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Difference between revisions of "2018 AMC 12A Problems/Problem 20"

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(→‎Solution 4 (Coordinate Geometry))
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== Solution 4 (Coordinate Geometry) ==
 
== Solution 4 (Coordinate Geometry) ==
−
Let <math>A</math> lie on <math>(0,0)</math>, <math>E</math> on <math>(0,y)</math>, <math>I</math> on <math>(x,0)</math>, and <math>M</math> on <math>(\frac{3}{2},\frac{3}{2})</math>. <math>{AIME}</math> is cyclic, so <math>\angle EMI</math> is a right angle. Then, <math>\vec{ME}</math> and <math>\vec{MI}</math> are orthogonal; in other words, <math>\vec{ME} \cdot \vec{MI} = <\frac{-3}{2}, y-\frac{3}{2}> \cdot <x-\frac{3}{2}, -\frac{3}{2}> = 0</math>. Multiply out and simplify to arrive at the relation <math>y=3-x</math>. We can set up another equation for the area of <math>\triangle EMI</math> using the [[Shoelace Theorem]]. This is <math>2=(\frac{1}{2})[(\frac{3}{2})(y-\frac{3}{2})+(x)(-y)+(x+\frac{3}{2})(\frac{3}{2})]</math>. Multiplying, substituting <math>3-x</math> for <math>y</math> and simplifying, we arrive at <math>x^2 -3x + \frac{1}{2}=0</math>. Thus, the solution set is <math>(x,y)=</math> <math>(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2})</math>. But <math>AI>AE</math>, meaning <math>x=AI=\frac{3 + \sqrt{7}}{2} \rightarrow CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}</math>, and the final answer is <math>3+7+2=\boxed{12}</math>.
+
Let <math>A</math> lie on <math>(0,0)</math>, <math>E</math> on <math>(0,y)</math>, <math>I</math> on <math>(x,0)</math>, and <math>M</math> on <math>(\frac{3}{2},\frac{3}{2})</math>. <math>{AIME}</math> is cyclic, so <math>\angle EMI</math> is a right angle; therefore <math>\vec{ME} \cdot \vec{MI} = <\frac{-3}{2}, y-\frac{3}{2}> \cdot <x-\frac{3}{2}, -\frac{3}{2}> = 0</math>. Multiply out and simplify to arrive at the relation <math>y=3-x</math>. We can set up another equation for the area of <math>\triangle EMI</math> using the [[Shoelace Theorem]]. This is <math>2=(\frac{1}{2})[(\frac{3}{2})(y-\frac{3}{2})+(x)(-y)+(x+\frac{3}{2})(\frac{3}{2})]</math>. Multiplying, substituting <math>3-x</math> for <math>y</math> and simplifying, we arrive at <math>x^2 -3x + \frac{1}{2}=0</math>. Thus, the solution set is <math>(x,y)=</math> <math>(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2})</math>. But <math>AI>AE</math>, meaning <math>x=AI=\frac{3 + \sqrt{7}}{2} \rightarrow CI = 3-\frac{3 + \sqrt{7}}{2}=\frac{3 - \sqrt{7}}{2}</math>, and the final answer is <math>3+7+2=\boxed{12}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2018|ab=A|num-b=19|num-a=21}}
 
{{AMC12 box|year=2018|ab=A|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:14, 29 July 2018

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 $a+b+c$?

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

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 $\boxed{12}$. (trumpeter)


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 $(\frac{3}{2},\frac{3}{2})$. ${AIME}$ is cyclic, so $\angle EMI$ is a right angle; therefore $\vec{ME} \cdot \vec{MI} = <\frac{-3}{2}, y-\frac{3}{2}> \cdot <x-\frac{3}{2}, -\frac{3}{2}> = 0$. Multiply out and simplify to arrive at the relation $y=3-x$. We can set up another equation for 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 arrive at $x^2 -3x + \frac{1}{2}=0$. Thus, the solution set is $(x,y)=$ $(\frac{3 \pm \sqrt{7}}{2},\frac{3 \mp \sqrt{7}}{2})$. 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 $3+7+2=\boxed{12}$.

See Also

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AMC 12 Problems and Solutions

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