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 1

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 $D\left(\frac{1}{2}, 0\right)$ , $E\left(-\frac{\sqrt{3}}{2}, 0\right)$ , and $R\left(0, \frac{\sqrt{3}}{2}\right).$ $M =$ midpoint $(D, R) = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$ 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

Let $MP = x.$ Meanwhile, since $\triangle R PM$ 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 $ED - x$

Using Law of Sine in $\triangle RED$ , we find $RE$ = $\frac{\sqrt{6}}{2}$ .

We got the three sides of $\triangle AER$ . Now using the Law of Cosines on $\angle AER$ . There we can equate $x$ and solve for it. We got $AE=x=\frac{\sqrt{3}-1}{4\sqrt{3}+2}$ . Then rationalize the denominator, we get $AE = \frac{7 - \sqrt{27}}{22}$ .

Solution 3

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 by midpoint theorem $\overline{PM}\parallel\overline{CD}$ . Thus, $APME$ is a parallelogram and therefore $AE = PM = \tfrac 12 CD$ . $[asy] unitsize(8cm); pair a, d, r, e, m, cm, c,p; d=origin; r=dir(60); e=extension(d,left,r,r+dir(75)*(d-r)); m = midpoint(d--r); cm = foot(r, e, m); c=extension(r,cm,d,e); p=midpoint(r--c); a=p+(e-m); draw(e--m); draw(L(r, cm,1, 1)); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); draw(a--r, dashed); draw(a--c, dashed); draw(p--m, dashed); draw(a--p, dashed); pair[] PPAP = {a, d, r, e, m, c, p}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label("$A$", a, E); label("$E$", e, S); label("$C$", c, S); label("$D$", d, SW); label("$M$", m, NW); label("$R$", r, N); label("$P$", p, NW); MA("60^\circ",black,c,d,m,0.07, black); [/asy]$ We can now use coordinates with $D(0,0)$ as origin and $DE$ along the $x$ -axis.

Let $RD=4$ instead of $1$ (in the end we will scale down by $4$ ). Since $\angle D = 60^\circ$ , we get $R(2,2\sqrt{3})$ , and therefore $M(1, \sqrt{3})$ .

We use sine-law in $\triangle RED$ to find the coordinates $E(2+2\sqrt{3}, 0)$ : $DE =4\cdot \frac{\sin 75^\circ}{\sin 45^\circ} = 4(\sin 30^\circ + \cos 30^\circ) = 2+2\sqrt{3}.$ Since slope $(ME)= -\sqrt{3}/(1+2\sqrt{3})$ , and $RC\perp ME$ , it follows that slope $(RC)=(1+2\sqrt{3})/\sqrt{3}$ . If $C(c,0)$ then we have $\frac{2\sqrt{3}}{2-c}=\frac{1+2\sqrt{3}}{\sqrt{3}}\qquad \Longrightarrow\qquad c=\frac{4\sqrt{3}-4}{1+2\sqrt{3}}$ Now $\tfrac 12 CD = \tfrac 12c =(2\sqrt{3}-2)/(1+2\sqrt{3})= \tfrac 1{11}(14-6\sqrt{3})$ .

Scaling down by $4$ , we get $AE=\tfrac 1{22}(7-3\sqrt{3})$ , so our answer is $7+27+22=056$ .

Video Solution

https://youtu.be/muM8UcGKjHo?si=C6o7-C4DgB5i4yKv

~MathProblemSolvingSkills.com

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