AIME 2011 II · 第 10 题

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 $\triangle EPF$:

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

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

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 $\angle PEO+\angle PFO=180^{\circ}$, Quadrilateral $EPFO$ is inscribed in a Circle. Assume point $I$ is the center of this circle.

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

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

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

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,

$$\begin{aligned} \sin{\angle MNO} &= \frac{2[MNO]}{ON \cdot MN} \\ &= \frac{2}{9}\sqrt{14} \\ \sin{\angle MNO} &= \frac{OM}{OP} \\ &= \frac{2}{9}\sqrt{14} \\ OP &= \frac{90\sqrt{14}}{14} \\ OP^2 &= \frac{4050}{7} \implies \boxed{057}. \end{aligned}$$ ~ 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 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

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 $\overline{CD}$,

$P_y = E_y -\frac{1}{\tan \angle EOF} (OF - E_x) = - \frac{3\sqrt{14}}{7}$ (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,

Solution 8 (Analytic- Can use complex numbers or rotation matrix)

Solution 9

Let the midpoint of $AB$ be $M$, and the midpoint of $CD$ be $N$. Because $ON$ is a perpendicular bisector of $CD$, we know that $CN=7$ and $OC=25$ (the radius of the circle). Therefore, by the Pythagorean Theorem, $ON=24$.

Now that we have all the side lengths of $\triangle{OMN}$, we use Law of Cosines to find $\angle{OMN}$. \begin{align*} 20^2+12^2-(2)(20)(12)(\cos\angle{OMN})&=24^2\\ 544-480\cos\angle{OMN}&=576\\ \cos\angle{OMN}&=-\dfrac{1}{15}. \end{align*} Because $\angle{OMP}$ is right, we know that $\angle{PMN}=\angle{OMN}-90^\circ$.

We then use the cosine difference identity to find that (note that we use $\cos 90^\circ =0$ and $\sin 90^\circ = 1$) \begin{align*} \cos\angle{PMN}&=\cos(\angle{OMN}-90^\circ)\\ &=\cos\angle{OMN}\cos 90^\circ + \sin\angle{OMN}\sin 90^\circ\\ &=\sin{OMN}. \end{align*} Using the identity $\cos^2 \theta+\sin^2 \theta=1$, and since it is an acute angle it must be positive, we find that $\cos\angle{PMN}=\dfrac{4\sqrt{14}}{15}$.

Let $PM=x$, and $PN=y$. Using Law of Sines on $\triangle{PMN}$, we have the equation

$$x^2+144-(2)(x)(12)(\cos\angle{PMN})=y^2.$$ Simplifying, we get

$$x^2-y^2+144=\dfrac{96x\sqrt{14}}{15}.$$ Using Power of a Point on point $P$, we find that $(15-x)(15+x)=(7-y)(7+y)$, or $x^2-y^2=176$. Substituting this in and simplifying, we find that $x=\dfrac{50}{\sqrt{14}}$. Finally, using the Pythagorean Theorem on $\triangle{OMP}$, we find that

$$OP^2=400+x^2=400+\dfrac{2500}{14}=\dfrac{2800+1250}{7}=\dfrac{4050}{7}.$$ Thus, the answer to the problem is $4050+7=4057\rightarrow \boxed{057}.$

- The_Other_Guy

Solution 10

Place the circle of radius $25$ at the origin, $O=(0,0)$. Let chord $\overline{AB}$ be vertical with midpoint $M=(-20,0)$, so that $AB$ has length $30$. Then the endpoints are $A=(-20,-15), \quad B=(-20,15)$.

Let chord $\overline{CD}$ have length $14$, so its half-length is $7$. The distance from the origin to the midpoint of $\overline{CD}$ is $\sqrt{25^2 - 7^2} = \sqrt{625 - 49} = 24.$

Let the midpoint of $\overline{CD}$ be $N=(x,y)$. Since the distance between the midpoints of the chords is $12$, we have $(x+20)^2 + y^2 = 12^2 = 144$, and from the distance to the origin: $x^2 + y^2 = 24^2 = 576$.

Subtracting these equations eliminates $y^2$ and gives a simple linear equation:

Then, $y^2 = 144 - (x+20)^2 = 144 - \left(-\frac{4}{5}\right)^2 = \frac{3600}{25} - \frac{16}{25} = \frac{3584}{25} \Rightarrow y = \frac{16\sqrt{14}}{5}.$

Thus the midpoint of $\overline{CD}$ is $N = \left(-\frac{104}{5}, \frac{16\sqrt{14}}{5}\right)$.

The slope of $\overline{CD}$ is perpendicular to the radius at its midpoint: $\text{slope} = \frac{-(-104/5)}{16\sqrt{14}/5} = \frac{104/5}{16\sqrt{14}/5} = \frac{13}{2\sqrt{14}}$.

Equation of the line through $N$:$y - \frac{16\sqrt{14}}{5} = \frac{13}{2\sqrt{14}} \left(x + \frac{104}{5}\right)$.

Intersecting with $x=-20$ (the vertical line for AB): $y - \frac{16\sqrt{14}}{5} = \frac{13}{2\sqrt{14}} \cdot \frac{4}{5} \Rightarrow y = \frac{16\sqrt{14}}{5} + \frac{26}{5\sqrt{14}} = \frac{50}{\sqrt{14}}$.

So the intersection point $P$ is $P = \left(-20, \frac{50}{\sqrt{14}}\right)$, and $OP^2 = (-20)^2 + \left(\frac{50}{\sqrt{14}}\right)^2 = 400 + \frac{2500}{14} = \frac{4050}{7}$.

Thus $m+n = 4050 + 7 = 4057$, and the remainder when divided by $1000$ is $\boxed{57}.$

~MathKing555