1992 AIME Problems/Problem 9
Contents
[hide]Problem
Trapezoid has sides
,
,
, and
, with
parallel to
. A circle with center
on
is drawn tangent to
and
. Given that
, where
and
are relatively prime positive integers, find
.
Solution 1
Let be the base of the trapezoid and consider angles
and
. Let
and let
equal the height of the trapezoid. Let
equal the radius of the circle.
Then
and
Let be the distance along
from
to where the perp from
meets
.
Then and
so
now substitute this into
to get
and
.
Solution 2
From above,
and
. Adding these equations yields
. Thus,
, and
.
from solution 1 we get from 1 that h/r = 70/x and h/r = 50/ (92-x)
this implies that 70/x =50/(92-x) so x = 161/3
See also
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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