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

Latest revision as of 02:12, 18 December 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 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 $\overline{CD}$ ,

$P_x = OF = 24$

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

@@ Line 33: / Line 33: @@
 This reduces to <math>x^2 = \frac{4050}{7} = (OP)^2</math>; <math>4050 + 7 \equiv \boxed{057} \pmod{1000}</math>.
-==Solution 2==
+==Solution 2 - Fastest==
-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:
+We begin as in the first solution.  Once we see that <math>\triangle EOF</math> has side lengths <math>12</math>, <math>20</math>, and <math>24</math>, we can compute its area with Heron's formula:
-<math>K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}</math>.
+<cmath>K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}.</cmath>
-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>.
+Thus, the circumradius of triangle <math>\triangle EOF</math> is <math>R = \frac{abc}{4K} = \frac{45}{\sqrt{14}}</math>.  Looking at <math>EPFO</math>, we see that <math>\angle OEP = \angle OFP = 90^\circ</math>, which makes it a cyclic quadrilateral. This means <math>\triangle EOF</math>'s circumcircle and <math>EPFO</math>'s inscribed circle are the same.
+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>.
 ==Solution 3==
@@ Line 69: / Line 71: @@
 We note that quadrilateral <math>EPFO</math> is cyclic and use Ptolemy's theorem to solve for <math>x</math>:
-<cmath>20\cdot \sqrt{x^2-24^2} + 24\cdot \sqrt{x^2-20^2} = 12\cdot x</cmath>
+<cmath>20\cdot \sqrt{x^2-24^2} + 12\cdot x = 24\cdot \sqrt{x^2-20^2}</cmath>
 Solving, we have <math>x^2=\frac{4050}{7}</math> so the answer is <math>\boxed{057}</math>.
 -Solution by blueberrieejam
+~bluesoul changes the equation to a right equation, the previous equation isn't solvable
+==Solution 5 (Quick Angle Solution)==
+Let <math>M</math> be the midpoint of <math>AB</math> and <math>N</math> of <math>CD</math>. As <math>\angle OMP = \angle ONP</math>, quadrilateral <math>OMPN</math> is cyclic with diameter <math>OP</math>. By Cyclic quadrilaterals note that <math>\angle MPO = \angle MNO</math>.
+The area of <math>\triangle MNP</math> can be computed by Herons as <cmath>[MNO] = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}.</cmath> The area is also <math>\frac{1}{2}ON \cdot MN \sin{\angle MNO}</math>. Therefore,
+<cmath>\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*}</cmath>
+~ Aaryabhatta1
+==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 contains no quadratic equations (unless you count LoC).
+==Solution 7 Analytic Geometry==
+[[Image:2011 AIMEII Problem 10 CASE 2.png|525px]]
+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>.
+Since <math>E</math> and <math>F</math> are midpoints, <math>BE = 15</math> and <math>CF = 7</math>.
+<math>B</math> and <math>C</math> are located on the circumference of the circle, so <math>OB = OC = 25</math>.
+Since <math>\overline{OE}\perp \overline{AB}</math> and <math>\overline{OF}\perp \overline{CD}</math>,
+<math>OE = \sqrt{OB^2-BE^2}=20</math> and <math>OF = \sqrt{OC^2-OF^2}=24</math>
+With law of cosines, <math>\cos \angle EOF = \frac{OE^2+OF^2-EF^2}{2\cdot OE\cdot OF} = \frac{13}{15}</math>
+Since <math>EF < OF</math>, <math>\angle EOF</math> is acute angle. <math>\sin \angle EOF = \sqrt{1-\cos^2 \angle EOF} = \frac{\sqrt{56}}{15}</math> and <math>\tan \angle EOF = \frac{\sqrt{56}}{13}</math>
+Let <math>\overline{OF}</math> line be <math>x</math> axis.
+Line <math>\overline{DC}</math> equation is <math>x = OF</math>.
+Since line <math>\overline{AB}</math> passes point <math>E</math> and perpendicular to <math>\overline{OD}</math>, its equation is <math>y - E_y = -\frac{1}{\tan \angle EOF} (x - E_x)</math>
+where <math>E_x = OE\cos{\angle EOF}</math> , <math>E_y = OE\sin{\angle EOF}</math>
+Since <math>P</math> is the intersection of <math>\overline{AB}</math> and <math>\overline{CD}</math>,
+<math>P_x = OF = 24</math>
+<math>P_y = E_y -\frac{1}{\tan \angle EOF} (OF - E_x) = - \frac{3\sqrt{14}}{7}</math> (Negative means point <math>P</math> is between point <math>F</math> and <math>C</math>)
+<math>OP^2 = P_x^2 + P_y^2 = \frac{4050}{7}</math> and the answer is <math>\boxed{057}</math>.
+Note: if <math>EF</math> was longer, point <math>P</math> would be between point <math>D</math> and <math>F</math>. Then, <math>OP</math> would be the diagonal of quadrilateral <math>OEPF</math> not the side. To apply Ptolemy's theorem like solution 4, it is critical to know whether <math>OP</math> is the diagonal or side of quadrilateral. Equation for wrong case cannot be solved. For example,
+[[Image:2011 AIMEII Problem 10 CASE 1.png|525px]]
 ==See also==

2011 AIME II (Problems • Answer Key • Resources)
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