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

(Solution)
(Added a new, more elementary solution (no trig or high-level geometry theorems))
Line 11: Line 11:
 
</math>
 
</math>
  
== Solution ==
+
== Solution 1==
  
 
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)
Line 27: Line 27:
  
 
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)
 
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)
 +
 +
== Solution 3 (More Elementary) ==
 +
 +
Like above, notice that <math>\triangle{EMI}</math> is isosceles and right, which means that <math>\dfrac{ME \cdot MI}{2} = 2</math>, so <math>MI^2=4</math> and <math>MI = 2</math>. Then construct <math>\overline{MF}\perp\overline{AB}</math> and <math>\overline{MG}\perp\overline{AC}</math> as well as <math>\overline{MI}</math>. It's clear that <math>MG^2+GI^2 = MI^2</math> by Pythagorean, so knowing that <math>MG = \dfrac{AB}{2} = \dfrac{3}{2}</math> allows one to solve to get <math>GI = \dfrac{\sqrt{7}}{2}</math>. By just looking at the diagram, <math>CI=AC-MF-GI=\dfrac{3-\sqrt{7}}{2}</math>. The answer is thus <math>3+7+2=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 19:49, 6 May 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 $3+7+2=12$.


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

Invalid username
Login to AoPS