1990 AHSME Problems/Problem 28
Problem
A quadrilateral that has consecutive sides of lengths and
is inscribed in a circle and also has a circle inscribed in it. The point of tangency of the inscribed circle to the side of length 130 divides that side into segments of length
and
. Find
.
Solution
Let,
,
, and
be the vertices of this quadrilateral such that
,
,
, and
. Let
be the center of the incircle. Draw in the radii from the center of the incircle to the points of tangency. Let these points of tangency
,
,
, and
be on
,
,
, and
, respectively. Using the right angles and the fact that the
is cyclic, we see that quadrilaterals
and
are similar.
Let have length
. Chasing lengths, we find that
. Using Brahmagupta's Formula we find that
has area
and from that we find, using that fact that
, where
is the inradius and
is the semiperimeter,
.
From the similarity we have
Or, after cross multiplying and writing in terms of the variables,
Plugging in the value of
and solving the quadratic gives
, and from there we compute the desired difference to get
.
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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