2008 iTest Problems/Problem 98
Problem
Convex quadrilateral has side-lengths , , , and there exists a circle, lying inside the quadrilateral and having center , that is tangent to all four sides of the quadrilateral. Points and are on the midpoints of and respectively. It can be proven that point always lies on segment . Supposing further that is the midpoint of , the area of quadrilateral may be expressed as , where and are positive integers and is not divisible by the square of any prime. Compute .
Solution (credit to official solution)
Let be the midpoint of and be the midpoint of . Draw lines from and to and , as seen in the diagram. By SAS Similarity, we find that and , so . Similarly, .
Because opposite sides of have equal length, is a parallelogram. Thus, the diagonals bisect each other, and .
Let be on where , and let be on where . Also, let be on where , and let be on where . By HL Congruency, , so . Now we have information we can use to determine side lengths to compute the inradius so we can determine the area of .
Let . That means , , and . That means . Additionally, since and , and . Since , we must have .
We know that , so
Let be the radius of the incircle. That means
Therefore, the area of the quadrilateral is , so .
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 97 |
Followed by: Problem 99 | |
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