2014 AIME II Problems/Problem 11

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

Call $MP$ $x$ . 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 $\overline A\overline E$ and $\overline P\overline M$ .

Now we just have to calculate $ED$ . Using the Law of Sines, or perhaps using altitude $\overline R\overline O$ , we get $ED = \frac{\sqrt{3}+1}{2}$ . $CA=RA$ , which equals $ED - x$

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. There we can equate $x$ and solve for it.

Taking $\triangle AER$ and using $\angle AER$ , of course, we find out (after some calculation) that $AE = \frac{7 - \sqrt{27}}{22}$ . The step before? $x=\frac{1-\sqrt{3}}{4\sqrt{3}+2}$ .

2014 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2014 AIME II Problems/Problem 11

Contents

Problem 11

Solution

Solution 2

See also