Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
m (→‎Solution 1)
Line 19: Line 19:
  
  
−
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>.
+
Extend <math>\overline{A_1 A_2}</math> and <math>\overline{A_3 A_4}</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>.
  
 
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>.
 
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>.

Revision as of 20:07, 14 January 2016

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 $\overline{A_3 A_4}$ 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 $M_1M_5 = 2+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}$,$\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}$, and $m+n=\boxed{037}$.

See also

2011 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_14&oldid=74555"