2018 AMC 12A Problems/Problem 20

Problem

Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$ . Let $M$ be the midpoint of hypotenuse $\overline{BC}$ . Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$ , respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$ , the length $CI$ can be written as $\frac{a-\sqrt{b}}{c}$ , where $a$ , $b$ , and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is the value of $a+b+c$ ?

$\textbf{(A) }9 \qquad \textbf{(B) }10 \qquad \textbf{(C) }11 \qquad \textbf{(D) }12 \qquad \textbf{(E) }13 \qquad$

Diagram

$[asy] import olympiad; size(200); pair A, B, C, I, M, E; A = (0, 0); B = (3, 0); C = (0, 3); M = (1.5, 1.5); I = (0, 1.5 + sqrt(2) / 2); E = (1.5 - sqrt(2) / 2, 0); draw(A -- B -- C -- cycle); draw(I -- M -- E -- cycle); draw(rightanglemark(I, A, E, 4)); dot(A); dot(B); dot(C); dot(I); dot(M); dot(E); label("$A$", A, SW); label("$B$", B, E); label("$C$", C, N); label("$I$", I, NE); label("$M$", M, NE); label("$E$", E + (0.1, 0.04), NE); label("$3$", (A + C) / 2, W); label("$3$", (A + B) / 2, S); [/asy]$

Solution 1

Observe that $\triangle{EMI}$ is isosceles right ( $M$ is the midpoint of diameter arc $EI$ ), so $MI=2,MC=\frac{3}{\sqrt{2}}$ . With $\angle{MCI}=45^\circ$ , we can use Law of Cosines to determine that $CI=\frac{3\pm\sqrt{7}}{2}$ . The same calculations hold for $BE$ also, and since $CI<BE$ , we deduce that $CI$ is the smaller root, giving the answer of $\boxed{12}$ .

Solution 2 (Using Ptolemy)

We first claim that $\triangle{EMI}$ is isosceles and right.

Proof: Construct $\overline{MF}\perp\overline{AB}$ and $\overline{MG}\perp\overline{AC}$ . Since $\overline{AM}$ bisects $\angle{BAC}$ , one can deduce that $MF=MG$ . Then by AAS it is clear that $MI=ME$ and therefore $\triangle{EMI}$ is isosceles. Since quadrilateral $AIME$ is cyclic, one can deduce that $\angle{EMI}=90^\circ$ . Q.E.D.

Since the area of $\triangle{EMI}$ is 2, we can find that $MI=ME=2$ , $EI=2\sqrt{2}$

Since $M$ is the mid-point of $\overline{BC}$ , it is clear that $AM=\frac{3\sqrt{2}}{2}$ .

Now let $AE=a$ and $AI=b$ . By Ptolemy's Theorem, in cyclic quadrilateral $AIME$ , we have $2a+2b=6$ . By Pythagorean Theorem, we have $a^2+b^2=8$ . One can solve the simultaneous system and find $b=\frac{3+\sqrt{7}}{2}$ . Then by deducting the length of $\overline{AI}$ from 3 we get $CI=\frac{3-\sqrt{7}}{2}$ , giving the answer of $\boxed{12}$ . (Surefire2019)

Solution 3 (More Elementary)

Like above, notice that $\triangle{EMI}$ is isosceles and right, which means that $\dfrac{ME \cdot MI}{2} = 2$ , so $MI^2=4$ and $MI = 2$ . Then construct $\overline{MF}\perp\overline{AB}$ and $\overline{MG}\perp\overline{AC}$ as well as $\overline{MI}$ . It's clear that $MG^2+GI^2 = MI^2$ by Pythagorean, so knowing that $MG = \dfrac{AB}{2} = \dfrac{3}{2}$ allows one to solve to get $GI = \dfrac{\sqrt{7}}{2}$ . By just looking at the diagram, $CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}$ . The answer is thus $3+7+2=12$ .

Solution 4 (Coordinate Geometry)

Let $A$ lie on $(0,0)$ , $E$ on $(0,y)$ , $I$ on $(x,0)$ , and $M$ on $(\frac{3}{2},\frac{3}{2})$ . Since ${AIME}$ is cyclic, $\angle EMI$ (which is opposite of another right angle) must be a right angle; therefore, $\vec{ME} \cdot \vec{MI} = <\frac{-3}{2}, y-\frac{3}{2}> \cdot <x-\frac{3}{2}, -\frac{3}{2}> = 0$ . Compute the dot product to arrive at the relation $y=3-x$ . We can set up another equation involving the area of $\triangle EMI$ using the Shoelace Theorem. This is $2=(\frac{1}{2})[(\frac{3}{2})(y-\frac{3}{2})+(x)(-y)+(x+\frac{3}{2})(\frac{3}{2})]$ . Multiplying, substituting $3-x$ for $y$ , and simplifying, we get $x^2 -3x + \frac{1}{2}=0$ . Thus, $(x,y)=$ $(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2})$ . But $AI>AE$ , meaning $x=AI=\frac{3 + \sqrt{7}}{2} \rightarrow CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}$ , and the final answer is $3+7+2=\boxed{12}$ .

Solution 5 (Quick)

From $AIME$ cyclic we get $\angle{MEI} = \angle{MAI} = 45^\circ$ and $\angle{MIE} = \angle{MAE} = 45^\circ$ , so $\triangle{EMI}$ is an isosceles right triangle.

From $[EMI]=2$ we get $EM=MI=2$ .

Notice $\triangle{AEM} \cong \triangle{CIM}$ , because $\angle{AEM}=180-\angle{AIM}=\angle{CIM}$ , $EM=IM$ , and $\angle{EAM}=\angle{ICM}=45^\circ$ .

Let $CI=AE=x$ , so $AI=3-x$ .

By Pythagoras on $\triangle{EAI}$ we have $x^2+(3-x)^2=EI^2=8$ , and solve this to get $x=CI=\dfrac{3-\sqrt{7}}{2}$ for a final answer of $3+7+2=\boxed{12}$ .

Solution 6(bash)

Let $CI=a$ , $BE=b$ . Because opposite angles in a cyclic quadrilateral are supplementary, we have $\angle EMI=90^{\circ}$ . By the law of cosines, we have $MI^2=a^2+\frac{9}{4}-\frac{3}{2}a$ , and $ME^2=b^2+\frac{9}{4}-\frac{3}{2}b$ . Notice that $EI=2MO$ , where $O$ is the origin of the circle mentioned in the problem. Thus $\frac{2MO*MO}{2}=2\implies MO=\sqrt{2}, EI=2\sqrt{2}$ . By the Pythagorean Theorem, we have $ME^2+MI^2=EI^2\implies a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b=(2\sqrt{2})^2=8$ . By the Pythagorean Theorem, we have $AE^2+AI^2=EI^2\implies (3-a)^2+(3-b)^2=(2\sqrt{2})^2=8\implies 18-6a-6b+a^2+b^2=8$ . Thus we have $18-6a-6b+a^2+b^2=a^2+\frac{9}{4}-\frac{3}{2}a+b^2+\frac{9}{4}-\frac{3}{2}b\implies 18-6a-6b=\frac{9}{2}-\frac{3}{2}a-\frac{3}{2}b\implies \frac{27}{2}$ $=\frac{9}{2}a+\frac{9}{2}b\implies a+b=3\implies 3-b=a$ . We know that $(3-a)^2+(3-b)^2=8\implies (3-a)^2+a^2=8\implies 2a^2-6a+9=8\implies 2a^2-6a+1=0\implies a=\frac{6\pm \sqrt{36-8}}{2}=\frac{3\pm\sqrt{7}}{2}$ . We take the smaller solution because we have $AI>AE\implies 3-AI<3-AE\implies CI<CE$ , and we want $CI$ , not $CE$ , thus $CI=\frac{3-\sqrt{7}}{2}$ . Thus our final answer is $3+7+2=\boxed{12\textbf{(D)}}$

-vsamc

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search