Difference between revisions of "2011 AIME I Problems/Problem 14"
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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>. | 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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==Diagram== | ==Diagram== | ||
<asy> | <asy> | ||
− | pair A,B,C,D,E,F,G,H,M,N,O,P,W,X,Y,Z; | + | size(250); |
− | A=(- | + | pair A,B,C,D,E,F,G,H,M,N,O,O2,P,W,X,Y,Z; |
− | B=( | + | A=(-76.537,184.776); |
− | C=( | + | B=(76.537,184.776); |
− | D=( | + | C=(184.776,76.537); |
− | E=( | + | D=(184.776,-76.537); |
− | F=(- | + | E=(76.537,-184.776); |
− | G=(- | + | F=(-76.537,-184.776); |
− | H=(- | + | G=(-184.776,-76.537); |
+ | H=(-184.776,76.537); | ||
M=(A+B)/2; | M=(A+B)/2; | ||
N=(C+D)/2; | N=(C+D)/2; | ||
O=(E+F)/2; | O=(E+F)/2; | ||
+ | O2=(A+E)/2; | ||
P=(G+H)/2; | P=(G+H)/2; | ||
− | W=( | + | W=(100,-41.421); |
− | X=(- | + | X=(-41.421,-100); |
− | Y=(- | + | Y=(-100,41.421); |
− | Z=( | + | Z=(41.421,100); |
draw(A--B--C--D--E--F--G--H--A); | draw(A--B--C--D--E--F--G--H--A); | ||
label("$A_1$",A,dir(112.5)); | label("$A_1$",A,dir(112.5)); | ||
Line 64: | Line 37: | ||
label("$M_5$",O,dir(270)); | label("$M_5$",O,dir(270)); | ||
label("$M_7$",P,dir(180)); | label("$M_7$",P,dir(180)); | ||
+ | label("$O$",O2,dir(152.5)); | ||
draw(M--W,red); | draw(M--W,red); | ||
draw(N--X,red); | draw(N--X,red); | ||
draw(O--Y,red); | draw(O--Y,red); | ||
draw(P--Z,red); | draw(P--Z,red); | ||
− | label("$B_1$",W,dir(292.5)); | + | draw(O2--(W+X)/2,red); |
+ | draw(O2--N,red); | ||
+ | label("$\textcolor{blue}{B_1}$",W,dir(292.5)); | ||
+ | label("$B_2$",(W+X)/2,dir(292.5)); | ||
label("$B_3$",X,dir(202.5)); | label("$B_3$",X,dir(202.5)); | ||
label("$B_5$",Y,dir(112.5)); | label("$B_5$",Y,dir(112.5)); | ||
label("$B_7$",Z,dir(22.5)); | label("$B_7$",Z,dir(22.5)); | ||
</asy> | </asy> | ||
+ | All distances are to scale. | ||
+ | |||
+ | ==Solution 1== | ||
+ | We use coordinates. Let the octagon have side length <math>2</math> and center <math>(0, 0)</math>. Then all of its vertices have the form <math>(\pm 1, \pm\left(1+\sqrt{2}\right))</math> or <math>(\pm\left(1+\sqrt{2}\right), \pm 1)</math>. | ||
+ | |||
+ | By symmetry, <math>B_{1}B_{3}B_{5}B_{7}</math> is a square. Thus lines <math>\overleftrightarrow{B_{1}B_{3}}</math> and <math>\overleftrightarrow{B_{5}B_{7}}</math> are parallel, and its side length is the distance between these two lines. However, this is given to be the side length of the octagon, or <math>2</math>. | ||
+ | |||
+ | Suppose the common slope of the lines is <math>m</math> and let <math>m=\tan\theta</math>. Then, we want to find <cmath>\cos 2\left(90-\theta\right)=2\cos^{2}\left(90-\theta\right)-1=2\sin^{2}\theta-1.</cmath> | ||
+ | |||
+ | It can easily be seen that the equations of the lines are <cmath>\begin{align*} B_{1}B_{3}: y-mx+m\left(1+\sqrt{2}\right)=0 \\ B_{5}B_{7}: y-mx-m\left(1+\sqrt{2}\right)=0.\end{align*}</cmath> By the [https://www.cuemath.com/geometry/distance-between-two-lines/ distance between parallel lines formula], a corollary of the [https://brilliant.org/wiki/dot-product-distance-between-point-and-a-line/ point to line distance formula], the distance between these two lines is <cmath>\frac{|c_{2}-c_{1}|}{\sqrt{a^{2}+b^{2}}}=\frac{2m\left(1+\sqrt{2}\right)}{\sqrt{m^{2}+1}}.</cmath> Since we want this to equal <math>2</math>, we have <cmath>\begin{align*}\frac{2m\left(1+\sqrt{2}\right)}{\sqrt{m^{2}+1}}&=2 \\ 4m^{2}\left(3+2\sqrt{2}\right)&=4m^{2}+4 \\ \left(12+8\sqrt{2}\right)m^{2}&=4m^{2}+4 \\ \left(8+8\sqrt{2}\right)m^{2}&=4 \\ m^{2}&=\frac{4}{8+8\sqrt{2}} \\ \Rightarrow m^{2}=\tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}&=\frac{1}{2+2\sqrt{2}}.\end{align*}</cmath> Since <math>\sin^{2}\theta+\cos^{2}\theta=1,</math> we have <math>\sin^{2}\theta=\frac{1}{3+2\sqrt{2}}</math>. Thus <cmath>2\sin^{2}\theta-1=\frac{2}{3+2\sqrt{2}}-1=\frac{-1-2\sqrt{2}}{3+2\sqrt{2}}=\frac{\left(-1-2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}=\frac{5-4\sqrt{2}}{1}=5-\sqrt{32}.</cmath> The answer is <math>\boxed{037}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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 \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta</math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>\cos\theta-\sin\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2</math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}</math>, <math>\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}</math>, and <math>m+n=\boxed{037}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | Notice that <math>R_3</math> and <math>R_7</math> are parallel (<math>B_1B_3B_5B_7</math> is a square by symmetry and since the rays are perpendicular) and <math>B_1B_3=B_3B_5=s=</math> the distance between the parallel rays. If the regular hexagon as a side length of <math>s</math>, then <math>M_3M_7</math> has a length of <math>s+s\sqrt{2}</math>. Let <math>X</math> be on <math>R_3</math> such that <math>M_7X</math> is perpendicular to <math>M_3X</math>, and <math>\phi=\angle M_7M_3X</math>. The distance between <math>R_3</math> and <math>R_7</math> is <math>s=M_7X</math>, so <math>\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}</math>. | ||
+ | |||
+ | Since we are considering a regular hexagon, <math>M_3</math> is directly opposite to <math>M_7</math> and <math>\angle A_3M_3B_1=90 ^\circ +\phi</math>. All that's left is to calculate <math>\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi</math>. By drawing a right triangle or using the Pythagorean identity, <math>\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}</math> and <math>\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}</math>, so <math>m+n=\boxed{037}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 5== | ||
+ | Assume that <math>A_1A_2=1.</math> | ||
+ | Denote the center <math>O</math>, and the midpoint of <math>B_1</math> and <math>B_3</math> as <math>B_2</math>. Then we have that<cmath>\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angle OM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{1/2}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.</cmath>Thus, by the cosine double-angle theorem,<cmath>\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},</cmath>so <math>m+n=\boxed{037}</math>. | ||
+ | |||
== See also == | == See also == | ||
{{AIME box|year=2011|n=I|num-b=13|num-a=15}} | {{AIME box|year=2011|n=I|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:39, 9 February 2023
Contents
Problem
Let be a regular octagon. Let
,
,
, and
be the midpoints of sides
,
,
, and
, respectively. For
, ray
is constructed from
towards the interior of the octagon such that
,
,
, and
. Pairs of rays
and
,
and
,
and
, and
and
meet at
,
,
,
respectively. If
, then
can be written in the form
, where
and
are positive integers. Find
.
Diagram
All distances are to scale.
Solution 1
We use coordinates. Let the octagon have side length and center
. Then all of its vertices have the form
or
.
By symmetry, is a square. Thus lines
and
are parallel, and its side length is the distance between these two lines. However, this is given to be the side length of the octagon, or
.
Suppose the common slope of the lines is and let
. Then, we want to find
It can easily be seen that the equations of the lines are By the distance between parallel lines formula, a corollary of the point to line distance formula, the distance between these two lines is
Since we want this to equal
, we have
Since
we have
. Thus
The answer is
.
Solution 2
Let . Thus we have that
.
Since is a regular octagon and
, let
.
Extend and
until they intersect. Denote their intersection as
. Through similar triangles & the
triangles formed, we find that
.
We also have that through ASA congruence (
,
,
). Therefore, we may let
.
Thus, we have that and that
. Therefore
.
Squaring gives that and consequently that
through the identities
and
.
Thus we have that . Therefore
.
Solution 3
Let . Then
and
are the projections of
and
onto the line
, so
, where
. Then since
,
, and
.
Solution 4
Notice that and
are parallel (
is a square by symmetry and since the rays are perpendicular) and
the distance between the parallel rays. If the regular hexagon as a side length of
, then
has a length of
. Let
be on
such that
is perpendicular to
, and
. The distance between
and
is
, so
.
Since we are considering a regular hexagon, is directly opposite to
and
. All that's left is to calculate
. By drawing a right triangle or using the Pythagorean identity,
and
, so
.
Solution 5
Assume that
Denote the center
, and the midpoint of
and
as
. Then we have that
Thus, by the cosine double-angle theorem,
so
.
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 |
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