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

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

@@ 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>.
-[b][color=f00f00]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>
-[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>
+==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
+==Solution 3 (Angle-Chasing)==
+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>.
+-xXINs1c1veXx
+==Solution 4 ==
+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>.
+Mathdummy
+==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>.
+==Solution 6==
+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>.
+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.
+<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=
+\angle{FEB}</math>, leads to <math>FE=FB</math>
+Notice that quadrilateral <math>AFED</math> is a cyclic quadrilateral since <math>\angle{ADE}+\angle{AFE}=\angle{ADE}+\angle{ABC}=180^{\circ}</math>.
+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>
+Let the circumcircle of <math>AFED</math> meets <math>AC</math> at <math>Q</math>
+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>.
+<math>\angle{FDE}=\alpha+\beta; \angle{FED}=\angle{BCD}=\alpha+\beta</math>, it leads to <math>FD=FE</math>.
+since <math>FE=FB;FD=FE, DF=BF</math> as desired
+~bluesoul
+==See Also==
+{{USAJMO newbox|year=2020|num-b=3|num-a=5}}
+{{MAA Notice}}

2020 USAJMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
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