1992 AIME Problems/Problem 9
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
.
you don't have to use trig nor angles A and B. From similar triangles,
and
this implies that so
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
Solution 3
Extend and
to meet at a point
. Since
and
are parallel,
. If
is further extended to a point
and
is extended to a point
such that
is tangent to circle
, we discover that circle
is the incircle of triangle
. Then line
is the angle bisector of
. By homothety,
is the intersection of the angle bisector of
with
. By the angle bisector theorem,
$
Let , then
.
. Thus,
.
Solution 4
The area of the trapezoid is , where
is the height of the trapezoid.
Draw lines CP and BP. We can now find the area of the trapezoid as the sum of the three triangles BPC, CPD, and PBA.
[BPC] = (where
is the radius of the tangent circle.)
[CPD] =
[PBA] =
[BPC] + [CPD] + [PBA] = = Trapezoid area =
From Solution 1 above,
Substituting , we get
-->
.
See also
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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