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

(Created page with "== Problem == Triangle <math>ABC</math> is an isosceles right triangle with <math>AB=AC=3</math>. Let <math>M</math> be the midpoint of hypotenuse <math>\overline{BC}</math>....")
 
(→‎Solution)
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Observe that <math>\triangle{EMI}</math> is isosceles right (<math>M</math> is the midpoint of diameter arc <math>EI</math>), so <math>MI=2,MC=\frac{3}{\sqrt{2}}</math>. With <math>\angle{MCI}=45^\circ</math>, we can use Law of Cosines to determine that <math>CI=\frac{3\pm\sqrt{7}}{2}</math>. The same calculations hold for <math>BE</math> also, and since <math>CI<BE</math>, we deduce that <math>CI</math> is the smaller root, giving the answer of <math>\boxed{12}</math>. (trumpeter)
 
Observe that <math>\triangle{EMI}</math> is isosceles right (<math>M</math> is the midpoint of diameter arc <math>EI</math>), so <math>MI=2,MC=\frac{3}{\sqrt{2}}</math>. With <math>\angle{MCI}=45^\circ</math>, we can use Law of Cosines to determine that <math>CI=\frac{3\pm\sqrt{7}}{2}</math>. The same calculations hold for <math>BE</math> also, and since <math>CI<BE</math>, we deduce that <math>CI</math> is the smaller root, giving the answer of <math>\boxed{12}</math>. (trumpeter)
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== Solution 2 (Using Ptolemy) ==
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We first claim that <math>\triangle{EMI}</math> is isosceles and right.
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Proof: Construct <math>\overline{MF}\perp\overline{AB}</math> and <math>\overline{MG}\perp\overline{AC}</math>. Since <math>\overline{AM}</math> bisects <math>\angle{BAC}</math>, one can deduce that <math>MF=MG</math>. Then by AAS it is clear that <math>MI=ME</math> and therefore <math>\triangle{EMI}</math> is isosceles. Since quadrilateral <math>AIME</math> is cyclic, one can deduce that <math>\angle{EMI}=90^\circ</math>. Q.E.D.
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Since the area of <math>\triangle{EMI}</math> is 2, we can find that <math>MI=ME=2</math>, <math>EI=2\sqrt{2}</math>
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Since <math>M</math> is the mid-point of <math>\overline{BC}</math>, it is clear that <math>AM=\frac{3\sqrt{2}}{2}</math>.
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Now let <math>AE=a</math> and <math>AI=b</math>. By Ptolemy's Theorem, in cyclic quadrilateral <math>AIME</math>, we have <math>2a+2b=6</math>. By Pythagorean Theorem, we have <math>a^2+b^2=8</math>. One can solve the simultaneous system and find <math>b=\frac{3+\sqrt{7}}{2}</math>. Then by deducting the length of <math>\overline{AI}</math> from 3 we get <math>CI=\frac{3-\sqrt{7}}{2}</math>, giving the answer of <math>\boxed{12}</math>. (Surefire2019)
  
 
==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 01:45, 9 February 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

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)

See Also

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

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