Difference between revisions of "2020 USOJMO Problems/Problem 4"
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Latest revision as of 13:27, 19 April 2021
Let be a convex quadrilateral inscribed in a circle and satisfying . Points and are chosen on sides and such that and . Prove that .
Let be the intersection of and and be the intersection of and .
By Pascal's on , we see that the intersection of and , , and are collinear. Since , we know that as well.
Note that since all cyclic trapezoids are isosceles, . Since and , we know that , from which we have that is an isosceles trapezoid and . It follows that , so is an isosceles trapezoid, from which , as desired.
Let , and let . Now let and .
From and , we have so . From cyclic quadrilateral ABCD, . Since , .
Now from cyclic quadrilateral ABC and we have . Thus F, A, D, and E are concyclic, and Let this be statement 1.
Now since , triangle ABC gives us . Thus , or .
Right triangle BHC gives , and implies
Now triangle BGE gives . But , so . Using triangle FGD and statement 1 gives
Thus, , so as desired.
Solution 3 (Angle-Chasing)
Proving that is equivalent to proving that . Note that because quadrilateral is cyclic. Also note that because . , which follows from the facts that and , implies that . Thus, we would like to prove that triangle is similar to triangle . In order for this to be true, then must equal which implies that must equal . In order for this to be true, then quadrilateral must be cyclic. Using the fact that , we get that , and that , and thus we have proved that quadrilateral is cyclic. Therefore, triangle is similar to isosceles triangle from AA and thus .
BE is perpendicular bisector of AC, so . FE is parallel to BC and ABCD is cyclic, so AFED is also cyclic. . Hence, , .
Let be on such that , and . Then is the orthic triangle of . Thus, is the midpoint of and lies on the bisector of .