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Difference between revisions of "2011 AIME II Problems/Problem 10"

m (→‎Solution 2 - Fastest)
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~ Aaryabhatta1
 
~ Aaryabhatta1
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==Solution 6==
 +
Define <math>M</math> and <math>N</math> as the midpoints of <math>AB</math> and <math>CD</math>, respectively. Because <math>\angle OMP = \angle ONP = 90^{\circ}</math>, we have that <math>ONPM</math> is a cyclic quadrilateral. Hence, <math>\angle PNM = \angle POM.</math> Then, let these two angles be denoted as <math>\alpha</math>.
 +
Now, assume WLOG that <math>PD = x < 7</math> and <math>PB = y < 15</math> (We can do this because one of <math>PD</math> or <math>PC</math> must be less than 7, and similarly for <math>PB</math> and <math>PA</math>). Then, by Power of a Point on P with respect to the circle with center <math>O</math>, we have that
 +
<cmath>(14-x)x = (30-y)y</cmath>
 +
<cmath>(7-x)^{2}+176=(15-y)^{2}.</cmath>
 +
Then, let <math>z = (7-x)^{2}</math>. From Law of Cosines on <math>\triangle NMP</math>, we have that
 +
<cmath>\textrm{cos } \angle MNP = \frac{NP^{2}+MN^{2}-MP^{2}}{2 \cdot NP \cdot MN} </cmath>
 +
<cmath>\textrm{cos } \alpha  = \frac{(7-x)^{2} + 12^{2} - (14-x)^{2}}{24 \cdot (7-x)}.</cmath>
 +
Plugging in <math>z</math> in gives
 +
<cmath>\textrm{cos } \alpha  = \frac{-32}{24 \cdot \sqrt{z}}</cmath>
 +
<cmath>\textrm{cos } \alpha  = \frac{-4}{3\sqrt{z}}</cmath>
 +
<cmath>\textrm{cos }^{2} \alpha  = \frac{16}{9z}.</cmath>
 +
Hence,
 +
<cmath>\textrm{tan }^{2} \alpha  = \frac{\frac{9z-16}{9z}}{\frac{16}{9z}} = \frac{9z-16}{16}.</cmath>
 +
Then, we also know that
 +
<cmath>\textrm{tan } \alpha  = \textrm{tan } \angle MOP  = \frac{MP}{OM} = \frac{14-y}{20}.</cmath>
 +
Squaring this, we get
 +
<cmath>\textrm{tan }^{2} \alpha  = \frac{z+176}{400}.</cmath>
 +
Equating our expressions for <math>z</math>, we get
 +
<math>\frac{z+176}{400} = \frac{9z-16}{16}.</math>
 +
Solving gives us that
 +
<math>z = \frac{18}{7}</math>.
 +
Since <math>\angle ONP = 90^{\circ}</math>, from the Pythagorean Theorem,
 +
<math>OP^{2} = ON^{2}+PN^{2} = 25^{2}-7^{2} + z = 576+z = \frac{4050}{7}</math>,
 +
and thus the answer is <math>4050+7 = 4057</math>, which when divided by a thousand leaves a remainder of <math>\boxed{57}.</math>
 +
 +
-Mr.Sharkman
 +
 +
 +
Note: my solution was very long and tedious. It was definitely was the least elegant solution. The only thing I like about it is it's the only solution without a quadratic equation.
 
==See also==
 
==See also==
 
{{AIME box|year=2011|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2011|n=II|num-b=9|num-a=11}}

Revision as of 06:13, 11 August 2023

Problem 10

A circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$. The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000.

Solution 1

Let $E$ and $F$ be the midpoints of $\overline{AB}$ and $\overline{CD}$, respectively, such that $\overline{BE}$ intersects $\overline{CF}$.

Since $E$ and $F$ are midpoints, $BE = 15$ and $CF = 7$.

$B$ and $C$ are located on the circumference of the circle, so $OB = OC = 25$.

The line through the midpoint of a chord of a circle and the center of that circle is perpendicular to that chord, so $\triangle OEB$ and $\triangle OFC$ are right triangles (with $\angle OEB$ and $\angle OFC$ being the right angles). By the Pythagorean Theorem, $OE = \sqrt{25^2 - 15^2} = 20$, and $OF = \sqrt{25^2 - 7^2} = 24$.

Let $x$, $a$, and $b$ be lengths $OP$, $EP$, and $FP$, respectively. OEP and OFP are also right triangles, so $x^2 = a^2 + 20^2 \to a^2 = x^2 - 400$, and $x^2 = b^2 + 24^2 \to b^2 = x^2 - 576$

We are given that $EF$ has length 12, so, using the Law of Cosines with $\triangle EPF$:

$12^2 = a^2 + b^2 - 2ab \cos (\angle EPF) = a^2 + b^2 - 2ab \cos (\angle EPO + \angle FPO)$

Substituting for $a$ and $b$, and applying the Cosine of Sum formula:

$144 = (x^2 - 400) + (x^2 - 576) - 2 \sqrt{x^2 - 400} \sqrt{x^2 - 576} \left( \cos \angle EPO \cos \angle FPO - \sin \angle EPO \sin \angle FPO \right)$

$\angle EPO$ and $\angle FPO$ are acute angles in right triangles, so substitute opposite/hypotenuse for sines and adjacent/hypotenuse for cosines:

$144 = 2x^2 - 976 - 2 \sqrt{(x^2 - 400)(x^2 - 576)} \left(\frac{\sqrt{x^2 - 400}}{x} \frac{\sqrt{x^2 - 576}}{x} - \frac{20}{x} \frac{24}{x} \right)$

Combine terms and multiply both sides by $x^2$: $144 x^2 = 2 x^4 - 976 x^2 - 2 (x^2 - 400) (x^2 - 576) + 960 \sqrt{(x^2 - 400)(x^2 - 576)}$

Combine terms again, and divide both sides by 64: $13 x^2 = 7200 - 15 \sqrt{x^4 - 976 x^2 + 230400}$

Square both sides: $169 x^4 - 187000 x^2 + 51,840,000 = 225 x^4 - 219600 x^2 + 51,840,000$

This reduces to $x^2 = \frac{4050}{7} = (OP)^2$; $4050 + 7 \equiv \boxed{057} \pmod{1000}$.

Solution 2 - Fastest

We begin as in the first solution. Once we see that $\triangle EOF$ has side lengths $12$, $20$, and $24$, we can compute its area with Heron's formula:

\[K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}.\]

Thus, the circumradius of triangle $\triangle EOF$ is $R = \frac{abc}{4K} = \frac{45}{\sqrt{14}}$. Looking at $EPFO$, we see that $\angle OEP = \angle OFP = 90^\circ$, which makes it a cyclic quadrilateral. This means $\triangle EOF$'s circumcircle and $EPFO$'s inscribed circle are the same.

Since $EPFO$ is cyclic with diameter $OP$, we have $OP = 2R = \frac{90}{\sqrt{14}}$, so $OP^2 = \frac{4050}{7}$ and the answer is $\boxed{057}$.

Solution 3

We begin as the first solution have $OE=20$ and $OF=24$. Because $\angle PEF+\angle PFO=180^{\circ}$, Quadrilateral $EPFO$ is inscribed in a Circle. Assume point $I$ is the center of this circle.

$\because \angle OEP=90^{\circ}$

$\therefore$ point $I$ is on $OP$

Link $EI$ and $FI$, Made line $IK\bot EF$, then $\angle EIK=\angle EOF$

On the other hand, $\cos\angle EOF=\frac{EO^2+OF^2-EF^2}{2\cdot EO\cdot OF}=\frac{13}{15}=\cos\angle EIK$

$\sin\angle EOF=\sin\angle EIK=\sqrt{1-\frac{13^2}{15^2}}=\frac{2\sqrt{14}}{15}$

As a result, $IE=IO=\frac{45}{\sqrt 14}$

Therefore, $OP^2=4\cdot \frac{45^2}{14}=\frac{4050}{7}.$

As a result, $m+n=4057\equiv \boxed{057}\pmod{1000}$

Solution 4

Let $OP=x$.

Proceed as the first solution in finding that quadrilateral $EPFO$ has side lengths $OE=20$, $OF=24$, $EP=\sqrt{x^2-20^2}$, and $PF=\sqrt{x^2-24^2}$, and diagonals $OP=x$ and $EF=12$.

We note that quadrilateral $EPFO$ is cyclic and use Ptolemy's theorem to solve for $x$:

\[20\cdot \sqrt{x^2-24^2} + 12\cdot x = 24\cdot \sqrt{x^2-20^2}\]

Solving, we have $x^2=\frac{4050}{7}$ so the answer is $\boxed{057}$.

-Solution by blueberrieejam

~bluesoul changes the equation to a right equation, the previous equation isn't solvable

Solution 5 (Quick Angle Solution)

Let $M$ be the midpoint of $AB$ and $N$ of $CD$. As $\angle OMP = \angle ONP$, quadrilateral $OMPN$ is cyclic with diameter $OP$. By Cyclic quadrilaterals note that $\angle MPO = \angle MNO$.

The area of $\triangle MNP$ can be computed by Herons as \[[MNO] = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}.\] The area is also $\frac{1}{2}ON \cdot MN \sin{\angle MNO}$. Therefore, \begin{align*} \sin{\angle MNO} &= \frac{2[MNO]}{ON \cdot MN} \\ &= \frac{2}{9}\sqrt{14} \\ \sin{\angle MNO} &= \frac{OM}{OP} \\ &= \frac{2}{9}\sqrt{14} \\ OP &= \frac{90\sqrt{14}}{14} \\ OP^2 &= \frac{4050}{7} \implies \boxed{057}. \end{align*}

~ Aaryabhatta1

Solution 6

Define $M$ and $N$ as the midpoints of $AB$ and $CD$, respectively. Because $\angle OMP = \angle ONP = 90^{\circ}$, we have that $ONPM$ is a cyclic quadrilateral. Hence, $\angle PNM = \angle POM.$ Then, let these two angles be denoted as $\alpha$. Now, assume WLOG that $PD = x < 7$ and $PB = y < 15$ (We can do this because one of $PD$ or $PC$ must be less than 7, and similarly for $PB$ and $PA$). Then, by Power of a Point on P with respect to the circle with center $O$, we have that \[(14-x)x = (30-y)y\] \[(7-x)^{2}+176=(15-y)^{2}.\] Then, let $z = (7-x)^{2}$. From Law of Cosines on $\triangle NMP$, we have that \[\textrm{cos } \angle MNP = \frac{NP^{2}+MN^{2}-MP^{2}}{2 \cdot NP \cdot MN}\] \[\textrm{cos } \alpha = \frac{(7-x)^{2} + 12^{2} - (14-x)^{2}}{24 \cdot (7-x)}.\] Plugging in $z$ in gives \[\textrm{cos } \alpha = \frac{-32}{24 \cdot \sqrt{z}}\] \[\textrm{cos } \alpha = \frac{-4}{3\sqrt{z}}\] \[\textrm{cos }^{2} \alpha = \frac{16}{9z}.\] Hence, \[\textrm{tan }^{2} \alpha = \frac{\frac{9z-16}{9z}}{\frac{16}{9z}} = \frac{9z-16}{16}.\] Then, we also know that \[\textrm{tan } \alpha = \textrm{tan } \angle MOP = \frac{MP}{OM} = \frac{14-y}{20}.\] Squaring this, we get \[\textrm{tan }^{2} \alpha = \frac{z+176}{400}.\] Equating our expressions for $z$, we get $\frac{z+176}{400} = \frac{9z-16}{16}.$ Solving gives us that $z = \frac{18}{7}$. Since $\angle ONP = 90^{\circ}$, from the Pythagorean Theorem, $OP^{2} = ON^{2}+PN^{2} = 25^{2}-7^{2} + z = 576+z = \frac{4050}{7}$, and thus the answer is $4050+7 = 4057$, which when divided by a thousand leaves a remainder of $\boxed{57}.$

-Mr.Sharkman


Note: my solution was very long and tedious. It was definitely was the least elegant solution. The only thing I like about it is it's the only solution without a quadratic equation.

See also

2011 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AIME Problems and Solutions

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