problem_id stringlengths 32 32 | link stringlengths 75 84 | problem stringlengths 14 5.33k | solution stringlengths 15 6.63k | letter stringclasses 5
values | answer stringclasses 957
values |
|---|---|---|---|---|---|
3f1557bfb02dd983288cd405dfb9cce7 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13 | For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+... | Consider a given subset $T$ of $S$ that contains $7$ ; then there is a subset $T'$ which contains all the elements of $T$ except for $7$ , and only those elements . Since each element of $T'$ has one fewer element preceding it than it does in $T$ , their signs are opposite. Thus the sum of the alternating sums of $T$ a... | null | 448 |
3f1557bfb02dd983288cd405dfb9cce7 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13 | For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+... | Denote the desired total of all alternating sums of an $n$ -element set as $S_n$ . We are looking for $S_7$ . Notice that all alternating sums of an $n$ -element set are also alternating sums of an $n+1$ -element set. However, when we go from an $n$ to $n+1$ element set, for each subset with the new element, we are add... | null | 448 |
3f1557bfb02dd983288cd405dfb9cce7 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13 | For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+... | We analyze all the numbers from 1 to 7 separately to see where the number contributes its positive or negative to the sum of the alternating sums. Whenever 7 appears, which it does 64 times, it contributes a positive because it is always first. This gives a net gain of $7 \cdot 64=448$
If we look at when 6 appears, whi... | null | 448 |
3f1557bfb02dd983288cd405dfb9cce7 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_13 | For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+... | Let $\mathbb{N}_n := \{1, 2, 3, \dots n\}$ . Let the alternating sum of a certain subset of $S$ of $\mathbb{N}_n$ be $\xi(S),$ and let \[\mathcal{A}(\mathbb{N}_n) := \sum_{S \subseteq \mathbb{N}_n} \xi(S).\] We see that \[\mathcal{A}(\mathbb{N}_n) = \sum_{S \subseteq \mathbb{N}_n} \xi(S) = \sum_{n \in S, S \subseteq \m... | null | 448 |
f92f27a02f4888f14d9e66e599de31e8 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | Firstly, notice that if we reflect $R$ over $P$ , we get $Q$ . Since we know that $R$ is on circle $B$ and $Q$ is on circle $A$ , we can reflect circle $B$ over $P$ to get another circle (centered at a new point $C$ , and with radius $6$ ) that intersects circle $A$ at $Q$ . The rest is just finding lengths, as follows... | null | 130 |
f92f27a02f4888f14d9e66e599de31e8 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | [asy] size(0,5cm); pair a=(8,0),b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--r--n--b--a--m--n); draw(a--q--m); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); draw(rightanglemark(b,n,r,24)); label("$A$",a,S);... | null | 130 |
f92f27a02f4888f14d9e66e599de31e8 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | Let $QP=PR=x$ . Angles $QPA$ $APB$ , and $BPR$ must add up to $180^{\circ}$ . By the Law of Cosines $\angle APB=\cos^{-1}\left(\frac{{-11}}{24}\right)$ . Also, angles $QPA$ and $BPR$ equal $\cos^{-1}\left(\frac{x}{16}\right)$ and $\cos^{-1}\left(\frac{x}{12}\right)$ . So we have
Taking the cosine of both sides, and sim... | null | 130 |
f92f27a02f4888f14d9e66e599de31e8 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | Let $QP = PR = x$ . Extend the line containing the centers of the two circles to meet $R$ , and to meet the other side of the large circle at a point $S$
The part of this line from $R$ to the point nearest to $R$ where it intersects the larger circle has length $6+(12-8)=10$ .
The length of the diameter of the large... | null | 130 |
f92f27a02f4888f14d9e66e599de31e8 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | [asy] size(0,5cm); pair a=(8,0),b=(20,0),t=(14,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0); draw(b--a--m--n--cycle); draw(p--t); draw(q--m); draw(n--r); draw(circumcircle(origin,q,p)); draw(circumcircle((14,0),p,r)); draw(rightanglemark(a,m,n,24)); label("$A$",a,S); lab... | null | 130 |
f92f27a02f4888f14d9e66e599de31e8 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | Looking at Drawing 2 (by the way, we don't need point $R$ ), we set $AM=a$ and $BN=b$ , and the desired length $QP=x=PR$ . We know that a radius perpendicular to a chord bisects the chord, so $MP=\frac{x}{2}$ and $PN=\frac{x}{2}$ . Draw line $AP$ and $PB$ , and we see that they are radii of Circles $A$ and $B$ , respec... | null | 130 |
f92f27a02f4888f14d9e66e599de31e8 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | The centers are collinear, you can prove it (but that is already given in the later section [Proof that R,A, and B are collinear]). Drop a perpendicular from $P$ to the lines that the centers are on. You then have 2 separate segments, separated by the foot of the altitude of $P$ . Call them $a$ and $b$ respectively. Ca... | null | 130 |
f92f27a02f4888f14d9e66e599de31e8 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | We use coordinate geometry to approach this problem. Let the center of larger circle be the origin $O_1$ , the smaller circle be $O_2$ , and the x-axis be $O_1O_2$ . Hence, we can get the two circle equations: $x^2+y^2 = 64$ and $(x-12)^2+y^2=36$
Let point $P$ be $(a, b)$ . Noting that it lies on both circles, we can p... | null | 130 |
f92f27a02f4888f14d9e66e599de31e8 | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_14 | In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$ , one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | Let the center of the circle with radius $8$ be $A,$ and let the center of the one with radius $6$ be $B.$ Also, let $QP = PR = x.$ Using law of cosines on triangle $APB,$ we have that $\cos {APB} = \left(-\frac{{11}}{24}\right).$ Angle chasing gives that $\angle{QAR} = \angle{APB},$ so its cosines must be the same. Ap... | null | 130 |
f3de69d8561599d36cbe34c94c55a6ad | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central an... | Let $M$ be the midpoint of the chord $BC$ . From right triangle $OMB$ , we have $OM = \sqrt{OB^2 - BM^2} =4$ . This gives $\tan \angle BOM = \frac{BM}{OM} = \frac 3 4$
Notice that the distance $OM$ equals $PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)$ , where $r$ is the radius of circle $P$
Hence \[\cos \angle AOM ... | null | 175 |
f3de69d8561599d36cbe34c94c55a6ad | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central an... | This solution, while similar to Solution 2, is arguably more motivated and less contrived.
Firstly, we note the statement in the problem that " $AD$ is the only chord starting at $A$ and bisected by $BC$ " – what is its significance? What is the criterion for this statement to be true?
We consider the locus of midpoint... | null | 175 |
f3de69d8561599d36cbe34c94c55a6ad | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central an... | Let the center of the circle be $O$ . Fix $B,C,$ and $A$ . Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$ , it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$
Now, ... | null | 175 |
f3de69d8561599d36cbe34c94c55a6ad | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central an... | Let I be the intersection of AD and BC.
Lemma: $AI = ID$ if and only if $\angle AIO = 90$
Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If $\angle AIO = 90$ , We can get $\triangle AIO \cong \triangle OID$
Let be this the circle with diameter AO.
Thus, we get $\angle AIO = 90$ , imp... | null | 175 |
f3de69d8561599d36cbe34c94c55a6ad | https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_15 | The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$ . Suppose that the radius of the circle is $5$ , that $BC=6$ , and that $AD$ is bisected by $BC$ . Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$ . It follows that the sine of the central an... | Let $O$ be the center of the circle. The locus of midpoints of chords with $A$ as a endpoint is a circle with diameter $\overline{AO}$ . Additionally, this circle must be tangent to $\overline{BC}$ . Let the center of this circle be $P$ . Let $M$ be the midpoint of $BC$ $N$ be the foot of the perpendicular from $P$ to ... | null | 175 |
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