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

(Created page with 'Let <math> A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}A_{8} </math> be a regular octagon. Let <math> M_{1},M_{3},M_{5},M_{7} </math> be the midpoints of sides <math> \overline{A_{1}A_{2}…')
 
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Let <math> A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}A_{8} </math> be a regular octagon. Let <math> M_{1},M_{3},M_{5},M_{7} </math> be the midpoints of sides <math> \overline{A_{1}A_{2}},\overline{A_{3}A_{4}},\overline{A_{5}A_{6}} </math>, and <math> \overline{A_{7}A_{8}} </math>, respectively. For <math> i=1,3,5,7 </math>, ray <math> R_{i} </math> is suspended such that <math> R_{1}\perp R_{3}, R_{3}\perp R_{5}, R_{5}\perp R_{7}, R_{7}\perp R_{1} </math>. Pairs of rays <math>R_{1}</math> and <math>R_{3}</math>,  <math>R_{3}</math> and <math>R_{5}</math>, <math>R_{5}</math> and <math>R_{7}</math>, and <math>R_{7}</math> and <math>R_{1}</math> meet at <math> B_{1},B_{3},B_{5},B_{7} </math>, respectively. If <math> B_{1}B_{3}=A_{1}A_{2} </math> , then <math> \cos{2\angle{A_{3}M_{3}B_{1}} </math> can be written as <math> m-\sqrt{n} </math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m+n</math>.
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== Problem ==
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Let <math>A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8</math> be a regular octagon. Let <math>M_1</math>, <math>M_3</math>, <math>M_5</math>, and <math>M_7</math> be the midpoints of sides <math>\overline{A_1 A_2}</math>, <math>\overline{A_3 A_4}</math>, <math>\overline{A_5 A_6}</math>, and <math>\overline{A_7 A_8}</math>, respectively. For <math>i = 1, 3, 5, 7</math>, ray <math>R_i</math> is constructed from <math>M_i</math> towards the interior of the octagon such that <math>R_1 \perp R_3</math>, <math>R_3 \perp R_5</math>, <math>R_5 \perp R_7</math>, and <math>R_7 \perp R_1</math>. Pairs of rays <math>R_1</math> and <math>R_3</math>, <math>R_3</math> and <math>R_5</math>, <math>R_5</math> and <math>R_7</math>, and <math>R_7</math> and <math>R_1</math> meet at <math>B_1</math>, <math>B_3</math>, <math>B_5</math>, <math>B_7</math> respectively. If <math>B_1 B_3 = A_1 A_2</math>, then <math>\cos 2 \angle A_3 M_3 B_1</math> can be written in the form <math>m - \sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers. Find <math>m + n</math>.
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== Solution ==
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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>.
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Since <math>A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8</math> is a regular octagon and <math>B_1 B_3 = A_1 A_2</math>, let <math>k=A_1 A_2 = A_2 A_3 = B_1 B_3</math>.
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Extend <math>\overline{A_1 A_2}</math> and <math>\overline{A_5 A_6}</math> until they intersect. Denote their intersection as <math>I_1</math>. Through similar triangles & the <math>45-45-90</math> triangles formed, we find that <math>M_1 M_3=\frac{k}{2}(2+\sqrt2)</math>.
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We also have that<math>\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3</math> through ASA congruence (<math>\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3</math>, <math>M_7 M_1 = M_1 M_3</math>, <math>\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1</math>). Therefore, we may let <math>n=M_1 B_7 = M_3 B_1</math>.
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Thus, we have that <math>\sin\theta=\frac{n+k}{\frac{k}{2}(2+\sqrt2)}</math> and that <math>\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}</math>. Therefore <math>\sin\theta-\cos\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2</math>.
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Squaring gives that <math>\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2</math> and consequently that <math>-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta</math> through the identities <math>\sin^2\theta + \cos^2\theta = 1</math> and <math>\sin2\theta = 2\sin\theta\cos\theta</math>.
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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>.

Revision as of 14:40, 20 March 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

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 $m+n=5+32=\boxed{037}$.

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