2004 USAMO Problems/Problem 6
(Zuming Feng) A circle is inscribed in a quadrilateral . Let be the center of . Suppose that
Prove that is an isosceles trapezoid.
Our proof is based on the following key Lemma.
Lemma: If a circle , centered at , is inscribed in a quadrilateral , then
Proof: Since circle is inscribed in , we get , , , , and . Construct a point outside of the quadrilateral such that is similar to . We obtain implying that the quadrilateral is cyclic. By Ptolemy's Theorem, we have , or Because is cyclic, it is not difficult to see that, as indicated in the figure, , , , and . Note that and are similar, implying that Substituting the above equalities into the identity , we arrive at or Note also that and are similar, implying that , or Substituting the above identity back into gives the desired relation , establishing the Lemma.
Now we prove our main result. By the Lemma and symmetry, we have Adding the two identities and gives By the AM-GM Inequality, we have . Thus, where the equality holds if and only if . Likewise, we have where the equality holds if and only if . Adding the last two identities gives because . (The latter equality is true because the circle is inscribed in the quadrilateral .)
By the given condition in the problem, all the equalities in the above discussion must hold, that is, and . Consequently, we have , , and so , implying that . It is not difficult to see that and are congruent, implying that . Thus, is an isosceles trapezoid.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
|2004 USAMO (Problems • Resources)|
|1 • 2 • 3 • 4 • 5 • 6|
|All USAMO Problems and Solutions|