13} = s_1... | We claim that the probability is $\frac{10^{12}+4}{11^{12}}$ . We can count this as follows.
First, note that if it was not in a circle (without the condition at $i=12$ ), then we can fix the first element ( $a_i$ ), and then each element thereafter has a $\frac{10}{11}$ chance of not being the mapping of the pr... | [
"<details><summary>Answer</summary>1000000000004</details>\n<details><summary>Solution</summary>Let's count the number $N$ of ordered pairs $(T, \\pi)$ , where $T$ is the 12-tuple, satisfying the given condition.\n\nWe will use inclusion-exclusion on the 12 constraints of the form $s_{i+1} \\neq \\pi(s_i)$ . ... | [
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] | {
"answer_score": 180,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804653.json"
} |
Find all natural numbers $n$ for which there is a permutation $\sigma$ of $\{1,2,\ldots, n\}$ that satisfies:
\[
\sum_{i=1}^n \sigma(i)(-2)^{i-1}=0
\] | Awesome problem, the hardest and nicest problem of the test, very similar to <details><summary>past IMO problem</summary>IMO 2012, P6</details> in terms of the ideas involved. Solved with $\textbf{Aryan-23}$ . $\textbf{Lemma 1:}$ $n \equiv 1 \pmod{3}$ do not work. $\textbf{Proof}$
Assume for the sake of contradi... | [
"It's easy to get $n \\equiv 0,2 \\pmod 3$ and then you die for the rest of the contest. ",
"NVT's problem!!\n\nThis was easily the best problem on the test according to me. \n\nSoln. Looking mod 3, we get that $3|n^2+n$ and thus, $n\\not\\equiv 1 \\pmod{3}$ . This is kinda a thing that's easy to miss and th... | [
"origin:aops",
"2022 Contests",
"2022 India National Olympiad"
] | {
"answer_score": 176,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2022 Contests/2022 India National Olympiad/2795409.json"
} |
Let $D$ be an interior point on the side $BC$ of an acute-angled triangle $ABC$ . Let the circumcircle of triangle $ADB$ intersect $AC$ again at $E(\ne A)$ and the circumcircle of triangle $ADC$ intersect $AB$ again at $F(\ne A)$ . Let $AD$ , $BE$ , and $CF$ intersect the circumcircle of triangle $... | Guys I found a solution heavily using the properties of miquel point !
<blockquote>Let $D$ be an interior point on the side $BC$ of an acute-angled triangle $ABC$ . Let the circumcircle of triangle $ADB$ intersect $AC$ again at $E(\ne A)$ and the circumcircle of triangle $ADC$ intersect $AB$ again at $... | [
"<details><summary>Thanks INMO :)</summary>\n\n- Angle chase to show $BE_1$ and $CF_1$ are angle bisectors in $D_1E_1F_1$ .\n- Chase more to show $\\angle EDF = \\angle E_1D_1F_1$ (infact the corresponding lines are parallel).\n- Apply $\\angle BIC = 90 + \\angle A/2$ in $DEF$ , $D_1E_1F_1$ to finish.\n... | [
"origin:aops",
"2022 Contests",
"2022 India National Olympiad"
] | {
"answer_score": 170,
"boxed": false,
"end_of_proof": true,
"n_reply": 40,
"path": "Contest Collections/2022 Contests/2022 India National Olympiad/2795412.json"
} |
For a positive integer $N$ , let $T(N)$ denote the number of arrangements of the integers $1, 2, \cdots N$ into a sequence $a_1, a_2, \cdots a_N$ such that $a_i > a_{2i}$ for all $i$ , $1 \le i < 2i \le N$ and $a_i > a_{2i+1}$ for all $i$ , $1 \le i < 2i+1 \le N$ . For example, $T(3)$ is $2$ , since ... | For the sake of completeness, here's a full solution.
Consider a graph with vertices as $1,2,\cdots,N$ and draw an arrow $i \rightarrow j$ if $j = 2i$ or $j = 2i+1$ . For $N = 2^n - 1$ , this actually is a binary tree with $n-1$ levels with all numbers in the range $[2^k, 2^{k+1})$ on the $k$ th level.
... | [
"Legendre Formula , and Counting works... I am not sure that my solution works , but I got a general formula for $\\nu _{2} ( T(2k+1))$ .. ( and this was the only problem I could solve in Test :stink:",
" $T(7)=80$ . Could not formally write the complete proof of second part but it required to be deal with pow... | [
"origin:aops",
"2022 Contests",
"2022 India National Olympiad"
] | {
"answer_score": 1178,
"boxed": false,
"end_of_proof": false,
"n_reply": 21,
"path": "Contest Collections/2022 Contests/2022 India National Olympiad/2795422.json"
} |
Given a monic quadratic polynomial $Q(x)$ , define \[ Q_n (x) = \underbrace{Q(Q(\cdots(Q(x))\cdots))}_{\text{compose $n$ times}} \]
for every natural number $n$ . Let $a_n$ be the minimum value of the polynomial $Q_n(x)$ for every natural number $n$ . It is known that $a_n > 0$ for every natural number $n$ ... | <details><summary>Official Solution (by Fajar Yuliawan)</summary><details><summary>(a)</summary>Let $Q(a) = a_1$ , then $Q(x) = (x - a)^2 + a_1$ .
Claim 1: $a_1 > a$ Proof. Suppose (FTSOC) $a \geq a_1$ . Notice that $Q_1(a) = a_1$ . Assume that $Q_n(b_n) = a_1$ where $b_n \geq a$ for some $n \geq 1$ . Let $b... | [
"Nice one. Who is the author?",
"We do this more generally for a strictly convex twice differentiable function $f$ with a single minimum in $x_0$ .\nWe classify $3$ possible behaviours for $a_n$ . $I$ ) $f(x)>x$ $\\forall x\\in\\mathbb{R}$ . Therefore we have $t=inf(im(f(x)-x))=min(im(f(x)-x))>0$ , or in... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 168,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2735502.json"
} |
Five numbers are chosen from $\{1, 2, \ldots, n\}$ . Determine the largest $n$ such that we can always pick some of the 5 chosen numbers so that they can be made into two groups whose numbers have the same sum (a group may contain only one number). | I think,answer is a $$ n
\leq15 $$ {1,2,4,8,16} for $$ n\geq 16 $$ | [
"<blockquote>I think,answer is a15 ${1,2,4,8,16}$ for $n>=16$ </blockquote>\n\nWrong:7,8,9,11,14",
"I think answer is 12\n6,9,11,12,13 for n >=13",
"Bump this ",
"Bump bump bump",
"bump bump\n",
"Soooo, this is actually a pretty nice problem. \nFirst we show that n>=13 is impossible. \nWe can easily che... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2735505.json"
} |
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