Shortlist 2017/G1

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27943 posts

#1 h Jul 10, 2018, 10:42 AM • 18 Y

Y by Durjoy1729, DiamondVillager, Carpemath, Davi-8191, blacksheep2003, qubatae, itslumi, son7, violin21, rayfish, megahertz13, megarnie, Ahmed102, Adventure10, Mango247, deplasmanyollari, lian_the_noob12, pythonmaster245

Let $ABCDE$ be a convex pentagon such that $AB=BC=CD$ , $\angle{EAB}=\angle{BCD}$ , and $\angle{EDC}=\angle{CBA}$ . Prove that the perpendicular line from $E$ to $BC$ and the line segments $AC$ and $BD$ are concurrent.

This post has been edited 3 times. Last edited by Amir Hossein, Jul 10, 2018, 3:59 PM

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USJL

535 posts

USJL

#2 h Jul 10, 2018, 10:53 AM • 7 Y

Y by Carpemath, thczarif, MountainC, PRMOisTheHardestExam, Adventure10, Mango247, Before-the-snow-melted

It's a really cute problem

Hide by request

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juckter

322 posts

juckter

#5 h Jul 10, 2018, 11:08 AM • 3 Y

Y by Carpemath, Adventure10, Mango247

Let $P$ be the intersection of $AC$ and $BD$ and let $\angle BAC = \angle BCA = \alpha, \angle CAD = \angle CDA = \beta$ . Through simple angle chasing we find that $\angle EAP = \angle ABP = 180^{\circ} - 2\beta - \alpha$ and $\angle EDP = \angle DCP = 180^{\circ} - 2\alpha - \beta$ . We may now note that $EP$ is perpendicular to $BC$ if and only if $\angle EPA = 90^{\circ} - \alpha$ , which happens if and only if $\angle AEP = 2\alpha + 2\beta - 90^{\circ}$ , if and only if $EP$ bisects angle $AED$ . We now prove that $P$ has the same distance to $EA$ and $ED$ , which implies the result.

Looking at triangles $EAP$ and $EDP$ we find that $d(P, EA) = AP\sin(\alpha + 2\beta)$ and $d(P, ED) = DP\sin(2\alpha + \beta)$ . We now wish to prove that

$\frac{AP}{\sin(2\alpha + \beta)} = \frac{DP}{\sin(\alpha + 2\beta)}$
However, this follows from the Law of Sines applied to triangles $APB, BPC$ and $CPD$ , because

$\frac{AP}{\sin(2\alpha + \beta)} = \frac{BP}{\sin \alpha} = \frac{CP}{\sin \beta} = \frac{DP}{\sin(\alpha + 2\beta)}.$

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Durjoy1729

221 posts

Durjoy1729

#6 h Jul 10, 2018, 11:15 AM • 5 Y

Y by Carpemath, potentialenergy, DonaldJ.Trump, Adventure10, Mango247

Diagram

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MarkBcc168

1594 posts

MarkBcc168

#7 h Jul 10, 2018, 11:15 AM • 9 Y

Y by fastlikearabbit, Carpemath, vivoloh, ONE_CALL_AWAY, MountainC, ike.chen, PRMOisTheHardestExam, Adventure10, Mango247

A bit hard for G1.

Let $P=AB\cap CD,\ Q=BC\cap ED,\ R=BC\cap EA,\ T=AC\cap BD$ . The angle conditions imply that $BDQP$ and $CARP$ is cyclic. Combining with $AB=BC=CD$ gives $PC=CQ$ and $PB=BR$ .

Since $AB=BC$ , line $AT$ is parallel to angle bisector of $\angle PBC$ so if we let $I$ be the incenter of $\Delta PBC$ , we get that $BICT$ is parallelogram. Thus the foot $K$ from $T$ to $BC$ is the contact point of $P$ -excircle of $\Delta PBC$ .

Since $\angle EQR=\angle BPC =\angle ERQ$ , we get that $\Delta EQR$ is isosceles. Furthermore, length chasing reveals that
$QK = QC + CK = PC + \frac{PB + BC - PC}{2} = \frac{QR}{2}$ hence $QK=KR\implies EK\perp QR$ . But $TK\perp QR$ so we are done.

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TheDarkPrince

3042 posts

TheDarkPrince

#8 h Jul 10, 2018, 11:28 AM • 5 Y

Y by Carpemath, karitoshi, PRMOisTheHardestExam, Adventure10, Mango247

Extend $AB$ and $CD$ to meet at $R$ , extend $AE$ and $BC$ to meet at $X$ and extend $ED$ and $BC$ to meet at $Y$ . We get from the conditions, $\triangle AXB\cong CRB\cong CYD$ . Suppose $T$ is the feet from $E$ on $BC$ , $XT=YT$ . Also suppose $Z=AC\cap BD$ and the feet from $Z$ on $BC$ is $T'$ . Angle chasing gives $T'$ is the $P-$ excircle touchpoint with $BC$ . Thus $BT'=\frac{BC+PC-PB}{2}$ and $BT=\frac{QR}{2}-BR=\frac{BC+PC+PB}{2}-BP=\frac{BC+PC-PB}{2}.$ Thus $T'\equiv T$ .

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Vrangr

1600 posts

Vrangr

#9 h Jul 10, 2018, 12:22 PM • 3 Y

Y by Carpemath, Adventure10, Mango247

Let $P = AC\cap BD$ . Let $H$ be the orthocenter of triangle $BCP$ .
Claim 1: $D$ is the reflection of $B$ in $\overline{CH}$ and $A$ is the reflection of $C$ in $\overline{BH}$ .
Proof: $AB = BC$ and $AC \perp BH$ . So, we are pretty much done.

Claim 2: $H$ lies on the angle bisector of $\angle AED$ .
Proof: By the above, $\triangle ABH \cong \triangle CBH \cong \triangle CDH$ .
So,
$\angle EDH = \angle EDC - \angle CDH = \angle ABC - \angle ABH = \angle ABH = \angle CDH$ Thus, $H$ lies on the angle bisector $\angle CDE$ .
Similarly, for $\angle BAE$ .
Thus, by the property of angle bisectors,
$d(H, \overline{AE}) = d(H, \overline{AB}) = d(H, \overline{BC}) = d(H, \overline{CD}) = d(H, \overline{DE}).$ Therefore, $H$ lies on the angle bisector of $\angle AED$ .

Claim 3: $P$ lies on the angle bisector of $\angle E$ .
By Sine Rule,
$\frac{AP}{\sin\angle PDE} = \frac{AP}{\sin\angle ABP} = \frac{BP}{\sin \angle BAP} = \frac{BP}{\sin \angle BCP} = \frac{CP}{\sin\angle CBP} = \frac{CP}{\sin\angle CDP} = \frac{DP}{\sin\angle PCD} = \frac{DP}{\sin\angle PAE}$ $\iff d(P, \overline{AB}) = AP\sin\angle PAE = DP\sin\angle PDE = d(P, \overline{ED})$
Main Proof:
Since, $P, H, E$ are collinear and $PH\perp BC$ , we are done.

Any non-trig solutions of claim 3?

This post has been edited 1 time. Last edited by Vrangr, Jul 10, 2018, 12:23 PM

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v_Enhance

6870 posts

v_Enhance

#10 h Jul 10, 2018, 12:27 PM • 25 Y

Y by Durjoy1729, adihaya, Kayak, ibadat, magicarrow, vivoloh, karitoshi, JustKeepRunning, Cindy.tw, aops29, Kamran011, v4913, AllanTian, rayfish, IceWolf10, megahertz13, michaelwenquan, PRMOisTheHardestExam, IAmTheHazard, Adventure10, channing421, MELSSATIMOV40, smileapple, ehuseyinyigit, Akram_Lahlou

We present three solutions.

$[asy] pair T = dir(270); pair S = dir(215); pair U = dir(10); pair R = S*T/U; pair V = T/S*U; pair A = 2*R*S/(R+S); pair B = 2*S*T/(S+T); pair C = 2*T*U/(T+U); pair D = 2*U*V/(U+V); pair E = 2*V*R/(V+R); draw(A--B--C--D--E--cycle, heavygreen); pair X = extension(D, E, A, B); pair Y = extension(D, C, A, E); draw(A--C, lightblue); draw(B--D, lightblue); pair P = extension(A, C, B, D); pair H = orthocenter(B, P, C); draw(B--Y, red); draw(C--X, red); draw(A--X--E--Y--D, dashed+heavygreen); draw(unitcircle, blue+dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$P$", P, dir(P)); dot("$H$", H, dir(180)); /* TSQ Source: T := dir 270 S := dir 215 U := dir 10 R := S*T/U V := T/S*U A = 2*R*S/(R+S) B = 2*S*T/(S+T) C = 2*T*U/(T+U) D = 2*U*V/(U+V) E = 2*V*R/(V+R) A--B--C--D--E--cycle 0.1 lightgreen / heavygreen X = extension D E A B Y = extension D C A E A--C lightblue B--D lightblue P = extension A C B D H = orthocenter B P C R180 B--Y red C--X red A--X--E--Y--D dashed heavygreen unitcircle blue dotted */ [/asy]$

First solution by Brianchon Draw the circle tangent to $AB$ , $BC$ , $CD$ inside the pentagon; then the angle conditions guarantee the incircle is tangent to $EA$ and $ED$ as well. Now apply Brianchon theorem.

Second solution by Pappus Theorem (mine) Let the diagonals $AC$ and $BD$ meet at $P$ . Let $H$ be the orthocenter of $\triangle BCP$ . As $BH$ and $CH$ are perpendicular bisectors of $AC$ and $BD$ , it follows that $BH$ , $AE$ , $CD$ are concurrent at $Y$ , say. Similarly, $CH$ , $DE$ , and $AB$ are concurrent at a point $X$ .

By Pappus theorem on $\overline{XAB}$ and $\overline{YCD}$ , it follows that $H$ , $P$ , $E$ are collinear, as desired.

Third solution by trig (with Ishika Shah) Again let $P = \overline{AC} \cap \overline{BD}$ . Let $T$ be the foot from $P$ to $\overline{BC}$ . Let $\alpha = \angle BAC = \angle BCA$ , and $\beta = \angle CBD = \angle BDC$ .

$[asy] pair T = dir(270); pair S = dir(215); pair U = dir(10); pair R = S*T/U; pair V = T/S*U; pair A = 2*R*S/(R+S); pair B = 2*S*T/(S+T); pair C = 2*T*U/(T+U); pair D = 2*U*V/(U+V); pair E = 2*V*R/(V+R); draw(A--B--C--D--E--cycle, heavygreen); draw(A--C, lightblue); draw(B--D, lightblue); pair P = extension(A, C, B, D); draw(unitcircle, blue+dotted); pair U = extension(A, E, B, C); pair V = extension(D, E, B, C); draw(A--U--B, dotted+blue); draw(D--V--C, dotted+blue); pair T = foot(P, B, C); draw(P--T, red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$P$", P, dir(95)); dot("$U$", U, dir(U)); dot("$V$", V, dir(V)); dot("$T$", T, dir(T)); /* TSQ Source: T := dir 270 S := dir 215 U := dir 10 R := S*T/U V := T/S*U A = 2*R*S/(R+S) B = 2*S*T/(S+T) C = 2*T*U/(T+U) D = 2*U*V/(U+V) E = 2*V*R/(V+R) A--B--C--D--E--cycle 0.1 lightgreen / heavygreen A--C lightblue B--D lightblue P = extension A C B D R95 unitcircle blue dotted U = extension A E B C V = extension D E B C A--U--B dotted blue D--V--C dotted blue T = foot P B C P--T red */ [/asy]$

Let $E_1$ be a point on $\overline{PT}$ satisfying $\angle E_1AB = \angle C = \pi - 2 \beta$ . Similarly, let $E_2$ be a point on $\overline{PT}$ satisfying $\angle E_2DC = \angle B = \pi - 2 \alpha$ . Let $U = \overline{AE_1} \cap \overline{BC}$ and $V = \overline{DE_2} \cap \overline{BC}$ . By angle chasing we see that $\angle BUA = \angle DVC = 180^{\circ} - (2\alpha + 2\beta)$ .

We just need to show that $UT = TV$ now. Then, that will imply $E = E_1 = E_2$ .

WLOG, scale to let $AB = BC = CD = \sin(2\alpha+2\beta).$ Then, by sine law on $\triangle AUB$ , we get $BU = \sin(2\beta)$ . Meanwhile, by the law of sines on $\triangle BPC$ we have $BP = \frac{BC}{\sin \angle BPC} \cdot \sin \angle PCB = \frac{\sin(2\alpha+2\beta)}{\sin(\alpha+\beta)} \cdot \sin\alpha.$ Then, $BT = BP \cos \beta = \frac{\sin(2\alpha+2\beta) \sin \alpha \cos \beta}{\sin (\alpha+\beta)}$ and hence we have $\begin{align*} TU &= TB + BU \\ &= \frac{\sin(2\alpha+2\beta) \sin \alpha \cos \beta} {\sin(\alpha+\beta)} + \sin(2\beta) \\ &= 2\cos(\alpha+\beta) \sin \alpha \cos \beta + \sin(2\beta) \\ &= 2\cos(\alpha+\beta) \sin \alpha \cos \beta + 2\sin\beta\cos\beta \\ &= 2[\cos\alpha\cos\beta-\sin\alpha\sin\beta] \sin \alpha \cos \beta + 2\sin\beta\cos\beta \\ &= 2[\cos^2\beta \cos\alpha\sin\alpha + (1-\sin^2\alpha)\sin\beta\cos\beta] \\ &= 2[\cos^2\beta \cos\alpha\sin\alpha + \cos^2\alpha \sin\beta\cos\beta]. \end{align*}$ This is symmetric in $\alpha$ and $\beta$ , so $TV = 2[\cos^2\beta \cos\alpha\sin\alpha + \cos^2\alpha \sin\beta\cos\beta]$ as well. This finishes the proof.

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psi241

49 posts

psi241

#11 h Jul 10, 2018, 1:31 PM • 2 Y

Y by Adventure10, Mango247

Let $F=AE\cap BC,G=DE\cap BC,H=AC\cap BD,P=AB\cap CD$ and I be a excenter of $\Delta BCP$ opposite P.
Angle chasing reveals that $\Delta EFG$ is isoscele.Because $AB=BC$ , $\angle BAE=\angle BCD$ and $\angle ABF =\angle CBP$ so $\Delta ABF \cong \Delta BCP$ .Similarly $\Delta ABF \cong \Delta CBP \cong \Delta CDG$ . $IB$ bisects $\angle ABC$ so $IB$ is axis of reflection $\Delta ABF$ to $\Delta CBP$ .We get $IP$ bisects $\angle BPC$ \rightarrow $IF$ bisects $\angle AFB$ .In the same way, $IG$ bisects $\angle CGD$ .
$\therefore I$ is incenter of $\Delta EFG$ .This mean $EI\perp BC$ .

Because $\Delta ABC ,\Delta BCD$ are isoscele so $IB\perp AC$ and $IC\perp BD$ . $\therefore H$ is orthocenter of $\Delta BIC$ so $IH\perp BC$ .So we get that $E,I,H$ are collinear and this line perpendicular to $BC$ . $\square$

This post has been edited 1 time. Last edited by psi241, Sep 9, 2018, 8:15 AM

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sa2001

281 posts

sa2001

#12 h Jul 10, 2018, 5:16 PM • 3 Y

Y by guptaamitu1, Adventure10, Mango247

Let $P = AC \cap BD, X = ED \cap BC, Y = EA \cap CB$

Note that triangle $EXY$ is $E-isosceles$ .

Now, it would suffice to show that $P$ is equidistant from $YE$ and $XE$ .

Consider the triangles $ABP$ and $DCP$ .
Using the extended sine law, they have equal circumradii, and thus $DP sin ABP = AP sin PCD$ which means $DP sin PDE = AP sin PAE$ .

And we're done. $\blacksquare$

This post has been edited 2 times. Last edited by sa2001, Jul 11, 2018, 5:39 AM

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Generic_Username

1088 posts

Generic_Username

#13 h Jul 10, 2018, 6:50 PM • 2 Y

Y by Adventure10, Mango247

Let $X,Y$ be the intersections of lines $\overleftrightarrow{EA},\overleftrightarrow{ED}$ with $\overleftrightarrow{BC},$ respectively. Further let $K=\overline{AC}\cap \overline{BD},$ and let $T$ be the spiral center sending $\overline{CD}\to \overline{AB}.$

Lemma: $E,T$ lie on the perpendicular bisector of $\overline{XY}.$

Proof: The angle conditions as well as $CD=AB$ imply that $\triangle CDY\cong \triangle ABX.$ Thus $\angle EYX=\angle DYC=\angle BXA=\angle EXY,$ implying the first part. For the second part, note that as $T$ is the spiral center that sends $\overline {CD}\to \overline {AB},$ it also takes the corresponding triangles $\triangle CDY\to \triangle ABX,$ whence $TX=TY$ follows.

Now, notice that as $\overline{TB}$ is the perpendicular bisector of $\overline{AC}$ and $TC$ is the perpendicular bisector of $\overline{BD},$ $K$ is the orthocenter of $\triangle TBC\implies \overline{TK}\perp \overline{XY}.$

We may now conclude, since $E,T,K$ lie on the perpendicular bisector of $\overline{XY}\implies \overline{EK}\perp \overline{BC}.$

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william122

1576 posts

william122

#14 h Jul 11, 2018, 1:37 AM • 2 Y

Y by Adventure10, Mango247

Suppose $CD$ intersects $AB$ at $P$ and $AE$ at $Q$ , and let the foot of the perpendicular from $A$ be $R$ . Now, let $BC=CD=DE=c$ , $BP=DQ=a$ , $PC=EQ=b$ , $\angle PBC=\alpha$ , and $\angle PCB=\beta$ . We wish to show that $\frac{b+c-a}{2}\tan\frac{\alpha}{2}=\frac{c+a-b}{2}\tan\frac{\beta}{2}$ , or $\frac{b+c-a}{2}\frac{\sin\alpha}{1+\cos\alpha}=\frac{c+a-b}{2}\frac{\sin\beta}{1+\cos\beta}$ . Using law of sines as well as law of cosines, this is equivalent to $\frac{b+c-a}{2}\frac{2abc}{b^2+2bc+c^2-a^2}=\frac{a+c-b}{2}\frac{2abc}{a^2+c^2-2ac-b^2}\iff \frac{abc}{a+b+c}=\frac{abc}{a+b+c}$ , which is true.

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Vrangr

1600 posts

Vrangr

#16 h Jul 11, 2018, 6:46 PM • 2 Y

Y by Adventure10, Mango247

Here's another:

Let $I$ be the intersection point of the angle bisector $\angle ABC$ and $\angle BCD$ .
Note that $B$ and $D$ are symmetric in $\overline{CI}$ and, $A$ and $C$ are symmetric in $\overline{AI}$ .
Claim 1: $I$ lies on the angle bisector of $\angle BAE$ and $\angle CDE$ .
Proof.
$2\angle CDI = 2\angle IBC = \angle ABC = \angle CDE.$ Similarly for $\angle BAE$ .

Claim 2: $I$ lies on the angle bisector of $\angle AED$ .
Proof. By the property of angle bisectors,
$d(I, \overline{AE}) = d(I, \overline{AB}) = d(I, \overline{BC}) = d(I, \overline{CD}) = d(I, \overline{DE}).$ Claim 3: $\overline{EI}\perp \overline{BC}$
Proof.
Let $\overline{EI}\cap\overline{BC} = T$ .
$\angle BTE = 360^{\circ} - \angle TBA - \angle BAE - \angle BEI = 360^{\circ} - \frac12 \sum_{\text{cyc }\in\{A,B,C,D,E\}} \angle ABC = 90^{\circ}.$
Now we can proceed from here in 3 ways.
Proof 1: Since, $I$ is the concurrency point of the angle bisectors of the 5 angles and $IT\perp BC$ , pentagon $ABCDE$ is circumscribed about $\odot(I, \overline{IT})$ .
Thus, by Brianchon's Theorem on the degenerate hexagon $ATBCDE$ , $\overline{AC}$ , $\overline{BD}$ and $\overline{ET}$ are concurrent and we are done.

Let $P = \overline{AC}\cap \overline{BD}$ .
Proof 2: Since, $B$ and $D$ are reflections of each other in $\overline{CI}$ , $\overline{BD}\perp\overline{CI}$ . Similarly, $\overline{AC}\perp\overline{BI}$ .
Thus, $P$ is the orthocenter of $\triangle BCI$ . This implies that $\overline{IP}\perp\overline{BC}$ . However, we have $\overline{IE}\perp\overline{BC}$ .
Therefore, $\overline{EP}\perp\overline{BC}$ and we are done.

Proof 3: By Law of Sines,
$\begin{align*} \frac{AP}{\sin\angle PDE} = \frac{AP}{\sin\angle ABP} = \frac{BP}{\sin \angle BAP} = \frac{BP}{\sin \angle BCP} &= \frac{CP}{\sin\angle CBP} = \frac{CP}{\sin\angle CDP} = \frac{DP}{\sin\angle PCD} = \frac{DP}{\sin\angle PAE}\\ \iff d(P, \overline{AB}) = AP\sin\angle PAE &= DP\sin\angle PDE = d(P, \overline{ED}) \end{align*}$ Thus, $P$ lies on the angle bisector of $\angle AED$ , thus, $E, I, P$ are collinear.
Therefore, $\overline{EP}\perp\overline{BC}$ and we are done.

This post has been edited 1 time. Last edited by Vrangr, Jul 11, 2018, 6:49 PM

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Vfire

1354 posts

Vfire

#17 h Jul 11, 2018, 7:09 PM • 2 Y

Y by Adventure10, Mango247

$[asy] size(8cm); pair A = (1,1); pair B = (-1,1); pair C=B+2*dir(-140); pair D = C+2*dir(-20); pair X = A+dir(-120); pair Y = D+dir(20); pair E = extension(A,X,D,Y); draw(A--B--C--D--E--cycle, blue); pair P = extension(D,E,A,B); pair Q = extension(A,E,C,D); pair Z = extension(B,C,A,E); draw(B--Z^^A--Z, red); draw(A--P^^E--P^^E--Q^^D--Q, red); pair F = foot(E,B,C); pair H = A + dir(-150); pair O = extension(A,H,E,F); draw(circle(O, abs(O-F)), heavygreen); draw(B--D^^A--C, orange); draw(E--F, orange+dotted); dot("$A$", A, NE); dot("$B$", B, NW); dot("$C$", C, W); dot("$D$", D, SW); dot("$P$", P, NE); dot("$Q$", Q, S); dot("$Z$", Z, NE); dot("$F$", F, W); dot("$E$", E, SE); dot("$X$", extension(B,D,A,C), SE); [/asy]$
Let $P = AB \cap DE$ and $Q = AE \cap CD$ . Since $CB = CD$ and $\angle ABC = \angle CDE$ , we have $PB=PD$ so there exists a circle tangent to lines $\overleftrightarrow{AB}, \overleftrightarrow{BC}, \overleftrightarrow{CD}$ , and $\overleftrightarrow{DE}$ . Similarly, we have $AQ=CQ$ so there exists a circle tangent to $\overleftrightarrow{AB}, \overleftrightarrow{BC}, \overleftrightarrow{CD}$ , and $\overleftrightarrow{AE}$ . Hence, $ABCDE$ is a tangential pentagon with incircle $\omega$ .

Let $\angle BCD = \angle AEB = \alpha$ , $\angle ABC = \angle CDE = \beta$ , $Z = CB \cap AE$ , and $F$ be the foot from $E$ to $BC$ . Then
$\begin{align*} \angle FEA &= 90^\circ - \angle FZA\\ &= \angle ZBA + \angle ZAB - 90^\circ \\&= 270^\circ - \alpha - \beta \end{align*}$ but $\angle DEA = 540^\circ - (2\alpha + 2\beta)$ so $EF$ bisects $\angle DEA$ . Thus, the center of $\omega$ lies on $EF$ so $\omega$ is tangent to $BC$ at $F$ . Then by Brianchon's theorem on degenerate hexagon $ABFCDE$ , lines $AC$ , $BD$ , and $EF$ are concurrent. $\square$

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HeOmmBC

12 posts

HeOmmBC

#18 h Jul 14, 2018, 3:45 AM • 1 Y

Y by Adventure10

Let $AB \cap DC = P$ , $EA \cap CB = Q$ , $BA \cap DE = R$ , $AE \cap CD = S$ , $ED \cap BC = T$

Since $\angle SAC = \angle ACS$ it's easy to note that $SA = SC$ , thus, $BASC$ is a kite, and $SB \perp AC$ more important, $SB$ is the perpendicular bisector of $AC$ . Analogously, $RC$ is the perpendicular bisector of $BD$ . Let $Z = RC \cap SB$ .

Since $\angle EAB = \angle BCD$ , it's clear that $\angle PCQ = \angle PAQ$ then $PQAC$ is cyclic and $\angle AQC = \angle APC$
Since $\angle ABC = \angle CDE$ , it's clear that $\angle CBP = \angle TDC$ then $PBDT$ is cyclic and $\angle BTD = \angle BPD = APC$ ,
Then $\angle EQT = \angle AQC = \angle BTD = \angle QTE$ , therefore $EQ=ET$ , hence $E$ is part of the perpendicular bisector of $QT$

Since $BA=BC$ and $PQAC$ is cyclic, $\angle QAC = \angle QPA = \angle CAP$ therefore $PQ \parallel AC$ and since $ZB$ is perpendicular bisector of $AC$ , then is perpendicular of $PQ$ too.
Analogously, $ZC$ is perpendicular bisector of $PT$ .

Then $Z$ is circumcenter of $\triangle PQT$ , then $Z$ is on the perpendicular bisector of $QT$ , therefore $ZE$ is part of the perpendicular bisector of $QT$ , then $EZ \perp QT$ ,

Now, notice what happens in $\triangle ZBC$ , we have line $EZ \perp BC$ , we have $BZ \perp AC$ and $CZ \perp BD$ , then $BD$ , $AC$ and $EZ$ concur at the orthocentre of $\triangle BCZ$ , so we are done $Q.E.D.$

This post has been edited 2 times. Last edited by HeOmmBC, Jul 14, 2018, 3:48 AM
Reason: Correction

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shendrew7

793 posts

shendrew7

#93 h Apr 1, 2024, 6:00 AM

Y by

Let $\angle A = 180-2x$ , $\angle B = 180-2y$ , $AB=BC=CD=1$ , and suppose $F = EA \cap BC$ and $G = ED \cap BC$ . We find that $\triangle FAB \sim \triangle GCD$ and that $\triangle EFG$ is isosceles.

If we define $K = AC \cap BD$ and $L$ as the projection of $K$ onto $FG$ , it suffices to show $FL = GL$ , or
$FB+BL = \frac{\sin 2x}{\sin (2x+2y)} + \frac{\cos x \sin y}{\sin (x+y)} = \frac{\sin 2y}{\sin (2x+2y)} + \frac{\cos y \sin x}{\sin (x+y)} = GC+CL. \quad \blacksquare$

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alexgsi

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alexgsi

#94 h Apr 18, 2024, 10:54 PM

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AngeloChu

470 posts

AngeloChu

#95 h Apr 19, 2024, 8:56 PM

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let $AA'$ be a line parallel through $BD$ passing through $A$ , and let $DD'$ be a line parallel through $AC$ passing through $D$
pretty easily, $ABD'=CBD=AD'B=ADB$ , and $DCA'=BCA=DA'C=BAC$
we get that $ACA'$ and $DBD'$ are similar, and we can let $BD'$ intersect $AC$ at $F$ , and let $BD$ intersect $AC'$ at $G$
since $FBG=FCG$ , $BFGC$ is cyclic.
$BGA'=CGD=180-GCD-GDC$ , so $FGA'=180-2GCD-GDC$ and by previous similarities, $FGA'=BCA'=180-2GCD-GDC$ and $FG$ is parallel to $BC$ , but since $BFGC$ is cyclic $BF=GC$
this implies $180-2GCD-GDC=180-GCD-2GDC$ , and that $GCD=GDC$ , so $ABC=BCD=CDE=EAB$ , so then the conclusion is obvious since the perpendicular line from $E$ to $BC$ is a line of symmetry

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EpicBird08

1741 posts

EpicBird08

#96 h Apr 29, 2024, 3:02 PM

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Finally got around to solving this!

Let $I$ be the intersection of the internal angle bisectors of $\angle ABC$ and $\angle BCD,$ and let $X = AC \cap BD.$ Then the length conditions translate into $B$ and $D$ being reflections across $IC,$ so $BD \perp IC.$ Similarly, $AC \perp IB,$ so $X$ is the orthocenter of $\triangle BIC.$ Thus $IX \perp BC.$
Moreover, $\angle BCI = \angle BAI$ because of the reflections. Multiplying this by $2$ gives $2 \angle BAI = 2 \angle BCI = \angle BCD.$ But $\angle BCD = \angle EAB$ by the angle condition, so $\angle BAE = 2 \angle BAI.$ Thus $AI$ bisects $\angle BAE$ and similarly $DI$ bisects $\angle CDE.$ Therefore, there exists a circle centered at $I$ tangent to all sides of the pentagon $ABCDE.$ Thus if $T$ is the tangency point of this circle with $BC,$ a.k.a. the foot of the altitude from $I$ to $BC,$ Brianchon on $ABTCDE$ implies $E,X,T$ are collinear. However, since $X$ is the orthocenter of $\triangle BIC,$ we get $I,X,T$ are collinear, so $\overline{EX} = \overline{IT} \perp \overline{BC}$ as desired.

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Ritwin

155 posts

Ritwin

#97 h Aug 28, 2024, 6:42 AM

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Remark: The main idea of this solution is that the given conditions look a lot nicer once $P$ is defined (from which $H$ naturally follows), and then adding points $A'$ and $D'$ makes the problem completely determined by five congruent triangles. Here's the restatement:

Problem: Suppose five congruent triangles have the similarities $PD'A \overset{-}\sim PBA \overset{-}\sim PBC \overset{-}\sim PDC \overset{-}\sim PDA'.$ Then $AD' \cap A'D$ lies on the perpendicular from $P$ to $BC$ .

Solution

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AngeloChu

470 posts

AngeloChu

#98 h Aug 30, 2024, 7:34 PM

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let the perpendicular bisectors of $AC$ and $BD$ intersect at $Q$ and let $AC$ and $BD$ intersect at $P$
we easily prove that $APQY$ and $DPQC$ are cyclic
let $BQ$ intersect $AC$ at $R$ , let $CQ$ intersect $BD$ at $S$ , and let $PQ$ intersect $BC$ at $T$
by radical axes and stuff, we get that $BRPT$ and $CSRT$ are cyclic, and $PQ$ is perpendicular to $BC$
(not done, later gonna prove $E$ lies on $PQ$ but I'm lazy rn)

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ezpotd

1251 posts

ezpotd

#99 h Sep 22, 2024, 9:56 PM

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Let $EA, ED$ meet $BC$ at $M,N$ . Let $K$ be $AC \cap BD$ , let $F$ be the foot from $K$ to $BC$ , Clearly, $EMN$ is isosceles, so it suffices to prove $FM = FN$ . Take $BK = BC \frac{\sin x}{\sin x + y}, BF = BC \frac{\sin x \cos y}{\sin x + y}$ . Now $BM = AB \frac{\sin 2y}{\sin 2x + 2y} = BC \frac {\sin y \cos y}{\sin x + y \cos x + y}$ . Now $FM = BM + BF = \frac{BC \cos y}{\sin x + y} (\sin x +\frac{\sin y}{\cos x + y}) = \frac{BC \cos y}{\sin x + y} (\frac{\sin x \cos x \cos y - \sin x \sin x \sin y + \sin y}{\cos x + y}) =\frac{BC \cos y}{\sin x + y} (\frac{\sin x \cos x \cos y - (1 - \cos^2 x) \sin y + \sin y}{\cos x + y}) = \frac{BC \cos y}{\sin x + y} (\frac{\sin x \cos x \cos y + \cos x \cos x \sin y}{\cos x + y}) = \frac{BC \cos y}{\sin x + y} (\frac{\cos x \sin x + y}{\cos x + y}) = BC \frac{\cos x \cos y}{\cos x + y}$ , which is symmetric in $x,y$ , so we are done.

This post has been edited 2 times. Last edited by ezpotd, Sep 27, 2024, 8:58 PM

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Markas

105 posts

Markas

#100 h Sep 22, 2024, 10:03 PM

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Let $AC \cap BD = F$ . Let $EF \cap BC = G$ . We need to prove that $\angle EGC = 90^{\circ}$ . Let $\angle EAB = \angle BCD = \alpha$ and $\angle ABC = \angle EDC = \beta$ . Also $\angle AED = 540 - 2\alpha - 2\beta$ $\Rightarrow$ if EG is the angle bisector of $\angle AED$ it follows that $\angle GED = 270 - \alpha - \beta$ $\Rightarrow$ $\angle EGC = 360 - (270 - \alpha - \beta) - \alpha - \beta = 90^{\circ}$ $\Rightarrow$ we need to prove that EG is the angle bisector of $\angle AED$ and we would be ready. By law of sines on $\triangle FED$ we get that $\frac{FD}{\sin \angle FED} = \frac{FE}{\sin \angle FDE}$ $\Rightarrow$ $FE \cdot \sin \angle FED = FD \cdot \sin \angle FDE$ . By law of sines on $\triangle FAE$ we get that $\frac{FE}{\sin \angle FAE} = \frac{AF}{\sin \angle AEF}$ $\Rightarrow$ $FE \cdot \sin \angle AEF = AF \cdot \sin \angle FAE$ $\Rightarrow$ to show that $\angle AEF = \angle FED$ it is enough to show that $FD \cdot \sin \angle FDE = AF \cdot \sin \angle FAE$ . Now by law of sines starting from $\triangle FBA$ we get $\frac{FA}{\sin \angle FDE} = \frac{FA}{\sin \angle FBA} = \frac{AB}{\sin \angle AFB} = \frac{DC}{\sin \angle DFC} = \frac{FD}{\sin \angle FCD} = \frac{FD}{\sin\angle FAE}$ $\Rightarrow$ $\frac{FA}{\sin \angle FDE} = \frac{FD}{\sin\angle FAE}$ $\Rightarrow$ $FD \cdot \sin \angle FDE = FA \cdot \sin \angle FAE$ $\Rightarrow$ EF is the angle bisector of $\angle AED$ $\Rightarrow$ $\angle EGC = 90^{\circ}$ and we are ready.

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cursed_tangent1434

565 posts

cursed_tangent1434

#101 h Oct 8, 2024, 3:51 AM

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Let $P = \overline{AE} \cap \overline{CD}$ and $Q= \overline{AB}\cap \overline{DE}$ . Now, let $X = \overline{CQ} \cap \overline{BP}$ . The entirety of the solution is the following key claim.

Claim : Point $X$ is the incenter of tangential pentagon $ABCDE$ .

Proof : Since $X$ is the intersection of the diagonals of kites $PABC$ and $QBCD$ , it is clear that it is the incenter of both these kites. Thus, $X$ lies on the $\angle B-$ bisector and $\angle D-$ bisector of $ABCDE$ . Further, due to symmetry in kite $QBCD$ , $BX=DX$ . Thus in $\triangle BCX$ and $\triangle CDX$ we have $BX=XD$ , $BC = CD$ and $2\measuredangle XBC = \measuredangle ABC = \measuredangle CDE = 2\measuredangle CDX$ so $\measuredangle XBC = \measuredangle CDX$ . This implies that $\triangle XBC \cong \triangle XDC$ and thus $X$ lies on the $\angle C-$ bisector of $ABCDE$ as well. A similar argument shows that $X$ lies on the $\angle A-$ bisector of $ABCDE$ , which allows us to conclude that pentagon $ABCDE$ is tangential with $X$ being its incenter.

Now, this implies that the $\angle E-$ bisector must also pass thru $X$ . Let $M$ be the foot of the altitude from $E$ to $\overline{BC}$ . Note that since quadrilaterals $ABME$ and $CDEM$ have 3 pairs of equal angles, the other pair must also be equal, and thus $\measuredangle MEA = \measuredangle DEM$ . So, $M$ is the tangency point of the incircle of $ABCDE$ with side $BC$ . Now, by Brianchon's theorem on $ABMCDE$ , it follows that diagonals $\overline{AC}$ and $\overline{BD}$ are concurrent with the perpendicular line from $E$ to $\overline{BC}$ .

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Mathandski

734 posts

Mathandski

#102 h Oct 21, 2024, 3:56 PM

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Faster triglen sol exploiting symmetry.

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SimplisticFormulas

85 posts

SimplisticFormulas

#103 h Nov 23, 2024, 7:44 PM

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Let $AE \cap BC=X,AB \cap CD=Y,BC \cap DE=Z$
Let $\angle EAB=\angle BCD= \alpha, \angle ABC=\angle ECD=\beta$

Observe that
$\angle AXB=\angle BYC=\angle CZD=\alpha +\beta-180$ ALos noticing that $AB=BC=CD=a$ , we get the fact that
$AXYC$ and $BYZD$ are isosceles trapeziums (is it trapezia? idk)
which also implies that
$BX=BY=ZD=b$ and $AX=CY=CZ=c$ .
Also, $\angle FXZ=\angle FZX$ implies that $FX=FZ$ .
Let $\frac{a+b+c}{2}=s$ .
Let $Y$ -excircle and $Y$ -excentre of $\triangle YCB$ be $\omega$ and $O$ respectively.
Let $\omega$ touch $BC, AB, CD$ in $P,Q,R$ . Note that $P$ is midpoint of $XZ$ , hence $FP \perp XZ \implies F-O-P$ .
Note that $BP=s-b=BQ$ and $CP=s-c=CR$ , therefore $\omega$ is the $X$ -excircle of $\triangle AXB$ and the $Z$ -excircle of $\triangle ZCD$ . Hence, $\omega$ is inscribed in $ABCDE$ .

Let $AC \cap OP=M$ . let $I$ be the incentre of $\triangle YBC$ and $K$ be the foot of the altitude from $I$ on $BC$ .
Note that $\angle PCM= 90-\frac{\beta}{2}=\angle IBK$ , $\angle IKB=90=\angle MPC$ and $BK=CP$
$\implies \triangle IBK\cong \triangle MPC$ , so $IB=MC$ . Hence $BICM$ is a parallelogram and $\angle CBM=\angle IBC=90-\frac{\alpha}{2}=\angle DBC$ .
Hence, $AC\cap BD=M \in OP \in FP \perp BC$ , and we are done. $\blacksquare$

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eibc

598 posts

eibc

#104 h Dec 28, 2024, 10:42 PM

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Let $P = \overline{AE} \cap \overline{BC}$ , $Q = \overline{DE} \cap \overline{BC}$ , and $H$ be the foot of the perpendicular from $E$ to $\overline{BC}$ .

If we let $X_1 = \overline{AC} \cap \overline{EH}$ and $X_2 = \overline{BD} \cap \overline{EH}$ , we find that $HX_1 = \tfrac{CH}{\tan(B/2)}$ and $HX_2 = \tfrac{BH}{\tan(C/2)}$ . It then suffices to show that $HX_1 = HX_2$ , or $\tfrac{BH}{CH} = \tfrac{\tan(C/2)}{\tan(B/2)}$ .

WLOG assume $AB = BC = CD = 1$ . From the law of sines we have $PB = \tfrac{\sin C}{-\sin(B + C)}$ and $CQ = \tfrac{\sin B}{-\sin(B + C)}$ . Since $\angle P = \angle Q = B + C - 180^{\circ}$ , we have $PB + BH = CQ + CH$ , so
$BH - \frac{\sin C}{\sin(B + C)} = CH - \frac{\sin B}{\sin(B + C)}.$ Combining the above equation with $BH + CH = 1$ , we find that $BH = \tfrac{1}{2} \cdot \tfrac{\sin(B + C) + \sin C - \sin B}{\sin(B + C)}$ and $CH = \tfrac{1}{2} \cdot \tfrac{\sin(B + C) - \sin C + \sin B}{\sin(B + C)}$ . Thus,
$\frac{BH}{CH} = \frac{\sin(B + C) + \sin C - \sin B}{\sin(B + C) - \sin C + \sin B}.$ By using sum to product identities, this equals
$\begin{align*} \frac{\sin(B + C) + \sin C - \sin B}{\sin(B + C) - \sin C + \sin B} &= \frac{2 \sin(\tfrac{B + C}{2})\cos(\tfrac{B + C}{2}) + 2 \sin(\tfrac{C - B}{2})\cos(\tfrac{B + C}{2})}{2 \sin(\tfrac{B + C}{2})\cos(\tfrac{B + C}{2}) + 2 \sin(\tfrac{B - C}{2})\cos(\tfrac{B + C}{2})} \\ &= \frac{\sin(\tfrac{B + C}{2}) + \sin(\tfrac{C - B}{2})}{\sin(\tfrac{B + C}{2}) + \sin(\tfrac{B - C}{2})} \\ &= \frac{2\sin(\tfrac{C}{2})\cos(\tfrac{B}{2})}{2\sin(\tfrac{B}{2})\cos(\tfrac{C}{2})} \\ &= \frac{\tan(\tfrac{C}{2})}{\tan(\tfrac{B}{2})}, \end{align*}$ so we are done.

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smileapple

1010 posts

smileapple

#105 h Jan 2, 2025, 3:47 AM

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Set $X= AE\cap BC$ and $Y=DE\cap BC$ . Then $\triangle EXY$ is isosceles with $\triangle ABX\cong\triangle CDY$ . In particular, let us set $AX=a$ , $BX=b$ , $AB=c$ , $s=\frac{a+b+c}2$ , $\alpha=\frac{\angle XBA}2$ , and $\beta=\frac{\angle XAB}2$ .

Let $M$ be the midpoint of $XY$ and let $P=AC\cap BD$ . It suffices to show that $P$ lies on $EM$ . Considering $\triangle BCP$ , this reduces to showing that $\frac{\tan\angle ACB}{\tan\angle DBC}=\frac{BM}{CM}$ or equivalently that $\frac{\tan\alpha}{\tan\beta}=\frac{s-b}{s-a}$ . But in fact this identity always holds true for any triangle with angles $2\alpha$ , $2\beta$ , and $180^\circ-2(\alpha+\beta)$ and sides $a$ , $b$ , and $c$ , as one can verify computationally (use Law of Sines to obtain the ratios $a/c$ and $b/c$ to write $\frac{s-b}{s-a}$ in terms of $\alpha$ and $\beta$ , and then apply sum-to-product). $\blacksquare$

edit: just look at @above

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Bonime

28 posts

Bonime

#106 h Jan 21, 2025, 4:04 PM

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Pretty easy for a G1

Here´s my synthetic solution:
Initially, define the points $X=AB\cap CD$ , $Y=BC\cap AE$ and $Z=BC\cap DE$ , as well $T=AC\cap BD$ . If $I$ is the $X$ exincenter of the $\triangle XCB$ , we´re going to prove that:

a) $IT\bot BC$
b) $EI\bot BC$

And hence, by proving a) and b), we´ll get that $E$ and $T$ are in the perpendicular from $I$ to $BC$ , and then $ET\bot BC$ which´s actually what we´re trying to prove.

For a), see that $CI\bot BT$ and $BI\bot CT$ , so $T$ is the orthocenter of the $\triangle CIB$ , and $IT\bot BC$ . $\blacksquare$
For b), it´s easy to see that $\triangle BAI \equiv \triangle BCI \equiv DCI$ So $DI$ and $AI$ are, respectively the external bissectors of $\angle D$ and $\angle A$ , therefore $\triangle XCB$ , $\triangle ZDC$ and $\triangle YBA$ share the exincircle, which is tangent to all the sides of our pentagon. And then, $EI$ bissects $\angle E$ , so the rest is just angle chasing. $\blacksquare$

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zaidova

84 posts

zaidova

#107 h Mar 31, 2025, 5:33 PM

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Not so hard , i solved the problem only with angle chasings and some constructions like drawing $AE \cap BC=M$ , $ED \cap BC=N$ .Then $EM=EN$ . Angle chasing gives us the problem is equivalent to prove for $AC \cap BD=K$ , $EK$ is angle bisector of $\angle AED$ . And from isosceles triangle we found , let $EK \cap BC=T$ , we have to find $MT=NT$ . Triangles $MAB$ , $NCD$ are congruent. Then trigo bashing .

This post has been edited 1 time. Last edited by zaidova, Mar 31, 2025, 5:42 PM

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