AIME 2014 II · 第 11 题

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 $\boxed{056}$.

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 $AE = PM = \tfrac 12 CD$.

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 $7+27+22=056$.

Solution 4 (No intuition)

Define the foot of the altitude from $R$ to $ED$ as $X.$ Suppose the median $EM$ is equal to $m,$ and the angle $\angle{MED}$ as $\theta.$ Using the Law of Cosines in triangle $EDM,$ we get $m^2=1+\frac{\sqrt{3}}{4}.$ By Law of Sines in $EMD,$ $\frac{m}{\sin{60^{\circ}}} = \frac{\frac{1}{2}}{\sin\theta}.$ This gives $\sin\theta = \frac{\sqrt{3}}{4m},$ and solving for $\cos\theta$ gives $\cos\theta = \sqrt{1-\frac{3}{16m^2}} = \frac{\sqrt{16m^2-3}}{4m}.$ Using the established value of $m^2,$ we have $\cos{\theta} = \frac{\sqrt{13+4\sqrt{3}}}{4m} = \frac{1+2\sqrt{3}}{4m}.$ Thus, we have $\tan\theta = \frac{\sqrt{3}}{1+2\sqrt{3}}.$ Noting that $\angle{XRC}=\theta$ as well, this gives $CX = RX \tan \theta = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{1+2\sqrt{3}} = \frac{3}{2+4\sqrt{3}}.$ Suppose $AE=n.$ Now, using Pythagorean in triangle $AXR$ gives

$$\left(n + \frac{\sqrt{3}}{2}\right)^2 + \frac{3}{4} = \left(n + \frac{\sqrt{3}}{2} + \frac{3}{2+4\sqrt{3}}\right)^2.$$ Solving, we have

$$\frac{3}{4} = \left(\frac{3}{2+4\sqrt{3}}\right) \left(2n + \sqrt{3} + \frac{3}{2+4\sqrt{3}}\right),$$ $$2n + \sqrt{3} + \frac{3}{2+4\sqrt{3}} = \frac{1+2\sqrt{3}}{2},$$ $$2n + \frac{3}{2+4\sqrt{3}} = \frac{1}{2}.$$ Isolating $2n$ and rationalizing the denominator returns the final answer of $n=\frac{7-\sqrt{27}}{22},$ thus giving an answer of $\boxed{056}.$

~anduran

Video Solution

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

~MathProblemSolvingSkills.com