1991 AHSME Problems/Problem 19
Problem
Triangle has a right angle at
and
. Triangle
has a right angle at
and
. Points
and
are on opposite sides of
. The line through
parallel to
meets
extended at
. If
where
and
are relatively prime positive integers, then
Solution 1
Solution by e_power_pi_times_i
Let be the point such that
and
are parallel to
and
, respectively, and let
and
. Then,
. So,
. Simplifying
, and
. Therefore
, and
. Checking,
is the answer, so
. The answer is
.
Solution 2
Solution by Arjun Vikram
Extend lines and
to meet at a new point
. Now, we see that
. Using this relationship, we can see that
, (so
), and the ratio of similarity between
and
is
. This ratio gives us that
. By the Pythagorean Theorem,
. Thus,
, and the answer is
.
Solution 3 (Trig)
We have and
Now we are trying to find
Now we use the
angle sum identity, which states
Using this identity yields $\sin\left(\arcsin\left(\frac{3}{5}\right)\right)\cos\left(\arcsin\left(\frac{12}{13}\right)\right)+\cos\right(\arcsin\left(\frac{3}{5}\right)right)+\sin\left(\arcsin\left(\frac{12}{13}\right)\right).$ (Error compiling LaTeX. Unknown error_msg)
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.