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:

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

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 $\boxed{57}.$


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 contains no quadratic equations (unless you count LoC).

Solution 7 Analytic Geometry

2011 AIMEII Problem 10 CASE 2.png

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 $OB = OC = 25$.

Since $\overline{OE}\perp \overline{AB}$ and $\overline{OF}\perp \overline{CD}$, $OE = \sqrt{OB^2-BE^2}=20$ and $OF = \sqrt{OC^2-OF^2}=24$

With law of cosines, $\cos \angle EOF = \frac{OE^2+OF^2-EF^2}{2\cdot OE\cdot OF} = \frac{13}{15}$

Since $EF < OF$, $\angle EOF$ is acute angle. $\sin \angle EOF = \sqrt{1-\cos^2 \angle EOF} = \frac{\sqrt{56}}{15}$ and $\tan \angle EOF = \frac{\sqrt{56}}{13}$

Let $\overline{OF}$ line be $x$ axis.

Line $\overline{DC}$ equation is $x = OF$.

Since line $\overline{AB}$ passes point $E$ and perpendicular to $\overline{OD}$, its equation is $y - E_y = -\frac{1}{\tan \angle EOF} (x - E_x)$

where $E_x = OE\cos{\angle EOF}$ , $E_y = OE\sin{\angle EOF}$

Since $P$ is the intersection of $\overline{AB}$ and $\overline{CD}$,

$P_x = OF = 24$

$P_y = E_y -\frac{1}{\tan \angle EOF} (OF - E_x) = - \frac{3\sqrt{14}}{7}$ (Negative means point $P$ is between point $F$ and $C$)

$OP^2 = P_x^2 + P_y^2 = \frac{4050}{7}$ and the answer is $\boxed{057}$.

Note: if $EF$ was longer, point $P$ would be between point $D$ and $F$. Then, $OP$ would be the diagonal of quadrilateral $OEPF$ not the side. To apply Ptolemy's theorem like solution 4, it is critical to know whether $OP$ is the diagonal or side of quadrilateral. Equation for wrong case cannot be solved. For example,

2011 AIMEII Problem 10 CASE 1.png

See also

2011 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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