Difference between revisions of "2011 AIME I Problems/Problem 14"

Revision as of 15:54, 3 July 2011

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

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 $m+n=\boxed{037}$ .

@@ Line 3: / Line 3: @@
 == Solution ==
+===Solution 1===
 Let <math>\theta=\angle M_1 M_3 B_1</math>. Thus we have that <math>\cos 2 \angle A_3 M_3 B_1=\cos(2\theta + \frac{\pi}{2})=-\sin2\theta</math>.
@@ Line 18: / Line 19: @@
 Thus we have that <math>\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}</math>. Therefore <math>m+n=5+32=\boxed{037}</math>.
-Alternate Solution:
+===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 = 2/(2+sqrt 2)= sqrt 2 -1, cos 2x = 2*(cos x)^2 -1 = 5 - 4 sqrt 2 = 5- sqrt 32, m+n=37.
+Let <math>A_1A_2 = 2</math>.  Then <math>B_1</math> and <math>B_3</math> are the projections of <math>M_1</math> and <math>M_5</math> onto the line <math>B_1B_3</math>, so <math>2=B_1B_3=M_1M_5\cos x</math>, where <math>x = \angle A_3M_3B_1</math>.  Then since <math>M_1M_5 = 2+\sqrt{2}, \cos x = \dfrac{2}{2+\sqrt{2}}= \sqrt{2} -1</math>,<math>\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}</math>, and <math>m+n=\boxed{037}</math>.
 == See also ==
 {{AIME box|year=2011|n=I|num-b=13|num-a=15}}

2011 AIME I (Problems • Answer Key • Resources)
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