Difference between revisions of "2011 AIME I Problems/Problem 14"
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==Diagram== | ==Diagram== | ||
+ | <asy> | ||
+ | pair A,B,C,D,E,F,G,H,M,N,O,P,W,X,Y,Z; | ||
+ | A=(-38.268,92.388); | ||
+ | B=(38.268,92.388); | ||
+ | C=(92.388,38.268); | ||
+ | D=(92.388,-38.268); | ||
+ | E=(38.268,-92.388); | ||
+ | F=(-38.268,-92.388); | ||
+ | G=(-92.388,-38.268); | ||
+ | H=(-92.388,38.268); | ||
+ | M=(A+B)/2; | ||
+ | N=(C+D)/2; | ||
+ | O=(E+F)/2; | ||
+ | P=(G+H)/2; | ||
+ | W=(50,-20.711); | ||
+ | X=(-20.711,-50); | ||
+ | Y=(-50,20.711); | ||
+ | Z=(20.711,50); | ||
+ | draw(A--B--C--D--E--F--G--H--A); | ||
+ | label("$A_1$",A,dir(112.5)); | ||
+ | label("$A_2$",B,dir(67.5)); | ||
+ | label("$\textcolor{blue}{A_3}$",C,dir(22.5)); | ||
+ | label("$A_4$",D,dir(337.5)); | ||
+ | label("$A_5$",E,dir(292.5)); | ||
+ | label("$A_6$",F,dir(247.5)); | ||
+ | label("$A_7$",G,dir(202.5)); | ||
+ | label("$A_8$",H,dir(152.5)); | ||
+ | label("$M_1$",M,dir(90)); | ||
+ | label("$\textcolor{blue}{M_3}$",N,dir(0)); | ||
+ | label("$M_5$",O,dir(270)); | ||
+ | label("$M_7$",P,dir(180)); | ||
+ | draw(M--W,red); | ||
+ | draw(N--X,red); | ||
+ | draw(O--Y,red); | ||
+ | draw(P--Z,red); | ||
+ | label("$B_1$",W,dir(292.5)); | ||
+ | label("$B_3$",X,dir(202.5)); | ||
+ | label("$B_5$",Y,dir(112.5)); | ||
+ | label("$B_7$",Z,dir(22.5)); | ||
+ | </asy> | ||
== 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:01, 25 February 2016
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
.
Diagram
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.