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Difference between revisions of "2020 USOJMO Problems/Problem 4"
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==Solution 5==
 
==Solution 5==
 
Let <math>G</math> be on <math>AB</math> such that <math>GD \perp DB</math>, and <math>H = GD \cap EB</math>. Then <math>\angle{ADB} = \angle{EDB} = 90^{\circ} - \angle{ABE} \implies \triangle{DAE}</math> is the orthic triangle of <math>\triangle{HGB}</math>. Thus, <math>F</math> is the midpoint of <math>GB</math> and lies on the <math>\perp</math> bisector of <math>DB</math>.
 
Let <math>G</math> be on <math>AB</math> such that <math>GD \perp DB</math>, and <math>H = GD \cap EB</math>. Then <math>\angle{ADB} = \angle{EDB} = 90^{\circ} - \angle{ABE} \implies \triangle{DAE}</math> is the orthic triangle of <math>\triangle{HGB}</math>. Thus, <math>F</math> is the midpoint of <math>GB</math> and lies on the <math>\perp</math> bisector of <math>DB</math>.
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==Solution 6==
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Let <math>FE</math> meet <math>AC</math> at <math>J</math>, <math>BE</math> meet <math>AC</math> at <math>S</math>, connect <math>AE, SD</math>.
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Denote that <math>\angle{BCA}=\alpha; AB=BC, \angle{BAC}=\angle{BCA}=\alpha</math>, since <math>EF</math> is parallel to <math>BC</math>, <math>\angle{AJF}=\angle{ACB}=\alpha</math>. <math>\angle{AJF}</math>and <math>\angle{EJS}</math> are vertical angle, so they are equal to each other.
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<math>BE\bot{AC}</math>,<math>\angle{JES}=90^{\circ}-\alpha</math>, since <math>\angle{EFB}=\angle{AJF}+\angle{FAJ}=2\alpha</math>, we can express <math>\angle{FBE}=180^{\circ}-2\alpha-(90^{\circ}-\alpha)=90^{\circ}-\alpha=
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\angle{FEB}</math>, leads to <math>FE=FB</math>
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Notice that quadrilateral <math>AFED</math> is a cyclic quadrilateral since <math>\angle{ADE}+\angle{AFE}=\angle{ADE}+\angle{ABC}=180^{\circ}</math>.
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Assume <math>\angle{ECA}=\beta</math>, <math>\triangle{AES}</math> is congruent to <math>\triangle{CES}</math> since <math>AS=AS,\angle{ASE}=\angle{BSE}, SE=SE(SAS)</math>, so we can get <math>\angle{EAS}=\beta</math>
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Let the circumcircle of <math>AFED</math> meets <math>AC</math> at <math>Q</math>
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Now notice that <math>\widehat{QE}=\widehat{QE}, \angle{QAE}=\angle{QDE}=\beta</math>; similarly, <math>\widehat{FQ}=\widehat{FQ}; \angle{FDQ}=\angle{FAQ}=\alpha</math>.
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<math>\angle{FDE}=\alpha+\beta; \angle{FED}=\angle{BCD}=\alpha+\beta</math>, it leads to <math>FD=FE</math>.
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since <math>FE=FB;FD=FE, DF=BF</math> as desired
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~bluesoul
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==See Also==
 +
{{USAJMO newbox|year=2020|num-b=3|num-a=5}}
 +
{{MAA Notice}}

Latest revision as of 18:15, 6 October 2023

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

Solution 5

Let $G$ be on $AB$ such that $GD \perp DB$, and $H = GD \cap EB$. Then $\angle{ADB} = \angle{EDB} = 90^{\circ} - \angle{ABE} \implies \triangle{DAE}$ is the orthic triangle of $\triangle{HGB}$. Thus, $F$ is the midpoint of $GB$ and lies on the $\perp$ bisector of $DB$.

Solution 6

Let $FE$ meet $AC$ at $J$, $BE$ meet $AC$ at $S$, connect $AE, SD$. Denote that $\angle{BCA}=\alpha; AB=BC, \angle{BAC}=\angle{BCA}=\alpha$, since $EF$ is parallel to $BC$, $\angle{AJF}=\angle{ACB}=\alpha$. $\angle{AJF}$and $\angle{EJS}$ are vertical angle, so they are equal to each other. $BE\bot{AC}$,$\angle{JES}=90^{\circ}-\alpha$, since $\angle{EFB}=\angle{AJF}+\angle{FAJ}=2\alpha$, we can express $\angle{FBE}=180^{\circ}-2\alpha-(90^{\circ}-\alpha)=90^{\circ}-\alpha= \angle{FEB}$, leads to $FE=FB$

Notice that quadrilateral $AFED$ is a cyclic quadrilateral since $\angle{ADE}+\angle{AFE}=\angle{ADE}+\angle{ABC}=180^{\circ}$.

Assume $\angle{ECA}=\beta$, $\triangle{AES}$ is congruent to $\triangle{CES}$ since $AS=AS,\angle{ASE}=\angle{BSE}, SE=SE(SAS)$, so we can get $\angle{EAS}=\beta$ Let the circumcircle of $AFED$ meets $AC$ at $Q$ Now notice that $\widehat{QE}=\widehat{QE}, \angle{QAE}=\angle{QDE}=\beta$; similarly, $\widehat{FQ}=\widehat{FQ}; \angle{FDQ}=\angle{FAQ}=\alpha$.

$\angle{FDE}=\alpha+\beta; \angle{FED}=\angle{BCD}=\alpha+\beta$, it leads to $FD=FE$.

since $FE=FB;FD=FE, DF=BF$ as desired ~bluesoul

See Also

2020 USAJMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

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