2011 AIME II Problems/Problem 10

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

2011 AIME II (Problems • Answer Key • Resources)
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

Page

Toolbox

Search