Difference between revisions of "2014 AIME II Problems/Problem 14"
(→Solution) |
|||
Line 39: | Line 39: | ||
− | Then using right triangle <math>AHB</math>, we have HB=10 sin | + | We can also use Law of Sines: |
+ | |||
+ | <cmath>\frac{BC}{AB}=\frac{\sin\angle A}{\sin\angle C}</cmath> | ||
+ | <cmath>\frac{BC}{10}=\frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}}\implies BC=5\sqrt{2}</cmath> | ||
+ | |||
+ | Then using right triangle <math>AHB</math>, we have HB=10 sin 15^\circ | ||
Revision as of 16:29, 14 February 2016
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
http://data.artofproblemsolving.com/images/wiki/5/59/AOPS_wiki.PNG ( This is the 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 HB=10 sin 15^\circ
So HB=10 sin (15∘)=.
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 sin 15.
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.