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"

m (Undo revision 126249 by Donguri (talk))
(Tag: Undo)
Line 1: Line 1:
 +
==Problem==
 +
 +
Let <math>ABCD</math> be a convex quadrilateral inscribed in a circle and satisfying <math>DA < AB = BC < CD</math>. Points <math>E</math> and <math>F</math> are chosen on sides <math>CD</math> and <math>AB</math> such that <math>BE \perp AC</math> and <math>EF \parallel BC</math>. Prove that <math>FB = FD</math>.
 +
 +
==Solution==
 
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>Claim: <math>GH || FE || BC</math></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]
+
 
 
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>
 +
 +
==Solution 2==
 +
Let <math>G=\overline{FE}\cap\overline{BC}</math>, and let <math>G=\overline{AC}\cap\overline{BE}</math>. Now let <math>x=\angle ACE</math> and <math>y=\angle BCA</math>.
 +
 +
From <math>BA=BC</math> and <math>\overline{BE}\perp \overline{AC}</math>, we have <math>AE=EC</math> so <math>\angle EAC =\angle ECA = x</math>. From cyclic quadrilateral ABCD, <math>\angle ABD = \angle ACD = x</math>. Since <math>BA=BC</math>, <math>\angle BCA = \angle BAC = y</math>.
 +
 +
Now from cyclic quadrilateral ABC and <math>\overline{FE}\parallel \overline{BC}</math> we have <math>\angle FAC = \angle BAC = \pi - \angle BCD = \pi - \angle FED</math>. Thus F, A, D, and E are concyclic, and <math>\angle DFG = \angle DAE = \angle DAC - \angle EAC = \angle DBC - x</math> Let this be statement 1.
 +
 +
Now since <math>\overline{AH}\perp \overline {BH}</math>, triangle ABC gives us <math>\angle BAH + \angle ABG = \frac{\pi}{2}</math>. Thus <math>y+x+\angle GBE=\frac{\pi}{2}</math>, or <math>\angle GBE = \frac{\pi}{2}-x-y</math>.
 +
 +
Right triangle BHC gives <math>\angle HBC = \frac{\pi}{2}-y</math>, and <math>\overline{BC}\parallel \overline{FE}</math> implies <math>\angle BEG=\angle HBC = \frac{\pi}{2}-y.</math>
 +
 +
Now triangle BGE gives <math>\angle BGE = \pi - \angle BEG - \angle GBE = \pi - (\frac{\pi}{2}-y)-(\frac{\pi}{2}-x-y)=x+2y</math>. But <math>\angle FGB = \angle BGE</math>, so <math>\angle FGB=x+2y</math>. Using triangle FGD and statement 1 gives <cmath>\begin{align*}\angle FDG &= \pi - \angle DFG - \angle FGB \\
 +
&= \pi - (\angle DBC - x) - (x + 2y) \\
 +
&= \pi - (\angle GBE + \angle EBC - x) - (x + 2y) \\
 +
&= \pi - ([\frac{\pi}{2}-x-y]+[\frac{\pi}{2}-y]-x)-(x+2y) \\
 +
&= x \\
 +
&= \angle FBD\end{align*}</cmath>
 +
 +
Thus, <math>\angle FDB = \angle FBD</math>, so <math>\boxed{FB=FD}</math> as desired.<math>\blacksquare</math>
 +
 +
~MortemEtInteritum

Revision as of 12:53, 7 July 2020

Problem

Let $ABCD$ be a convex quadrilateral inscribed in a circle and satisfying $DA < AB = BC < CD$. Points $E$ and $F$ are chosen on sides $CD$ and $AB$ such that $BE \perp AC$ and $EF \parallel BC$. Prove that $FB = FD$.

Solution

Let $G$ be the intersection of $AE$ and $(ABCD)$ and $H$ be the intersection of $DF$ and $(ABCD)$.

Claim: $GH || FE || BC$

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. $\blacksquare$

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. $\blacksquare$

Solution 2

Let $G=\overline{FE}\cap\overline{BC}$, and let $G=\overline{AC}\cap\overline{BE}$. Now let $x=\angle ACE$ and $y=\angle BCA$.

From $BA=BC$ and $\overline{BE}\perp \overline{AC}$, we have $AE=EC$ so $\angle EAC =\angle ECA = x$. From cyclic quadrilateral ABCD, $\angle ABD = \angle ACD = x$. Since $BA=BC$, $\angle BCA = \angle BAC = y$.

Now from cyclic quadrilateral ABC and $\overline{FE}\parallel \overline{BC}$ we have $\angle FAC = \angle BAC = \pi - \angle BCD = \pi - \angle FED$. Thus F, A, D, and E are concyclic, and $\angle DFG = \angle DAE = \angle DAC - \angle EAC = \angle DBC - x$ Let this be statement 1.

Now since $\overline{AH}\perp \overline {BH}$, triangle ABC gives us $\angle BAH + \angle ABG = \frac{\pi}{2}$. Thus $y+x+\angle GBE=\frac{\pi}{2}$, or $\angle GBE = \frac{\pi}{2}-x-y$.

Right triangle BHC gives $\angle HBC = \frac{\pi}{2}-y$, and $\overline{BC}\parallel \overline{FE}$ implies $\angle BEG=\angle HBC = \frac{\pi}{2}-y.$

Now triangle BGE gives $\angle BGE = \pi - \angle BEG - \angle GBE = \pi - (\frac{\pi}{2}-y)-(\frac{\pi}{2}-x-y)=x+2y$. But $\angle FGB = \angle BGE$, so $\angle FGB=x+2y$. Using triangle FGD and statement 1 gives \begin{align*}\angle FDG &= \pi - \angle DFG - \angle FGB \\ &= \pi - (\angle DBC - x) - (x + 2y) \\ &= \pi - (\angle GBE + \angle EBC - x) - (x + 2y) \\ &= \pi - ([\frac{\pi}{2}-x-y]+[\frac{\pi}{2}-y]-x)-(x+2y) \\ &= x \\ &= \angle FBD\end{align*}

Thus, $\angle FDB = \angle FBD$, so $\boxed{FB=FD}$ as desired.$\blacksquare$

~MortemEtInteritum

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_USOJMO_Problems/Problem_4&oldid=127678"