Difference between revisions of "2014 AIME II Problems/Problem 14"
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==Diagram== | ==Diagram== | ||
− | + | <asy> | |
+ | unitsize(20); | ||
+ | pair A = MP("A",(-5sqrt(3),0)), B = MP("B",(0,5),N), C = MP("C",(5,0)), M = D(MP("M",0.5(B+C),NE)), D = MP("D",IP(L(A,incenter(A,B,C),0,2),B--C),N), H = MP("H",foot(A,B,C),N), N = MP("N",0.5(H+M),NE), P = MP("P",IP(A--D,L(N,N-(1,1),0,10))); | ||
+ | D(A--B--C--cycle); | ||
+ | D(B--H--A,blue+dashed); | ||
+ | D(A--D); | ||
+ | D(P--N); | ||
+ | markscalefactor = 0.05; | ||
+ | D(rightanglemark(A,H,B)); | ||
+ | D(rightanglemark(P,N,D)); | ||
+ | MP("10",0.5(A+B)-(-0.1,0.1),NW); | ||
+ | </asy> | ||
==Solution== | ==Solution== |
Revision as of 23:55, 27 October 2017
Contents
Problem
In , and
. Let
and
be points on the line
such that
,
, and
. Point
is the midpoint of the segment
, and point
is on ray
such that
. Then
, where
and
are relatively prime positive integers. Find
.
Diagram
Solution
As we can see,
is the midpoint of
and
is the midpoint of
is a
triangle, so
.
is
triangle.
and
are parallel lines so
is
triangle also.
Then if we use those informations we get and
and
or
Now we know that , we can find for
which is simpler to find.
We can use point to split it up as
,
We can chase those lengths and we would get
, so
, so
, so
We can also use Law of Sines:
Then using right triangle , we have
So .
And we know that .
Finally if we calculate .
. So our final answer is
.
Thank you.
-Gamjawon
Solution 2.
Here's a solution that doesn't need .
As above, get to . As in the figure, let
be the foot of the perpendicular from
to
. Then
is a 45-45-90 triangle, and
is a 30-60-90 triangle. So
and
; also,
,
, and
. But
and
are parallel, both being orthogonal to
. Therefore
, or
, and we're done.
See also
2014 AIME II (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.