1991 AHSME Problems/Problem 19
Contents
[hide]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
Solution 4 (Really simple similar triangles)
By Pythagoras on right triangle ,
. Draw the line through
parallel to
. Let this line intersect
at
.
[asy]
unitsize(0.3 cm);
pair A, B, C, D, E, F;
A = (0,3); B = (4,0); C = (0,0); D = scale(12/5)*rotate(90)*(B - A) + A; E = (D + reflect(B,C)*(D))/2; F = extension(A,C,D, D + C - E);
draw(A--C--E--D); draw(A--B--D--cycle); draw(A--F--D);
label("", A, W);
label("
", B, S);
label("
", C, SW);
label("
", D, NE);
label("
", E, SE);
label("
", F, NW);
[/asy]
Since
,
. Hence, right triangles
and
are similar. Then
so
.
Since quadrilateral is a rectangle,
. Therefore,
Then
,
, and
. The answer is (B).
~ math31415926535
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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