Difference between revisions of "2011 AIME I Problems/Problem 14"
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<asy> | <asy> | ||
size(250); | size(250); | ||
− | pair A,B,C,D,E,F,G,H,M,N,O,P,W,X,Y,Z; | + | pair A,B,C,D,E,F,G,H,M,N,O,O2,P,W,X,Y,Z; |
A=(-76.537,184.776); | A=(-76.537,184.776); | ||
B=(76.537,184.776); | B=(76.537,184.776); | ||
Line 58: | Line 58: | ||
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=(100,-41.421); | W=(100,-41.421); | ||
Line 76: | Line 77: | ||
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,NE); | ||
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); | ||
+ | draw(O2--(W+X)/2,red); | ||
+ | dot(O2); | ||
label("$\textcolor{blue}{B_1}$",W,dir(292.5)); | 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)); |
Revision as of 19:38, 10 March 2019
Contents
[hide]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
.
Solution
Solution 1
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 2
Let . Then
and
are the projections of
and
onto the line
, so
, where
. Then since
,
, and
.
Solution 3
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 4
Assume that
Denote the center
, and the midpoint of
and
as
. Then we have that
Thus, by the cosine double-angle theorem,
so
.
Diagram
All distances are to scale.
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.