2011 AMC 12A Problems/Problem 24
Problem
Consider all quadrilaterals such that
,
,
, and
. What is the radius of the largest possible circle that fits inside or on the boundary of such a quadrilateral?
Solution
Solution 1
Note as above that ABCD must be cyclic to obtain the circle with maximal radius. Let ,
,
, and
be the points on
,
,
, and
respectively where the circle is tangent. Let
and
. Since the quadrilateral is cyclic,
and
. Let the circle have center
and radius
. Note that
,
,
, and
are right angles.
Hence ,
,
, and
.
Therefore, and
.
Let . Then
,
,
, and
. Using
and
we have
, and
. By equating the value of
from each,
. Solving we obtain
so that
.
Solution 2
To maximize the radius of the circle, we also need to maximize the area. To maximize the area of the circle, the quadrilateral must be tangential (have an incircle). In a tangential quadrilateral, the sum of opposite sides is equal to the semiperimeter of the quadrilateral. So, . Therefore, it has an incircle. By definition, given
side lengths, a cyclic quadrilateral has the maximum area of any quadrilateral with those side lengths. Therefore, to maximize the area of the quadrilateral and thus the incircle, we assume that this quadrilateral is cyclic.
For cyclic quadrilaterals, the area is where
is the semiperimeter of the cyclic quadrilateral and
and
are the sides of the quadrilateral. Compute this area to get
. The area of a tangential quadrilateral is given by the
formula, where
. We can substitute
, the semiperimeter, and
, the area and solve for r to get
.
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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