Difference between revisions of "1992 AIME Problems/Problem 9"
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\frac{AX}{AP} &= \frac{XB}{BP}\\ | \frac{AX}{AP} &= \frac{XB}{BP}\\ | ||
\frac{AX}{AP} - \frac{XD}{AP} &= \frac{XB}{BP} - \frac{XC}{BP}\\ | \frac{AX}{AP} - \frac{XD}{AP} &= \frac{XB}{BP} - \frac{XC}{BP}\\ | ||
− | \frac{AD}{AP} &= \frac{ | + | \frac{AD}{AP} &= \frac{BC}{BP}\\ |
&=\frac{7}{5} | &=\frac{7}{5} | ||
\end{align*}</cmath> | \end{align*}</cmath> |
Revision as of 17:51, 20 September 2015
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
Let be the distance along
from
to where the perp from
meets
.
Then and
so
.
We can substitute this into
to find that
and
.
Remark: One can come up with the equations in without directly resorting to trig. From similar triangles,
and
. This implies that
, so
.
Solution 2
From above,
and
. Adding these equations yields
. Thus,
, and
.
We can use from Solution 1 to find that
and
.
This implies that so
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 and
. We can now find the area of the trapezoid as the sum of the areas of the three triangles
,
, and
.
(where
is the radius of the tangent circle.)
From Solution 1 above,
Substituting , we find
, hence the answer is
.
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.