2014 AIME II Problems/Problem 14
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 1
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.
Solution 3
Break our diagram into 2 special right triangle by dropping an altitude from to
we then get that
Since
is a 45-45-90,
We know that
and are 30-60-90.
Thus,
. So our final answer is
.
Solution 4
Draw the
. Now, take the perpendicular bisector of
to intersect the circumcircle of
and
at
as shown, and denote
to be the circumcenter of
. It is not difficult to see by angle chasing that
is cyclic, namely with diameter
. Then, by symmetry,
and as
are both subtended by equal arcs they are equal. Hence,
. Now, draw line
and intersect it at
at point
in the diagram. It is not hard to use angle chase to arrive at
a parallelogram, and from our length condition derived earlier,
. From here, it is clear that
; that is,
is just the intersection of the perpendicular from
down to
and
! After this point, note that
. It is easily derived that the circumradius of
is
. Now,
is a
triangle, and from here it is easy to arrive at the final answer of
. ~awang11's sol
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 |
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