2020 USOJMO Problems/Problem 4

Revision as of 18:12, 22 June 2020 by Aopsuser101 (talk | contribs) (Created page with "Let <math>G</math> be the intersection of <math>AE</math> and <math>(ABCD)</math> and <math>H</math> be the intersection of <math>DF</math> and <math>(ABCD)</math>. [b][color=...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Let $G$ be the intersection of $AE$ and $(ABCD)$ and $H$ be the intersection of $DF$ and $(ABCD)$. [b][color=#f00]Claim: $GH || FE || BC$[/color][/b] By Pascal's on $GDCBAH$, we see that the intersection of $GH$ and $BC$, $E$, and $F$ are collinear. Since $FE || BC$, we know that $HG || BC$ as well. $\blacksquare$ [b][color=#f00]Claim: $FB = FD$[/color][/b] Note that since all cyclic trapezoids are isosceles, $HB = GC$. Since $AB = BC$ and $EB \perp AC$, we know that $EA = EC$, from which we have that $DGCA$ is an isosceles trapezoid and $DA = GC$. It follows that $DA = GC = HB$, so $BHAD$ is an isosceles trapezoid, from which $FB = FD$, as desired. $\blacksquare$