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

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== Problem 10 ==
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A [[circle]] with center <math>O</math> has radius 25. [[Chord]] <math>\overline{AB}</math> of length 30 and chord <math>\overline{CD}</math> of length 14 intersect at point <math>P</math>. The distance between the [[midpoint]]s of the two chords is 12. The quantity <math>OP^2</math> can be represented as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find the remainder when <math>m + n</math> is divided by 1000.
  
A circle with center O has radius 25. Chord <math>\overline{AB}</math> of length 30 and chord <math>\overline{CD}</math> of length 14 intersect at point P. The distance between the midpoints of the two chords is 12. The quantity <math>OP^{2}</math> can be expressed as <math>\frac{m}{n}</math>, where m and n are relatively prime positive integers. Find the remainder when m + n is divided by 1000.
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== Solution 1 ==
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Let <math>E</math> and <math>F</math> be the midpoints of <math>\overline{AB}</math> and <math>\overline{CD}</math>, respectively, such that <math>\overline{BE}</math> intersects <math>\overline{CF}</math>.
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Since <math>E</math> and <math>F</math> are midpoints, <math>BE = 15</math> and <math>CF = 7</math>.
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<math>B</math> and <math>C</math> are located on the circumference of the circle, so <math>OB = OC = 25</math>.
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The line through the midpoint of a chord of a circle and the center of that circle is perpendicular to that chord, so <math>\triangle OEB</math> and <math>\triangle OFC</math> are right triangles (with <math>\angle OEB</math> and <math>\angle OFC</math> being the right angles).  By the Pythagorean Theorem, <math>OE = \sqrt{25^2 - 15^2} = 20</math>, and <math>OF = \sqrt{25^2 - 7^2} = 24</math>.
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Let <math>x</math>, <math>a</math>, and <math>b</math> be lengths <math>OP</math>, <math>EP</math>, and <math>FP</math>, respectively.  OEP and OFP are also right triangles, so <math>x^2 = a^2 + 20^2 \to a^2 = x^2 - 400</math>, and <math>x^2 = b^2 + 24^2 \to b^2 = x^2 - 576</math>
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We are given that <math>EF</math> has length 12, so, using the [[Law of Cosines]] with <math>\triangle EPF</math>:
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<math>12^2 = a^2 + b^2 - 2ab \cos (\angle EPF) = a^2 + b^2 - 2ab \cos (\angle EPO + \angle FPO)</math>
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Substituting for <math>a</math> and <math>b</math>, and applying the Cosine of Sum formula:
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<math>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)</math>
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<math>\angle EPO</math> and <math>\angle FPO</math> are acute angles in right triangles, so substitute opposite/hypotenuse for sines and adjacent/hypotenuse for cosines:
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<math>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)</math>
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Combine terms and multiply both sides by <math>x^2</math>: <math>144 x^2 = 2 x^4 - 976 x^2 - 2 (x^2 - 400) (x^2 - 576) + 960  \sqrt{(x^2 - 400)(x^2 - 576)}</math>
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Combine terms again, and divide both sides by 64: <math>13 x^2 = 7200 - 15 \sqrt{x^4 - 976 x^2 + 230400}</math>
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Square both sides: <math>169 x^4 - 187000 x^2 + 51,840,000 = 225 x^4 - 219600 x^2 + 51,840,000</math>
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This reduces to <math>x^2 = \frac{4050}{7} = (OP)^2</math>; <math>4050 + 7 \equiv \boxed{057} \pmod{1000}</math>.
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==Solution 2 - Fastest==
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We begin as in the first solution.  Once we see that <math>\triangle EOF</math> has side lengths 12,20, and 24, we can compute its area with Heron's formula:
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<math>K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}</math>.
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So its circumradius is <math>R = \frac{abc}{4K} = \frac{45}{\sqrt{14}}</math>.  Since <math>EPFO</math> is [[cyclic]] with diameter <math>OP</math>, we have <math>OP = 2R = \frac{90}{\sqrt{14}}</math>, so <math>OP^2 = \frac{4050}{7}</math> and the answer is <math>\boxed{057}</math>.
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==Solution 3==
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We begin as the first solution have <math>OE=20</math> and <math>OF=24</math>. Because <math>\angle PEF+\angle PFO=180^{\circ}</math>, Quadrilateral <math>EPFO</math> is inscribed in a Circle. Assume point <math>I</math> is the center of this circle.
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<math>\because \angle OEP=90^{\circ}</math>
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<math>\therefore</math> point <math>I</math> is on <math>OP</math>
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Link <math>EI</math> and <math>FI</math>, Made line <math>IK\bot EF</math>, then <math>\angle EIK=\angle EOF</math>
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On the other hand, <math>\cos\angle EOF=\frac{EO^2+OF^2-EF^2}{2\cdot EO\cdot OF}=\frac{13}{15}=\cos\angle  EIK</math>
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<math>\sin\angle EOF=\sin\angle EIK=\sqrt{1-\frac{13^2}{15^2}}=\frac{2\sqrt{14}}{15}</math>
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As a result, <math>IE=IO=\frac{45}{\sqrt 14}</math>
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Therefore, <math>OP^2=4\cdot \frac{45^2}{14}=\frac{4050}{7}.</math>
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As a result, <math>m+n=4057\equiv \boxed{057}\pmod{1000}</math>
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==Solution 4==
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Let <math>OP=x</math>.
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Proceed as the first solution in finding that quadrilateral <math>EPFO</math> has side lengths <math>OE=20</math>, <math>OF=24</math>, <math>EP=\sqrt{x^2-20^2}</math>, and <math>PF=\sqrt{x^2-24^2}</math>, and diagonals <math>OP=x</math> and <math>EF=12</math>.
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We note that quadrilateral <math>EPFO</math> is cyclic and use Ptolemy's theorem to solve for <math>x</math>:
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<cmath>20\cdot \sqrt{x^2-24^2} + 12\cdot x = 24\cdot \sqrt{x^2-20^2}</cmath>
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Solving, we have <math>x^2=\frac{4050}{7}</math> so the answer is <math>\boxed{057}</math>.
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-Solution by blueberrieejam
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~bluesoul changes the equation to a right equation, the previous equation isn't solvable
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==See also==
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{{AIME box|year=2011|n=II|num-b=9|num-a=11}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 08:17, 21 August 2021

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:

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

So its circumradius is $R = \frac{abc}{4K} = \frac{45}{\sqrt{14}}$. 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

See also

2011 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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