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

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==Problem==
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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>.
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==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>.
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[b][color=#f00]Claim: <math>GH || FE || BC</math>[/color][/b]
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<b>Claim: <math>GH || FE || BC</math></b>
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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>
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[b][color=#f00]Claim: <math>FB = FD</math>[/color][/b]
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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>
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==Solution 2==
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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>.
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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>.
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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.
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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>.
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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>
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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 \\
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&= \pi - (\angle DBC - x) - (x + 2y) \\
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&= \pi - (\angle GBE + \angle EBC - x) - (x + 2y) \\
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&= \pi - ([\frac{\pi}{2}-x-y]+[\frac{\pi}{2}-y]-x)-(x+2y) \\
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&= x \\
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&= \angle FBD\end{align*}</cmath>
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Thus, <math>\angle FDB = \angle FBD</math>, so <math>\boxed{FB=FD}</math> as desired.<math>\blacksquare</math>
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~MortemEtInteritum
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==Solution 3 (Angle-Chasing)==
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Proving that <math>FB=FD</math> is equivalent to proving that <math>\angle FBD= \angle FDB</math>. Note that <math>\angle FBD=\angle ACD</math> because quadrilateral <math>ABCD</math> is cyclic. Also note that <math>\angle BAC=\angle ACB</math> because <math>AB=BC</math>. <math>AE=EC</math>, which follows from the facts that <math>BE \perp AC</math> and <math>AB=AC</math>, implies that <math>\angle CAE= \angle ACE= \angle ACD= \angle FBD</math>. Thus, we would like to prove that triangle <math>FBD</math> is similar to triangle <math>AEC</math>. In order for this to be true, then <math>\angle BFD</math> must equal <math>\angle AEC</math> which implies that <math>\angle AFD</math> must equal <math>\angle AED</math>. In order for this to be true, then quadrilateral <math>AFED</math> must be cyclic. Using the fact that <math>EF \parallel BC</math>, we get that <math>\angle AFE= \angle ABC</math>, and that <math>\angle FED= \angle BCE</math>, and thus we have proved that quadrilateral <math>AFED</math> is cyclic. Therefore, triangle <math>FDB</math> is similar to isosceles triangle <math>AEC</math> from AA and thus <math>FB=FD</math>.
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-xXINs1c1veXx
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==Solution 4 ==
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BE is perpendicular bisector of AC, so <math>\angle ACE = \angle EAC</math>. FE is parallel to BC and ABCD is cyclic, so AFED is also cyclic. <math>\angle AFD = \angle AED = 2 \angle ACE = 2 \angle FBD</math>. Hence, <math>\angle FBD = \angle BDF</math>, <math> FD = FB</math>.
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Mathdummy

Revision as of 02:13, 29 October 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

Solution 3 (Angle-Chasing)

Proving that $FB=FD$ is equivalent to proving that $\angle FBD= \angle FDB$. Note that $\angle FBD=\angle ACD$ because quadrilateral $ABCD$ is cyclic. Also note that $\angle BAC=\angle ACB$ because $AB=BC$. $AE=EC$, which follows from the facts that $BE \perp AC$ and $AB=AC$, implies that $\angle CAE= \angle ACE= \angle ACD= \angle FBD$. Thus, we would like to prove that triangle $FBD$ is similar to triangle $AEC$. In order for this to be true, then $\angle BFD$ must equal $\angle AEC$ which implies that $\angle AFD$ must equal $\angle AED$. In order for this to be true, then quadrilateral $AFED$ must be cyclic. Using the fact that $EF \parallel BC$, we get that $\angle AFE= \angle ABC$, and that $\angle FED= \angle BCE$, and thus we have proved that quadrilateral $AFED$ is cyclic. Therefore, triangle $FDB$ is similar to isosceles triangle $AEC$ from AA and thus $FB=FD$.

-xXINs1c1veXx

Solution 4

BE is perpendicular bisector of AC, so $\angle ACE = \angle EAC$. FE is parallel to BC and ABCD is cyclic, so AFED is also cyclic. $\angle AFD = \angle AED = 2 \angle ACE = 2 \angle FBD$. Hence, $\angle FBD = \angle BDF$, $FD = FB$.

Mathdummy

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