Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 USOJMO Problems/Problem 4"

(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=...")
 
m
Line 1: Line 1:
 
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>.
 
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=#f00]Claim: <math>GH || FE || BC</math>[/color][/b]
+
[b][color=f00f00]Claim: <math>GH || FE || BC</math>[/color][/b]
 
By Pascal's on <math>GDCBAH</math>, we see that the intersection of <math>GH</math> and <math>BC</math>, <math>E</math>, and <math>F</math> are collinear. Since <math>FE || BC</math>, we know that <math>HG || BC</math> as well. <math>\blacksquare</math>
 
By Pascal's on <math>GDCBAH</math>, we see that the intersection of <math>GH</math> and <math>BC</math>, <math>E</math>, and <math>F</math> are collinear. Since <math>FE || BC</math>, we know that <math>HG || BC</math> as well. <math>\blacksquare</math>
−
[b][color=#f00]Claim: <math>FB = FD</math>[/color][/b]
+
[b][color=f00f00]Claim: <math>FB = FD</math>[/color][/b]
 
Note that since all cyclic trapezoids are isosceles, <math>HB = GC</math>. Since <math>AB = BC</math> and <math>EB \perp AC</math>, we know that <math>EA = EC</math>, from which we have that <math>DGCA</math> is an isosceles trapezoid and <math>DA = GC</math>. It follows that <math>DA = GC = HB</math>, so <math>BHAD</math> is an isosceles trapezoid, from which <math>FB = FD</math>, as desired. <math>\blacksquare</math>
 
Note that since all cyclic trapezoids are isosceles, <math>HB = GC</math>. Since <math>AB = BC</math> and <math>EB \perp AC</math>, we know that <math>EA = EC</math>, from which we have that <math>DGCA</math> is an isosceles trapezoid and <math>DA = GC</math>. It follows that <math>DA = GC = HB</math>, so <math>BHAD</math> is an isosceles trapezoid, from which <math>FB = FD</math>, as desired. <math>\blacksquare</math>

Revision as of 23:48, 22 June 2020

Let $G$ be the intersection of $AE$ and $(ABCD)$ and $H$ be the intersection of $DF$ and $(ABCD)$. [b][color=f00f00]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=f00f00]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. $\blacksquare$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_USOJMO_Problems/Problem_4&oldid=126249"