2011 AIME I Problems/Problem 14

Problem

Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$ , $M_3$ , $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ , $\overline{A_3 A_4}$ , $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively. For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ , $R_3 \perp R_5$ , $R_5 \perp R_7$ , and $R_7 \perp R_1$ . Pairs of rays $R_1$ and $R_3$ , $R_3$ and $R_5$ , $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ , $B_3$ , $B_5$ , $B_7$ respectively. If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers. Find $m + n$ .

Solution

$[asy] size(200); defaultpen(linewidth(0.8)); real dif = 45; pair A1=dir(22.5 + 3*dif)*15,A2=dir(22.5 + 2*dif)*15,A3=dir(22.5 + dif)*15,A4=dir(22.5)*15,A5=dir(22.5 + 7*dif)*15,A6=dir(22.5 + 6*dif)*15,A7=dir(22.5 + 5*dif)*15,A8=dir(22.5 + 4*dif)*15; pair M1=(A1+A2)/2,M3=(A3+A4)/2,M5=(A5+A6)/2,M7=(A7+A8)/2; pair B1=extension(M1,(A4.x-1,A4.y-1),M3,(A6.x-1,A6.y+1)),B3=extension(M3,(A6.x-1,A6.y+1),M5,(A8.x+1,A8.y+1)),B5=extension(M5,(A8.x+1,A8.y+1),M7,(A2.x+1,A2.y-1)),B7=extension(M7,(A2.x+1,A2.y-1),M1,(A4.x-1,A4.y-1)); draw(M1--B1^^M3--B3^^M5--B5^^M7--B7); draw(A1--A2--A3--A4--A5--A6--A7--A8--cycle); [/asy]$

Solution 1

Let $\theta=\angle M_1 M_3 B_1$ . Thus we have that $\cos 2 \angle A_3 M_3 B_1=\cos(2\theta + \frac{\pi}{2})=-\sin2\theta$ .

Since $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ is a regular octagon and $B_1 B_3 = A_1 A_2$ , let $k=A_1 A_2 = A_2 A_3 = B_1 B_3$ .

Extend $\overline{A_1 A_2}$ and $\overline{A_5 A_6}$ until they intersect. Denote their intersection as $I_1$ . Through similar triangles & the $45-45-90$ triangles formed, we find that $M_1 M_3=\frac{k}{2}(2+\sqrt2)$ .

We also have that $\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3$ through ASA congruence ( $\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3$ , $M_7 M_1 = M_1 M_3$ , $\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1$ ). Therefore, we may let $n=M_1 B_7 = M_3 B_1$ .

Thus, we have that $\sin\theta=\frac{n+k}{\frac{k}{2}(2+\sqrt2)}$ and that $\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}$ . Therefore $\sin\theta-\cos\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2$ .

Squaring gives that $\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2$ and consequently that $-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta$ through the identities $\sin^2\theta + \cos^2\theta = 1$ and $\sin2\theta = 2\sin\theta\cos\theta$ .

Thus we have that $\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}$ . Therefore $m+n=5+32=\boxed{037}$ .

Solution 2

Let $A_1A_2 = 2$ . Then $B_1$ and $B_3$ are the projections of $M_1$ and $M_5$ onto the line $B_1B_3$ , so $2=B_1B_3=M_1M_5\cos x$ , where $x = \angle A_3M_3B_1$ . Then since $M_1M_5 = 2+\sqrt{2}, \cos x = \dfrac{2}{2+\sqrt{2}}= \sqrt{2} -1$ , $\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}$ , and $m+n=\boxed{037}$ .

2011 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2011 AIME I Problems/Problem 14

Contents

Problem

Solution

Solution 1

Solution 2

See also