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 53: / Line 53: @@
 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>
@@ Line 61: / Line 62: @@
 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)
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