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Difference between revisions of "2014 AIME II Problems/Problem 11"

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dot(p);
 
dot(p);
 
</asy>
 
</asy>
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==Solution 2==
 +
 +
NOTE: Draw your own picture for this one! It requires some tenacity.
 +
 +
First, we are stuck, right? Nowhere to go. Wait a minute... there's medians and midpoints. Maybe we should draw auxiliary segment <math>MP</math>. For lack of better term, let's call the length <math>x</math>. Meanwhile, because <math>\triangle RPM</math> is similar to <math>\triangle RCD</math> (angle, side, and side- <math>RP</math> and <math>RC</math> ratio), <math>CD</math> must be 2<math>x</math>. Now, notice that <math>AE</math> is <math>x</math>, because of the parallel segments <math>\overline AE</math> and <math>\overline PM</math> (also using Midsegment Theorem).
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Now we just have to calculate <math>ED</math>. Using the Law of Sines, we get <math>ED = \frac{\sqrt{3}+1}{2}</math>, thus <math>CA = RA</math>, which must be equal to <math>ED - x</math>. (That's minus <math>2x</math> and plus <math>x</math>.)
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Finally, what is <math>RE</math>? It comes out to <math>\frac{\sqrt{6}}{2}</math>.
 +
 +
We got the three sides. Now all that is left is using the Law of Cosines.
 +
 +
Taking <math>\triangle AER</math> and using <math>\angle AER</math>, of course, we find out (after some calculation) that <math>AE = \frac{7 - \sqrt{27}}{22}</math>.
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== See also ==
 
== See also ==

Revision as of 17:51, 18 March 2017

Problem 11

In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $|RD|=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\frac{a-\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$.

Solution

Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and $AE = PM = \frac{CD}{2}$. We can then use coordinates. Let $O$ be the foot of altitude $RO$ and set $O$ as the origin. Now we notice special right triangles! In particular, $DO = \frac{1}{2}$ and $EO = RO = \frac{\sqrt{3}}{2}$, so $D(\frac{1}{2}, 0)$, $E(-\frac{\sqrt{3}}{2}, 0)$, and $R(0, \frac{\sqrt{3}}{2}).$ $M =$ midpoint$(D, R) = (\frac{1}{4}, \frac{\sqrt{3}}{4})$ and the slope of $ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + 2\sqrt{3}}$, so the slope of $RC = -\frac{1 + 2\sqrt{3}}{\sqrt{3}}.$ Instead of finding the equation of the line, we use the definition of slope: for every $CO = x$ to the left, we go $\frac{x(1 + 2\sqrt{3})}{\sqrt{3}} = \frac{\sqrt{3}}{2}$ up. Thus, $x = \frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{3}{4\sqrt{3} + 2} = \frac{3(4\sqrt{3} - 2)}{44} = \frac{6\sqrt{3} - 3}{22}.$ $DC = \frac{1}{2} - x = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{14 - 6\sqrt{3}}{22}$, and $AE = \frac{7 - \sqrt{27}}{22}$, so the answer is $\boxed{056}$.

[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label("$A$", a, W); label("$E$", e, SW); label("$C$", c, S); label("$O$", o, S); label("$D$", d, SE); label("$M$", m, NE); label("$R$", r, N); p = foot(a, r, c); label("$P$", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]

Solution 2

NOTE: Draw your own picture for this one! It requires some tenacity.

First, we are stuck, right? Nowhere to go. Wait a minute... there's medians and midpoints. Maybe we should draw auxiliary segment $MP$. For lack of better term, let's call the length $x$. Meanwhile, because $\triangle RPM$ is similar to $\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\overline AE$ and $\overline PM$ (also using Midsegment Theorem).

Now we just have to calculate $ED$. Using the Law of Sines, we get $ED = \frac{\sqrt{3}+1}{2}$, thus $CA = RA$, which must be equal to $ED - x$. (That's minus $2x$ and plus $x$.)

Finally, what is $RE$? It comes out to $\frac{\sqrt{6}}{2}$.

We got the three sides. Now all that is left is using the Law of Cosines.

Taking $\triangle AER$ and using $\angle AER$, of course, we find out (after some calculation) that $AE = \frac{7 - \sqrt{27}}{22}$.


See also

2014 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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