2018 AMC 12A Problems/Problem 20

Revision as of 04:03, 23 January 2023 by Pi is 3.14 (talk | contribs) (Video Solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)


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 $a+b+c$?

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


[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 $EI$ since $m\angle MEI = m\angle MAI = 45^\circ$), 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 $\boxed{\textbf{(D) }12}$.

Solution 2 (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 $\boxed{\textbf{(D) }12}$. (Surefire2019)

Solution 3 (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 $3+7+2=\boxed{\textbf{(D) }12}$.

Solution 4 (Coordinate Geometry)

Let $A$ lie on $(0,0)$, $E$ on $(0,y)$, $I$ on $(x,0)$, and $M$ on $\left(\frac{3}{2},\frac{3}{2}\right)$. Since ${AIME}$ is cyclic, $\angle EMI$ (which is opposite of another right angle) must be a right angle; therefore, $\overrightarrow{ME} \cdot \overrightarrow{MI} = \left<\frac{-3}{2}, y-\frac{3}{2}\right> \cdot \left<x-\frac{3}{2}, -\frac{3}{2}\right> = 0$. 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 \[2=\frac{1}{2}\left[\frac{3}{2}\left(y-\frac{3}{2}\right)-xy+\frac{3}{2}\left(x+\frac{3}{2}\right)\right].\] Multiplying, substituting $3-x$ for $y$, and simplifying, we get $x^2 -3x + \frac{1}{2}=0$. Thus, $(x,y)=\left(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2}\right)$. But $AI>AE$, meaning $x=AI=\frac{3 + \sqrt{7}}{2}$ and $CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}$, and the final answer is $3+7+2=\boxed{\textbf{(D) }12}$.

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 $3+7+2=\boxed{\textbf{(D) }12}$.

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 \begin{align*} (3-a)^2+(3-b)^2&=8 \\ (3-a)^2+a^2&=8 \\ 2a^2-6a+9&=8 \\ 2a^2-6a+1&=0 \\ a&=\frac{6\pm \sqrt{36-8}}{2}=\frac{3\pm\sqrt{7}}{2}. \end{align*} 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 $3+7+2=\boxed{\textbf{(D) }12}$.


Video Solution by OmegaLearn


~ pi_is_3.14

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png