problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
Suppose $A,B\in M_3(\mathbb C)$ , $B\ne0$ , and $AB=0$ . Prove that there exists $D\in M_3(\mathbb C)$ with $D\ne0$ such that $AD=DA=0$ . | We can easily prove a much more general result, as follows.
<u>Proposition</u>. Let $k$ be a field, and let $R$ be an algebraic $k$ -algebra. If $a \in R,$ then either $a \in U(R),$ the group of units of $R,$ or $ab=ba=0$ for some $0 \ne b \in R.$ <u>Proof</u>. Suppose that $a \notin U(R).$ Since $R$... | [] | [
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] | {
"answer_score": 46,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577589.json"
} |
Does there exist a twice differentiable function $f:\mathbb R\to\mathbb R$ such that $f'(x)=f(x+1)-f(x)$ for all $x$ and $f''(0)\ne0$ ? Justify your answer. | [] | [
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577590.json"
} | |
A walker and a jogger travel along the same straight line in the same direction. The walker walks at one meter per second, while the jogger runs at two meters per second. The jogger starts one meter in front of the walker. A dog starts with the walker, and then runs back and forth between the walker and the jogger with... | [] | [
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577785.json"
} | |
Given that $40!=\overline{abcdef283247897734345611269596115894272pqrstuvwx}$ , find $a,b,c,d,e,f,p,q,r,s,t,u,v,w,x$ .
| Indeed, the official solution is:
By examining the numbers $5,10,15,20,25,30,35,40$ (at most $40$ and divisible by $5$ ), we see that the power of $5$ dividing $40!$ is $9$ . Also it is easy to see that the power of $2$ dividing $40!$ is at least $9$ . Therefore $10^9$ divides $40!$ and we deduce th... | [
" $v_5(40!) = $ floor $(\\frac{40}{5}) + $ floor $(\\frac{40}{25}) + \\cdots$ $\\implies v_5(40!) = 9$ Observe there are exactly 9 variables at the end. \nSo $p = q = r = s = t = u = v = w = x = 0$ :):)\n<details><summary>please wait</summary>Still working on a,b,c,d,e,f.</details>",
"We already found the rig... | [
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] | {
"answer_score": 48,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577786.json"
} |
Two circles $\alpha,\beta$ touch externally at the point $X$ . Let $A,P$ be two distinct points on $\alpha$ different from $X$ , and let $AX$ and $PX$ meet $\beta$ again in the points $B$ and $Q$ respectively. Prove that $AP$ is parallel to $QB$ . | The homothety at $X$ transforming $\alpha$ to $\beta$ carries $AP$ to $BQ$ and hence the conclusion. | [] | [
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577954.json"
} |
Let $n$ be a nonzero integer. Prove that $n^4-7n^2+1$ can never be a perfect square. | Multiplying by four we want $(2n^2-7)^2-45$ to be a square. The only squares with difference $45$ are $23^2-22^2,9^2-6^2, 7^2-2^2$ . The equation $2n^2 - 7 = \pm 23, \pm 9, \pm 7$ has an integral solution only for $n=0$ , which is prohibited by the problem statement. | [
"Hint: just solve this: $(n^2-4)^2<n^4-7n^2+1<(n^2-3)^2$ "
] | [
"origin:aops",
"2009 VTRMC",
"Undergraduate Contests",
"VTRMC"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2009 VTRMC/2577955.json"
} |
Solve in $R$ the equation: $8x^3-4x^2-4x+1=0$ | <blockquote><blockquote>Solve in $R$ the equation: $8x^3-4x^2-4x+1=0$ </blockquote> $\left\{\cos\frac{\pi}{7},\cos\frac{3\pi}{7},\cos\frac{5\pi}{7}\right\}$ .</blockquote>
Because $\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{3\pi}{7}+... | [
"See here:\nhttps://www.wolframalpha.com/input/?i=8x%5E3-4x%5E2-4x%2B1%3D0",
"<blockquote>Solve in $R$ the equation: $8x^3-4x^2-4x+1=0$ </blockquote> $\\left\\{\\cos\\frac{\\pi}{7},\\cos\\frac{3\\pi}{7},\\cos\\frac{5\\pi}{7}\\right\\}$ .\n",
"For the collection.\nSolve the following equation. $$ x^5+x^4-12x... | [
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/1318258.json"
} |
Let $d$ be a positive integer and let $A$ be a $d\times d$ matrix with integer entries. Suppose $I+A+A_2+\ldots+A_{100}=0$ (where $I$ denotes the identity $d\times d$ matrix, and $0$ denotes the zero matrix, which has all entries $0$ ). Determine the positive integers $n\le100$ for which $A_n+A_{n+1}+... | [] | [
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564178.json"
} | |
Define a sequence by $a_1=1,a_2=\frac12$ , and $a_{n+2}=a_{n+1}-\frac{a_na_{n+1}}2$ for $n$ a positive integer. Find $\lim_{n\to\infty}na_n$ .
| It's clear $a_n>0$ and decreasing. It has limit so $\lim_{n\to \infty}\frac{a_n-a_{n+1}}{a_{n+1}-a_{n+2}}=1$ $\lim_{n\to\infty}na_n=\lim \frac{n}{\frac{1}{a_n}}=\lim \frac{n+1-n}{\frac{a_n-a_{n+1}}{a_n\cdot a_{n+1}}}=2$ | [] | [
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] | {
"answer_score": 6,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564179.json"
} |
Let $\sum_{n=1}^\infty a_n$ be a convergent series of positive terms (so $a_i>0$ for all $i$ ) and set $b_n=\frac1{na_n^2}$ for $n\ge1$ . Prove that $\sum_{n=1}^\infty\frac n{b_1+b_2+\ldots+b_n}$ is convergent. | One idea that I get is comparing the convergence nature to that of harmonic series. So $a_n $ must be smaller than some $\frac{1}{n^{1+\epsilon}}$ for some $ \epsilon > 0$ . $\Rightarrow b_n \ge n^{1+2\epsilon}$ .
Now the sum $\sum_{i=1}^n b_i $ has as its $n$ th sum proportional to $n^{2+\epsilon_2}$ , for ... | [
"Any ideas?",
"[https://artofproblemsolving.com/community/c7h562849p3285201](https://artofproblemsolving.com/community/c7h562849p3285201)"
] | [
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] | {
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564180.json"
} |
For $n$ a positive integer, define $f_1(n)=n$ and then for $i$ a positive integer, define $f_{i+1}(n)=f_i(n)^{f_i(n)}$ . Determine $f_{100}(75)\pmod{17}$ . Justify your answer. | [
"we have f1(75)mod17=7^7mod17=12\n12^12mod17=4\n4^4mod17=1\nso f100(75)mod17=1"
] | [
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564656.json"
} | |
Let $\triangle ABC$ be a triangle with sides $a,b,c$ and corresponding angles $A,B,C$ (so $a=BC$ and $A=\angle BAC$ etc.). Suppose that $4A+3C=540^\circ$ . Prove that $(a-b)^2(a+b)=bc^2$ . | $A=3B,C=180-4B, B<45$ $(a-b)^2(a+b)=bc^2 \to (\frac{ \sin 3B}{\sin B}-1)^2 (\frac{ \sin 3B}{\sin B}+1)=(\frac{\sin 4B}{\sin B})^2$ $(2-4 \sin^2 B)^2(4-4 \sin^2 B) = 16 cos^2 B \cos^2 2B$ $(1-2\sin^2 B)^2 (1- \sin^2B)=\cos^2B \cos^2 2B$ But $1-2\sin^2B=\cos 2B, 1-\sin^2B =\cos^2B$ or $ (\frac{ \sin 3B}{\sin B}-1)... | [] | [
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] | {
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564658.json"
} |
Let $A,B$ be two circles in the plane with $B$ inside $A$ . Assume that $A$ has radius $3$ , $B$ has radius $1$ , $P$ is a point on $A$ , $Q$ is a point on $B$ , and $A$ and $B$ touch so that $P$ and $Q$ are the same point. Suppose that $A$ is kept fixed and $B$ is rolled once round the ins... | [
"I thought something like this showed up on Mathical 2020."
] | [
"origin:aops",
"Undergraduate Contests",
"2010 VTRMC",
"VTRMC"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2010 VTRMC/2564660.json"
} | |
Evaluate $\int^4_1\frac{x-2}{(x^2+4)\sqrt x}dx$ | Let's bash this :blush:
set $u=\sqrt{x} \implies du=\frac{1}{2\sqrt{x}}$ so our integral changes to $I=\int_{1}^{2} \frac{2u^2-4}{u^{4}+4}du \implies I=\int_{1}^{2} \frac{2u^2-4}{(u^2-2u+2)(u^2+2u+2)}du$ $I=\int_{1}^{2} \frac{(u^2-2u+2)+(u^2+2u+2)-8}{(u^2-2u+2)(u^2+2u+2)}du \implies I=\int_{1}^{2} \frac{du}{u^2+2... | [
"first substitute $x=t^2$ and divide the numerator and denominator by t^2 (to make it visually appealing) after which substituting $t+\\frac{2}{t}$ as $y$ should do the trick",
"Or, $\\frac{2u^2-4}{u^{4}+4}=\\frac{u-1}{u^2-2u+2}-\\frac{u+1}{u^2+2u+2}$ . This means $\\int\\frac{x-2}{(x^2+4)\\sqrt{x}}\\ dx=... | [
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] | {
"answer_score": 1008,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561278.json"
} |
Find $\sum_{k=1}^\infty\frac{k^2-2}{(k+2)!}$ . | $\frac{k^2 - 2}{(k+2)!}=\frac{k^2 - 4+2}{(k+2)!}=\frac{k- 2}{(k+1)!} + \frac{2}{(k+2)!}=\frac{k+1-3}{(k+1)!} + \frac{2}{(k+2)!}=\frac{1}{k!}-\frac{3}{(k+1)!} + \frac{2}{(k+2)!}$ .
Taking the sum over $k=1,2,...$ we have by the help of natural number expansion, $\sum_{k=1} \frac{1}{k!} - 3\sum_{k=1} \frac{1}{(k+1)!} ... | [
"Isn't that $0$ ?",
"Obviously, and the proof?"
] | [
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] | {
"answer_score": 6,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561279.json"
} |
Find $\lim_{x\to\infty}\left((2x)^{1+\frac1{2x}}-x^{1+\frac1x}-x\right)$ .
| This just approaches to $2x-x-x=\boxed{0}$ . | [
"It's actually <details><summary>answer</summary>$\\ln2$</details>, according to WA and the answer key.\nFor instance, you can't say that $\\lim_{x\\to\\infty}\\left(1+\\frac1x\\right)^x=\\lim_{x\\to\\infty}1^x=1$ .",
"Oh ok, I just learned calculus, havent done limits with exponents yet",
"This is a principle... | [
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] | {
"answer_score": 1002,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561281.json"
} |
A sequence $(a_n)$ is defined by $a_0=-1,a_1=0$ , and $a_{n+1}=a_n^2-(n+1)^2a_{n-1}-1$ for all positive integers $n$ . Find $a_{100}$ . | [
" $a_0=-1,a_1=1,a_2=3,a_3=8,a_4=15,\\hdots$ . We see $a_n=n^2-1$ .\n"
] | [
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561538.json"
} | |
Let $m,n$ be positive integers and let $[a]$ denote the residue class $\pmod{mn}$ of the integer $a$ (thus $\{[r]|r\text{ is an integer}\}$ has exactly $mn$ elements). Suppose the set $\{[ar]|r\text{ is an integer}\}$ has exactly $m$ elements. Prove that there is a positive integer $q$ such that $q$ ... | [] | [
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561541.json"
} | |
Let $S$ be a set with an asymmetric relation $<$ ; this means that if $a,b\in S$ and $a<b$ , then we do not have $b<a$ . Prove that there exists a set $T$ containing $S$ with an asymmetric relation $\prec$ with the property that if $a,b\in S$ , then $a<b$ if and only if $a\prec b$ , and if $x,y\in T$ ... | [
"Is the relation transitive?\n"
] | [
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561543.json"
} | |
Let $P(x)=x^{100}+20x^{99}+198x^{98}+a_{97}x^{97}+\ldots+a_1x+1$ be a polynomial where the $a_i~(1\le i\le97)$ are real numbers. Prove that the equation $P(x)=0$ has at least one nonreal root. | Let be $x_1,x_2,\dots,x_{100}$ the roots of $P(x)$ .
Using the Vieta's relations, results: $\sum_{i=1}^{100}x_i=-20\quad(1)$ ; $\sum_{1\le i<j\le100}x_ix_j=198\quad(2)$ ; $\prod_{i=1}^{100}x_i=1\quad(3)$ .
Assume $x_i\in\mathbb{R},\;\forall i\in\{1,2,\dots,100\}$ .
Then: $\sum_{1\le i<j\le100}(x_i-x_j)^2=99\cdot\s... | [] | [
"origin:aops",
"Undergraduate Contests",
"2011 VTRMC",
"VTRMC"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2011 VTRMC/2561544.json"
} |
Evaluate $\sum_{n=1}^\infty \frac{n}{(2^n-2^{-n})^2}+\frac{(-1)^nn}{(2^n-2^{-n})^2}$ | For $|q|<1$ , we have $\sum_{k=1}^{\infty} q^k=\frac{q}{(q-1)}$ . Therefore for $|q|>1$ ,
\begin{align*}
\sum_{n=1}^{\infty} \frac{(-1)^n}{q^n-1}=\sum_{n=1}^{\infty} \frac{(-1)^nq^{-n}}{1-q^{-n}}
\end{align*}
\begin{align*}
\sum_{n=1}^{\infty} \frac{1}{q^n+1}&=\sum_{n=1}^{\infty} \frac{q^{-n}}{1+q^{-n}} &=\sum_{n=... | [
" $S=\\sum\\limits_{n=1}^{+\\infty }{\\left( \\frac{n}{\\left( 2^{n}-2^{-n} \\right)^{2}}+\\frac{n\\left( -1 \\right)^{n}}{\\left( 2^{n}-2^{-n} \\right)^{2}} \\right)}=4\\sum\\limits_{n=1}^{+\\infty }{\\frac{n}{\\left( 2^{2n}-2^{-2n} \\right)^{2}}}=4\\sum\\limits_{n=1}^{+\\infty }{\\frac{n}{\\left( 4^{n}-4^{-n} \\r... | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] | {
"answer_score": 112,
"boxed": false,
"end_of_proof": true,
"n_reply": 7,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/1279364.json"
} |
Prove that $$ \frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}
\leq\frac{3\sqrt{3}}{2} $$ for any positive real numbers $x, y,z$ such that $x+y+z = xyz.$ [2008 VTRMC #1](https://artofproblemsolving.com/community/c7h236610p10925499)
[here](http://www.math.vt.edu/people/plinnell/Vtregional/solut... | <blockquote>Prove that $$ \frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}
\leq\frac{3\sqrt{3}}{2} $$ for any positive real numbers $x, y,z$ such that $x+y+z = xyz.$ [2008 VTRMC #1](https://artofproblemsolving.com/community/c7h236610p10925499)
[here](http://www.math.vt.edu/people/plinnell/Vtre... | [
"<details><summary>Hint</summary>In a triangle we have $tanAtanBtanC=tanA+tanB+tanC$</details>\n<details><summary>If you still don't know</summary>So we let $x=tanA,y=tanB,z=tanC$ And $LHS=sinA+sinB+sinC$ Then we're left with a well-known ineq.\nJust use Jensen and we can get the result.</details>",
"<blockquo... | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/1700826.json"
} |
Let $I=3\sqrt2\int^x_0\frac{\sqrt{1+\cos t}}{17-8\cos t}dt$ . If $0<x<\pi$ and $\tan I=\frac2{\sqrt3}$ , what is $x$ ? | Rewrite $I$ as \[I=6\int_{0}^{x} \frac{\cos \frac{t}{2}}{9+16\sin ^2 \frac{t}{2}}\, dt.\] Then by letting $u=\sin \frac{t}{2}, du=\frac{1}{2}\cos \frac{t}{2} \, dt$ , we get \[I=12 \int_{0}^{\sin \frac{x}{2}} \frac{du}{9+16u^2}.\] We then perform another substitution, $u=\frac{3}{4}v, du=\frac{3}{4} \, dv$ , \[I=\i... | [] | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] | {
"answer_score": 1010,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/2559093.json"
} |
Define a sequence $(a_n)$ for $n\ge1$ by $a_1=2$ and $a_{n+1}=a_n^{1+n^{-3/2}}$ . Is $(a_n)$ convergent (i.e. $\lim_{n\to\infty}a_n<\infty$ )?
| Clearly $a_n$ is increasing so being convergent is equivalent to being bounded from above which is equivalent to $\log a_n$ being bounded from above. But
\[\log a_{n+1}=\log 2 \cdot \prod_{n=1}^N \left(1+n^{-3/2}\right) \le \log 2 \cdot \prod_{n=1}^N e^{n^{-3/2}} = \log 2 \cdot e^{\sum_{n=1}^N n^{-3/2}}\]
which is ... | [] | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] | {
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/2559094.json"
} |
Let
\begin{align*}X&=\begin{pmatrix}7&8&98&-9&-7-7&-7&9\end{pmatrix}Y&=\begin{pmatrix}9&8&-98&-7&77&9&8\end{pmatrix}.\end{align*}Let $A=Y^{-1}X$ and let $B$ be the inverse of $X^{-1}+A^{-1}$ . Find a matrix $M$ such that $M^2=XY-BY$ (you may assume that $A$ and $X^{-1}+A^{-1}$ are invertible). | [] | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/2559095.json"
} | |
Let $ABC$ be a right-angled triangle with $\angle ABC=90^\circ$ , and let $D$ be on $AB$ such that $AD=2DB$ . What is the maximum possible value of $\angle ACD$ ? | WLOG let $BD=1,AD=2$ . Consider a ray with vertex $B$ perpendicular to $AB$ . Let $O$ be a point on the perpendicular bisector of $AD$ , starting at its midpoint and moving in the direction of the ray. Considering the circle centered at $O$ through $A,D$ , all points on major arc $AD$ of that circle result ... | [] | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2013 VTRMC"
] | {
"answer_score": 1024,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2013 VTRMC/2559448.json"
} |
A positive integer $n$ is called special if it can be represented in the form $n=\frac{x^2+y^2}{u^2+v^2}$ , for some positive integers $x,y,u,v$ . Prove that
(a) $25$ is special;
(b) $2014$ is not special;
(c) $2015$ is not special. | (b) Notice that $19|2014$ and $19\equiv 3\text{(mod 4)}$ . We must have $19|x^2+y^2\implies 2|v_{19}(x^2+y^2)$ . This means that we also must have $19|u^2+v^2\implies 2|v_{19}(u^2+v^2)$ , but then $2|v_{19}(\frac{x^2+y^2}{u^2+v^2})$ , contradiction.
(c) Same thing as above, except for $31$ . | [
"(a) $\\frac{6^2+8^2}{2^2+2^2}=25$ ",
"<blockquote>(a) $\\frac{6^2+8^2}{2^2+2^2}=25$ </blockquote> $\\frac{6^2+8^2}{2^2+2^2}=\\frac{25}{2}$ Instead, consider $\\frac{5^2+5^2}{1^2+1^2}$ ",
"More universally, we can proof if for any p is 3(mod4),the power of such p in n is even,then n is special.\n(after revis... | [
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Find the least positive integer $n$ such that $2^{2014}$ divides $19^n-1$ . | LTE gives, if $n$ is even, $v_2(19^n-1)=2+v_2(n)\ge2014$ , so we must have $v_2(n)\ge2012$ , and the minimum for even $n$ is $2^{2012}$ . If $n$ is odd, then (again by LTE) we have $v_2(19^n-1)=1<2014$ , so the answer is $\boxed{2^{2012}}$ . (Can someone check this?) | [
"By LTE, $$ \\nu_2(19^n - 1) = \\nu_2(18) + \\nu_2(20) + \\nu_2(n) - 1 = 1 + 2 + \\nu_2(n) - 1 = 2 + \\nu_2(n). $$ We need this to be greater than $2014$ , so we must have $\\nu_2(n)\\geq 2012\\iff 2^{2012} | n$ . Since $n$ is positive, it follows that the smallest such $n$ is $\\boxed{2^{2012}}$ .\n\ned... | [
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} |
Find $\sum_{n=2}^\infty\frac{n^2-2n-4}{n^4+4n^2+16}$ .
| $\frac{n^2-2n-4}{n^4+4n^2+16}=\frac{n^2-2n-4}{(n^2-2n+4)(n^2+2n+4)}=\frac 12\times (\frac{n-2}{n^2-2n+4}-\frac{n}{n^2+2n+4})$ .
Note that $\frac{n+2}{n^2+2n+4}=\frac{n+2}{(n+2)^2-2(n+2)+4}$ , thus $\sum_{n=2}^\infty\frac{n^2-2n-4}{n^4+4n^2+16}=\frac 12\sum_{n=2}^\infty(\frac{n-2}{n^2-2n+4}-\frac{n}{n^2+2n+4})=\frac ... | [
"In the telescoping sum, it is not remain the first two members? ",
"<details><summary>solution</summary>The following telescoping works $$ \\sum_{n=2}^{\\infty} \\frac{n^2-2n-4}{(n^2-2n+4)(n^2+2n+4)} = \\frac{1}{2}\\sum_{n=2}^{\\infty}\\left(\\frac{n-2}{(n-1)^2+3}-\\frac{n}{(n+1)^2+3}\\right) = \\boxed{\\frac{... | [
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Evaluate $\int^2_0\frac{x(16-x^2)}{16-x^2+\sqrt{(4-x)(4+x)(12+x^2)}}dx$ .
| <details><summary>solution</summary>Divide both the numerator and the denominator by $(16-x^2)$ and perform the substitution $t=x^2$ . The required integral then becomes $$ I = \frac{1}{2}\int_{0}^{4}\frac{dt}{1+\sqrt{\frac{12+t}{16-t}}} $$ Making the change of variables $t\mapsto 4-t$ , we get $$ I = \frac{1}{... | [] | [
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Suppose we are given a $19\times19$ chessboard (a table with $19^2$ squares) and remove the central square. Is it possible to tile the remaining $19^2-1=360$ squares with $4\times1$ and $1\times4$ rectangles? (So that each of the $360$ squares is covered by exactly one rectangle.) Justify your answer. | It can't be done.
Mod a $4\times 4$ binary array, initially containing all zeroes, each $1\times 4$ rectangle inverts the bits in a row and each $4\times 1$ rectangle inverts the bits in a column. In either case, for any row $(a,b,c,d)$ in the array, any other is always $(a,b,c,d)$ or $(1-a,1-b,1-c,1-d)$ .... | [] | [
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} |
Let $n\ge1$ and $r\ge2$ be positive integers. Prove that there is no integer $m$ such that $n(n+1)(n+2)=m^r$ .
| <blockquote>Hint: because $(n,n+1,n+2)=1$ , we must have $n=a^r, n+1=b^r, n+2=c^r$ .</blockquote>
What! No, that is wrong on many levels. Take numbers $1, 7, 49$ . We have $\gcd(1, 7, 49) = 1$ but even though $1 \cdot 7 \cdot 49 = 7^3$ not all of $1, 7, 49$ are perfect cubes.
Here is a solution. We note that... | [
"Hint: because $n,n+1,n+2$ are relative prime two by two, we must have $n=a^r, n+1=b^r, n+2=c^r$ .",
"I used the same idea, butit was wrong:\n<blockquote> both $n+1$ and $(n+1)^2-1$ are perfect powers</blockquote>\nIn your case why it is true?",
"<blockquote>I used the same idea, butit was wrong:\n<block... | [
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} |
Let $A,B$ be two points in the plane with integer coordinates $A=(x_1,y_1)$ and $B=(x_2,y_2)$ . (Thus $x_i,y_i\in\mathbb Z$ , for $i=1,2$ .) A path $\pi:A\to B$ is a sequence of **down** and **right** steps, where each step has an integer length, and the initial step starts from $A$ , the last step ending at ... | [] | [
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Let $S$ denote the set of $2$ by $2$ matrices with integer entries and determinant $1$ , and let $T$ denote those matrices of $S$ which are congruent to the identity matrix $I\pmod3$ (so $\begin{pmatrix}a&bc&d\end{pmatrix}\in T$ means that $a,b,c,d\in\mathbb Z,ad-bc=1,$ and $3$ divides $b,c,a-1,d-1$... | [] | [
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Find all n such that $n^{4}+6n^{3}+11n^{2}+3n+31$ is a perfect square. | So I think it's safe to assume that $n$ is an integer
(otherwise, the expression is equal to each perfect square $\geq 36$ for one positive and one negative $n$ value)
<details><summary>Solution</summary>Let that be $f(n)$ $(n^{2}+3n+1)^{2}-f(n)=3n+30$ , positive for $n>-10$ $f(n)-(n^{2}+3n)^{2}=2n^{2}+... | [
"You made a mistake.\r $(n^{2}+3n+1)^{2}-f(n) = 3n-30$ \r\nThough it seems right other than that. Thanks.\r\n\r\nAnd n is integer."
] | [
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The planar diagram below, with equilateral triangles and regular hexagons, sides length $2$ cm, is folded along the dashed edges of the polygons, to create a closed surface in three-dimensional Euclidean spaces. Edges on the periphery of the planar diagram are identified (or glued) with precisely one other edge on th... | [
"Hint: It's a truncated tetrahedron."
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Let $(a_i)_{1\le i\le2015}$ be a sequence consisting of $2015$ integers, and let $(k_i)_{1\le i\le2015}$ be a sequence of $2015$ positive integers (positive integer excludes $0$ ). Let $$ A=\begin{pmatrix}a_1^{k_1}&a_1^{k_2}&\cdots&a_1^{k_{2015}}a_2^{k_1}&a_2^{k_2}&\cdots&a_2^{k_{2015}}\vdots&\vdots&\ddots&\v... | The first part itself is sufficient. For prime $p$ (we only care about $p\le2015$ ) and $k\in\mathbb N$ there are at least $\left\lceil\frac{2015}{p^k}\right\rceil$ of $a$ 's congruent $\bmod\,p^k$ by the pigeonhole principle, and so
\[v_p(\det A)
\ge v_p\left(\prod_{i<j}(a_i-a_j)\right)
\ge\sum_{k=1}^\infty\... | [
"If some $a_i=0$ or some $a_i=a_j,i\\not= j$ or some $k_i=k_j,i\\not= j$ , then $\\det(A)=0$ ; then we assume that it is not the case.\nAs for the Vandermonde determinant, by recurrence and linear combinations of columns, we show that $\\Pi_{i<j}(a_i-a_j)$ divides $\\det(A)$ .\nUnlike the Vandermonde case,... | [
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Consider the harmonic series $\sum_{n\ge1}\frac1n=1+\frac12+\frac13+\ldots$ . Prove that every positive rational number can be obtained as an unordered partial sum of this series. (An unordered partial sum may skip some of the terms $\frac1k$ .) | Well, for me a "partial sum" is always something finite. In any case, if we allow infinite sums, the problem becomes almost trivial (modulo the divergence of the harmonic series), while for finite sums one needs to be slightly more clever.
To put the problem in other words, we want to represent every positive rational ... | [
"If an infinite number of terms is allowed, this is true for any positive number, not just the rationals. Maybe you meant partial sum with finitely many terms?",
"All that I know is this. If $\\sum a_n$ is a series of real numbers that is converge but not absolutely, then for every $-\\infty \\le \\alpha \\le \... | [
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Evaluate $\int^\infty_0\frac{\operatorname{arctan}(\pi x)-\operatorname{arctan}(x)}xdx$ (where $0\le\operatorname{arctan}(x)<\frac\pi2$ for $0\le x<\infty$ ). | <details><summary>pretty standard approach if you've seen similar problems before</summary>This is equivalent to $$ \int_0^\infty \int_1^\pi \frac{da\, dx}{1+a^2x^2} $$ $$ \int_1^\pi\int_0^\infty \frac{dx\, da}{1+a^2x^2} $$ $$ \frac{\pi}{2}\int_1^\pi \frac{da}{a} $$ $$ \frac{\pi\ln\pi}{2} $$</details> | [
"Frullani integral will help"
] | [
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Let $(a_1,b_1),\ldots,(a_n,b_n)$ be $n$ points in $\mathbb R^2$ (where $\mathbb R$ denotes the real numbers), and let $\epsilon>0$ be a positive number. Can we find a real-valued function $f(x,y)$ that satisfies the following three conditions?
1. $f(0,0)=1$ ;
2. $f(x,y)\ne0$ for only finitely many $(x,... | [] | [
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Let $n$ be a positive integer and let $x_1,\ldots,x_n$ be $n$ nonzero points in $\mathbb R^2$ . Suppose $\langle x_i,x_j\rangle$ (scalar or dot product) is a rational number for all $i,j$ ( $1\le i,j\le n$ ). Let $S$ denote all points of $\mathbb R^2$ of the form $\sum_{i=1}^na_ix_i$ where the $a_i$ ... | [] | [
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} | |
Determine the number of real solutions to the equation $\sqrt{2 -x^2} = \sqrt[3]{3 -x^3}.$ | [
"Just write $(2-x^2)^3=(3-x^3)^2$ i.e. $2x^6-6x^4-6x^3+12x^2+1=0$ "
] | [
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Evaluate $ \int _ { 0 } ^ { a } d x / ( 1 + \cos x + \sin x ) $ for $ - \pi / 2 < a < \pi $ . Use your answer to show that $ \int _ { 0 } ^ { \pi / 2 } d x / ( 1 + \cos x + \sin x ) = \ln 2 $ . | $-\dfrac{\pi}{4}<\dfrac{a}{2}<\dfrac{\pi}{2}\Longrightarrow -\dfrac{\pi}{4}<x\le 0$ or $0\le x<\dfrac{\pi}{2}$ .
Using the substitution $\tan \dfrac{x}{2}=t$ results: $dx=\dfrac{2dt}{1+t^2}, \cos x=\dfrac{1-t^2}{1+t^2}, \sin x=\dfrac{2t}{1+t^2}$ , $ \int _ { 0 } ^ { a } \dfrac{d x}{ 1 + \cos x + \sin x}=\int_0^{\t... | [
" $ \\int _ { 0 } ^ { \\pi / 2 } d x / ( 1 + \\cos x + \\sin x ) = \\ln 2 $ To integrate this just write $(1+cosx)=2cos^{2}\\frac{x}{2}$ and\n $sinx=2sin\\frac{x}{2}cos\\frac{x}{2}$ Then we get $I=\\int\\frac {1}{2cos\\frac{x}{2}\\bigg(cos\\frac{x}{2}+sin\\frac{x}{2}\\bigg)}$ Now just divide the numerator and d... | [
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Let $ABC$ be a triangle and let $P$ be a point in its interior. Suppose $ \angle B A P = 10 ^ { \circ } , \angle A B P = 20 ^ { \circ } , \angle P C A = 30 ^ { \circ } $ and $ \angle P A C = 40 ^ { \circ } $ . Find $ \angle P B C $ . | Trigonometry does it. $\angle APB=150^{\circ}, \angle BPC=100^{\circ}$ . So, from sine law in triangle $ABP$ we have: $\frac{c}{\sin150^{\circ}}=\frac{AP}{\sin20^{\circ}}=\frac{BP}{\sin10^{\circ}}$ , so: $AP=2c\sin20^{\circ}$ and $BP=2c\sin10^{\circ}$ . From sine law in triangle $APC$ we have: $\frac{AP}{\sin30^... | [] | [
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Let $P$ be an interior point of a triangle of area $T$ . Through the point $P$ , draw lines parallel to the three sides, partitioning the triangle into three triangles and three parallelograms. Let $a$ , $b$ and $c$ be the areas of the three triangles. Prove that $ \sqrt { T } = \sqrt { a } + \sqrt { b } + \s... | Let the triangle be $ABC$ , and suppose the pints cut off from side $BC$ are $A_1, A_2$ and so on. Join $A$ to $P$ and extend it to meet $A_1A_2$ at $K$ ; then the triangles $PA_1A_2$ and $ABC$ being similar at ehomothetic w.r.t. point $K$ whence $\frac{a}{T} := \frac{[PA_1A_2]}{[ABC]} = (\frac{PL}... | [
"Let $AC \\parallel DE, BC \\parallel FG, AB \\parallel HI$ . Now, triangles $DFP, PIE, HPG, ABC$ are similar, so: $\\frac{DF}{PI}=\\frac{\\sqrt{a}}{\\sqrt{b}}$ and $\\frac{HP}{PI}=\\frac{\\sqrt{c}}{\\sqrt{b}}$ . Adding up these equations we get $\\frac{DF+HP}{PI}=\\frac{\\sqrt{a}+\\sqrt{c}}{\\sqrt{b}}$ Let's... | [
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Let $ f ( x , y ) = ( x + y ) / 2 , g ( x , y ) = \sqrt { x y } , h ( x , y ) = 2 x y / ( x + y ) $ , and let $$ S = \{ ( a , b ) \in \mathrm { N } \times \mathrm { N } | a \neq b \text { and } f( a , b ) , g ( a , b ) , h ( a , b ) \in \mathrm { N } \} $$ where $\mathbb{N}$ denotes the positive integers. Find... | Given $(a,b) \in \mathbb{N}^2$ , let $d:=gcd(a,b)$ Then $g(a,b)=d\sqrt{xy} \in \mathbb{N} \implies \exists m, n \in \mathbb{N}$ s.t. $gcd(m, n)=1, a=m^2, b=n^2$ . So $h(x, y) \in \mathbb{Z} \implies (m^2+n^2)|d \iff d:= (m^2+n^2)\lambda$ for some $\lambda \in \mathbb{N}$ , whence $a= m^2(m^2+n^2)\lambda, b= n... | [] | [
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Let $ f ( x ) \in \mathbb { Z } [ x ] $ be a polynomial with integer coefficients such that $ f ( 1 ) = - 1 , f ( 4 ) = 2 $ and $f ( 8 ) = 34 $ . Suppose $n\in\mathbb{Z}$ is an integer such that $ f ( n ) = n ^ { 2 } - 4 n - 18 $ . Determine all possible values for $n$ . | A slightly different solution:
Let $ g(x) = f(x) - x^2 + 4x + 18 = (x - n_i) h(x) $ , where $ n_i $ is a solution to the polynomial.
It's obvious that $h(x) \in \mathbb{Z}[x]$ , because $ x - n_i $ is monic.
And, $ \forall k \in \mathbb{Z}, g(k) \in \mathbb{Z} $ , because $ k^2 $ , $4k$ , and $18$ are all i... | [
"From the formula $ f ( n ) = n ^ { 2 } - 4 n - 18 $ how we get $f(1)=-1$ ?",
"Let $g(x)=x^2-4x+2$ then $g(1)=f(1),g(4)=f(4),g(8)=f(8)$ $f(x)=(x-1)(x-4)(x-8)h(x)+g(x)$ $n^2-4n-18=f(n)=(n-1)(n-4)(n-8)h(n)+n^2-4n+2$ $-20=(n-1)(n-4)(n-8)h(n)$ $n-1$ is divisor of $20$ so $n=-19,-9,-4,-3,-1,0,2,3,5,6,11,... | [
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} |
Find all pairs $(m, n)$ of nonnegative integers for which $ m ^ { 2 } + 2 \cdot 3 ^ { n } = m \left( 2 ^ { n + 1 } - 1 \right) $ . | [
"https://artofproblemsolving.com/community/c6h418640p2361999\nIMO Shortlist 2010 - Problem N2 "
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It is known that $\int_1^2x^{-1}\arctan (1+x)\ dx = q\pi\ln(2)$ for some rational number $q.$ Determine $q.$ Here, $0\leq\arctan(x)<\frac{\pi}{2}$ for $0\leq x <\infty.$ | You might be also interested in his "brother": https://artofproblemsolving.com/community/u380742h1735424p11265354 $$ I=\int_1^2 \frac{\arctan(1+x)}{x}dx $$ let us substitute $x=\frac{2} {t}\Rightarrow dx=-\frac{2} {t^2} dt $ $$ I=\int_1^2 \frac{\arctan\left(1+\frac{2 }{t}\right) } {\frac{2} {t}} \frac{2} {t^2} dt=... | [] | [
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"boxed": false,
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Prove that there is no function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $f(f(n))=n+1.$ Here $\mathbb{N}$ is the positive integers $\{1,2,3,\dots\}.$ | <blockquote>Prove that there is no function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $f(f(n))=n+1.$ Here $\mathbb{N}$ is the positive integers $\{1,2,3,\dots\}.$ </blockquote> $f(f(n))=n+1$ implies $f(n+1)=f(n)+1$ and so $f(n)=n+c$ And pluging this back in orginal equation, we get $c=\frac 12$ , impos... | [] | [
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"path": "Contest Collections/Undergraduate Contests/VTRMC/2018 VTRMC/1748083.json"
} |
Let $A, B \in M_6 (\mathbb{Z} )$ such that $A \equiv I \equiv B \text{ mod }3$ and $A^3 B^3 A^3 = B^3$ . Prove that $A = I$ . Here $M_6 (\mathbb{Z} )$ indicates the $6$ by $6$ matrices with integer entries, $I$ is the identity matrix, and $X \equiv Y \text{ mod }3$ means all entries of $X-Y$ are divis... | $\bullet$ Here $A,B\in M_n(\mathbb{Z})$ and the $U_i,V_i,W_i$ too. $A=I+3U_0,B=I+3V_0$ . Moreover $A^3=I+9U_1,B^3=I+9V_1$ . Then $A^3B^3A^3-B^3=0$ implies $2U_1+(9U_1^2+9U_1V_1+9V_1U_1+9^2U_1V_1U_1)=0$ . Thus $U_1=9U_2$ and therefore $2U_2+(9^2U_2^2+9U_2V_1+9V_1U_2+9^3U_2V_1U_2)=0$ . Thus $U_2=9U_3$ and so... | [] | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2018 VTRMC"
] | {
"answer_score": 50,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2018 VTRMC/2990632.json"
} |
Let $m, n$ be integers such that $n \geq m \geq 1$ . Prove that $\frac{\text{gcd} (m,n)}{n} \binom{n}{m}$ is an integer. Here $\text{gcd}$ denotes greatest common divisor and $\binom{n}{m} = \frac{n!}{m!(n-m)!}$ denotes the binomial coefficient. | For all positive integers $m$ and $n$ , there exists some integers $x$ and $y$ such that $mx + ny = \mathrm{gcd}(m, n)$ . You can continue from here :D | [
"Consider separately the cases m=1, m=n, and n>m>1.\n\nIf m=1 or n then the given expression equals 1.\n\nIf n>m>1 then simplify the expression by breaking apart the factorials like n! = n*(n-1)!. Then using some reasoning involving divisibility and gcd, you can concluded the expression is an integer.",
"Isn't th... | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2018 VTRMC"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2018 VTRMC/2990635.json"
} |
For $n \in \mathbb{N}$ , let $a_n = \int _0 ^{1/\sqrt{n}} | 1 + e^{it} + e^{2it} + \dots + e^{nit} | \ dt$ . Determine whether the sequence $(a_n) = a_1, a_2, \dots$ is bounded. | <blockquote>For $n \in \mathbb{N}$ , let $a_n = \int _0 ^{1/\sqrt{n}} | 1 + e^{it} + e^{2it} + \dots + e^{nit} | \ dt$ . Determine whether the sequence $(a_n) = a_1, a_2, \dots$ is bounded.</blockquote>
Note that $\sin (x) \geqslant \frac{2}{\pi}x, \forall x \in [0, \frac{\pi}{2}]$ . Therefore, $a_n = \int _0 ^{... | [] | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2018 VTRMC"
] | {
"answer_score": 1022,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2018 VTRMC/2990636.json"
} |
For $n \in \mathbb{N}$ , define $a_n = \frac{1 + 1/3 + 1/5 + \dots + 1/(2n-1)}{n+1}$ and $b_n = \frac{1/2 + 1/4 + 1/6 + \dots + 1/(2n)}{n}$ . Find the maximum and minimum of $a_n - b_n$ for $1 \leq n \leq 999$ . | [
"Quick solution: See the below.\n<details><summary>Click to expand</summary>[https://personal.math.vt.edu/plinnell/Vtregional/S18/index.html](https://personal.math.vt.edu/plinnell/Vtregional/S18/index.html)\nIt was shocking that I didn’t have to find generalized term.</details>"
] | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2018 VTRMC"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2018 VTRMC/2990640.json"
} | |
A continuous function $f : [a,b] \to [a,b]$ is called piecewise monotone if $[a, b]$ can be subdivided into finitely many subintervals $$ I_1 = [c_0,c_1], I_2 = [c_1,c_2], \dots , I_\ell = [ c_{\ell - 1},c_\ell ] $$ such that $f$ restricted to each interval $I_j$ is strictly monotone, either increasing or dec... | **Lemma:** For all piecewise monotone $f, g \colon [a,b] \to [a,b]$ we have $\ell ( f \circ g ) \le \ell ( f ) \ell ( g )$ .**Proof:** Let $I_1 = [c_0,c_1], I_2 = [c_1,c_2], \dots , I_\ell = [ c_{\ell - 1},c_\ell ]$ with $\ell = \ell ( g )$ be such that on each interval $I_i$ we have that $g \colon I_i \to [a... | [] | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2018 VTRMC"
] | {
"answer_score": 36,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2018 VTRMC/2990642.json"
} |
Give all possible representations of $2022$ as a sum of at least two consecutive positive integers and prove that these are the only representations. | <blockquote>Give all possible representations of $2022$ as a sum of at least two consecutive positive integers and prove that these are the only representations.</blockquote>
If $2022$ is the sum of the integers between $m$ and $n$ (inclusive), we have $$ 2022=\sum \limits _{k=m}^nk=\frac{(n+m)(n+1-m)}{2}, $$... | [] | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2022 VTRMC"
] | {
"answer_score": 36,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2022 VTRMC/2988732.json"
} |
Let $A$ and $B$ be the two foci of an ellipse and let $P$ be a point on this ellipse. Prove that the focal radii of $P$ (that is, the segments $\overline{AP}$ and $\overline{BP}$ ) form equal angles with the tangent to the ellipse at $P$ . | *For the sake of completing the solution:*
Wlog work with the standard ellipse i.e $\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$ and assume that $a \geq b$ .
The foci are given by $S=(ae,0)$ and $S'=(-ae,0)$ where $e$ (eccentricity) satisfies $b^2=a^2\left(1-e^2\right)$ .
Parametrize the points on the ellipse as $P=(a... | [
"<blockquote>Let $A$ and $B$ be the two foci of an ellipse and let $P$ be a point on this ellipse. Prove that the focal radii of $P$ (that is, the segments $\\overline{AP}$ and $\\overline{BP}$ ) form equal angles with the tangent to the ellipse at $P$ .</blockquote>\n\nMore general [equality](https://... | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2022 VTRMC"
] | {
"answer_score": 26,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2022 VTRMC/2988733.json"
} |
Find all positive integers $a, b, c, d,$ and $n$ satisfying $n^a + n^b + n^c = n^d$ and prove that these are the only such solutions. | <blockquote>Find all positive integers $a, b, c, d,$ and $n$ satisfying $n^a + n^b + n^c = n^d$ and prove that these are the only such solutions.</blockquote>
Clearly $n>1$ . WLOG let $a\leq b\leq c$ and write $b=a+x$ and $c=b+y=a+x+y$ , with $x,y\in \mathbb{N}_0$ . Then $n^d=n^a(1+n^x+n^{x+y})$ . If $x... | [
"Assume wlog that $a \\le b \\le c$ and write $b-a=q$ , $c-a=r$ , $d-a=s$ . Then $1 + n^q + n^r = n^s$ .\nSince $s > r$ we have $n^s \\le 3 n^r \\le 3 n ^{s-1}$ and so $n \\le 3$ . Also clearly $n>1$ .\n\nIf $n=3$ then we must have $q=r=0$ , since otherwise $1+3^q+3^r$ is not divisible by $3$ . So... | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2022 VTRMC"
] | {
"answer_score": 114,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2022 VTRMC/2988735.json"
} |
Calculate the exact value of the series $\sum _{n=2} ^\infty \log (n^3 +1) - \log (n^3 - 1)$ and provide justification. | <blockquote><blockquote>Calculate the exact value of the series $\sum _{n=2} ^\infty \log (n^3 +1) - \log (n^3 - 1)$ and provide justification.</blockquote>
We have\begin{align*}\sum \limits _{n=2}^\infty \log (n^3+1)-\log (n^3-1) & =\lim \limits _{m\to \infty}\sum \limits _{n=2}^m \log (n^3+1)-\log (n^3-1)
& =\lim... | [
"<blockquote>Calculate the exact value of the series $\\sum _{n=2} ^\\infty \\log (n^3 +1) - \\log (n^3 - 1)$ and provide justification.</blockquote>\n\nWe have\\begin{align*}\\sum \\limits _{n=2}^\\infty \\log (n^3+1)-\\log (n^3-1) & =\\lim \\limits _{m\\to \\infty}\\sum \\limits _{n=2}^m \\log (n^3+1)-\\log (n^... | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2022 VTRMC"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2022 VTRMC/2988736.json"
} |
Let $A$ be an invertible $n \times n$ matrix with complex entries. Suppose that for each positive integer $m$ , there exists a positive integer $k_m$ and an $n \times n$ invertible matrix $B_m$ such that $A^{k_m m} = B_m A B_m ^{-1}$ . Show that all eigenvalues of $A$ are equal to $1$ . | <blockquote>Let $A$ be an invertible $n \times n$ matrix with complex entries. Suppose that for each positive integer $m$ , there exists a positive integer $k_m$ and an $n \times n$ invertible matrix $B_m$ such that $A^{k_m m} = B_m A B_m ^{-1}$ . Show that all eigenvalues of $A$ are equal to $1$ .</bloc... | [
"Then $A=I+N$ , where $N$ is nilpotent. Yet, we cannot do better\nIndeed one has many $>0$ integers $k$ s.t. $A^k$ is similar to $A$ . But $(I+N)^k=I+kN+\\cdots$ is similar to $I+kN$ which is similar to $A$ .",
"Two things.\n<blockquote>\nLet $\\lambda$ be an arbitrary eigenvalue of $A$ and let... | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2022 VTRMC"
] | {
"answer_score": 98,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2022 VTRMC/2988738.json"
} |
Let $f : \mathbb{R} \to \mathbb{R}$ be a function whose second derivative is continuous. Suppose that $f$ and $f''$ are bounded. Show that $f'$ is also bounded. | Suppose $|f(x)|,|f''(x)|\le M$ for all $x$ . Let $k$ be a positive number and suppose that $f'(x)\ge k$ for some $x$ .
For $a\ge x$ we have $ |f'(a)-f'(x)|=|\int_x^a f''(t) dt|\le M(a-x)$ , hence $f'(a)\ge f'(x)-M(a-x)\ge k-M(a-x)$ .
Hence $f'(a)\ge k/2$ on the interval $[x,x+{k\over {2M}}]$ .
Hence $2M... | [
"By Taylor up to second order: $$ f(t+1)=f(t)+f'(t)+f''(t+u)/2 $$ some $u \\in ]0,1[$ Therefore $$ f'(t)=f(t+1)-f(t)-f''(t+u)/2 $$ Now if $|f(x)|\\leq a$ and $|f''(x)|\\leq b$ , $$ |f'(t)|=|f(t+1)-f(t)-f''(t+u)/2|\\leq 2a+b/2 $$ and hence is bounded\n\nRegards\nClaudio.",
"<blockquote>\n\nNow if $|f... | [
"origin:aops",
"Undergraduate Contests",
"VTRMC",
"2022 VTRMC"
] | {
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/Undergraduate Contests/VTRMC/2022 VTRMC/2988739.json"
} |
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