Parallel lines lead to similar triangles

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Source: USA TST for EGMO 2020, Problem 2, by Andrew Gu

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223 posts

#1 h Dec 16, 2019, 5:24 PM • 5 Y

Y by NJOY, kat1012, Adventure10, Mango247, Rounak_iitr

Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$ , $BC$ , or $CA$ . Distinct points $D$ , $E$ , and $F$ lie on lines $BC$ , $AC$ , and $AB$ , respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$ . Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$ .

Andrew Gu

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#2 h Dec 16, 2019, 5:25 PM • 11 Y

Y by NJOY, Feridimo, kat1012, Kagebaka, AllanTian, centslordm, ike.chen, Newmaths, aidan0626, Adventure10, Funcshun840

My problem. Not terribly original, though: you may notice that it is just a generalization of ELMO SL 2013 G3, USA TST 2008/7, and USA December TST 2012/1.

Solution 1

Solution 2

Solution 3

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#3 h Dec 16, 2019, 5:59 PM • 6 Y

Y by math_pi_rate, mira74, Pluto1708, Adventure10, Mango247, NCbutAN

a1267ab wrote:

Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$ , $BC$ , or $CA$ . Distinct points $D$ , $E$ , and $F$ lie on lines $BC$ , $AC$ , and $AB$ , respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$ . Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$ .

Andrew Gu

Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point. So, we show for $D=B$ and $D=C$ that $Q$ lies on this circle. Both cases are analogous, so effectively, we just check $D=B$ . Then $E=B$ and say $F=F'$ with $\overline{BF'} \parallel \overline{CP}$ . Then $\measuredangle BQA=\measuredangle PCA=\measuredangle BF'A$ so $Q$ lies on $\odot(ABF')$ . The result follows. $\blacksquare$

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#4 h Dec 16, 2019, 6:37 PM • 3 Y

Y by v4913, Adventure10, Mango247

anantmudgal09 wrote:

Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point.

You can also use barycentric coordinates to show this, which is what I did.

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#5 h Dec 16, 2019, 7:37 PM • 1 Y

Y by Adventure10

length bash

Attachments:

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#6 h Dec 16, 2019, 8:00 PM • 2 Y

Y by Adventure10, Mango247

a1267ab wrote:

My problem. Not terribly original, though: you may notice that it is just a generalization of ELMO SL 2013 G3, USA TST 2008/7, and USA December TST 2012/1.

Solution 1

Solution 2

Solution 3

In Solution 1. I think the correct is $\angle QFA=\angle QEC$ following the spiral similarity.

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#7 h Dec 18, 2019, 10:21 AM • 3 Y

Y by Kagebaka, Adventure10, Mango247

Perform $\sqrt{bc}$ inversion in $ABC$ . We have that $Q$ is mapped to $P$ . Hence if $E\rightarrow E', F\rightarrow F'$ then we need to show $E',P,F'$ collinear. Observe that $D\rightarrow E, D\rightarrow F$ are perspectivities while $E\rightarrow E', F\rightarrow F'$ are projectivities. Hence $E'\rightarrow F'$ is a projectivity. When $D$ is to infinity along $BC$ , $E,F$ are also at infinity along respective lines and hence $E',F'$ are at $A$ . Hence the projectivity fixes a point, hence it is a perspectivity. When $D=B$ line $E'F'$ is line $CP$ and when $D=C$ , line $E'F'$ is line $BP$ . Hence $P$ is the centre of perspectivity. We are done.

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#9 h Dec 18, 2019, 10:35 AM • 14 Y

Y by kapilpavase, AlastorMoody, Pluto1708, amar_04, SomeUser221104, riadok, Soundricio, euclides05, IAmTheHazard, Adventure10, Mango247, Aryan-23, math_comb01, Ritwin

Here is an amusing solution using Generalized Reim's theorem.

Lemma (Generalized Reim's): Let $\mathcal{C}$ be a conic and $A,B$ be points on it. Let $\omega_1$ and $\omega_2$ be circles through $A,B$ which intersect $\mathcal{C}$ again at $\{C,D\}$ and $\{E,F\}$ respectively. Then $CD\parallel EF$ .

Proof: Let $I,J$ be circle points. Consider a projective transformation which takes $A,B$ to $I,J$ respectively. Denote the image with primes. Then $\mathcal{C}', \omega_1', \omega_2'$ are all circles. Moreover, since $\omega_1, \omega_2$ intersect at $A,B,I,J$ , we get the other two intersection of $\omega_1', \omega_2'$ are $I',J'$ . Thus by radical axis theorem, $C'D', E'F', I'J'$ are concurrent. Taking the projective transformation back, we deduce the lemma. $\blacksquare$

Back to the main problem. Let $Q$ be the point such that $\triangle BAQ\stackrel{+}{\sim}\triangle PAC$ . Clearly $X=CQ\cap DF$ lie on $\odot(AQF)$ and $Y=BQ\cap DE$ lie on $\odot(AQE)$ . Moreover, by converse of Pascal's theorem on $AFXQYE$ , these six points lie on a conic.

Assume for the contradiction that this conic is not a circle. Then by Generalized Reim's theorem, $FX\parallel EY$ which is a contradiction as two lines intersect at $D$ !

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#10 h Dec 18, 2019, 1:13 PM • 2 Y

Y by anantmudgal09, Adventure10

10 min solve using moving points (Although I didn't realize the much easier solution given by Anant above

). Anyway, here goes nothing: Fix $A,B,C,D,F$ and animate $P$ on the line $\ell$ through $B$ parallel to $DF$ . Then $P \mapsto \infty_{CP} \mapsto D\infty_{CP} \mapsto E$ is a projective map. Also, it's easy to see that $P \mapsto Q$ is also a projective map (Try thinking about spiral similarities). In fact, $\angle AQC$ is fixed (since $\angle ABP$ is also fixed), amd so $Q$ moves on a fixed circle through $A$ and $C$ . Thus, it suffices to prove $A,E,F,Q$ concyclic for 3 positions of $P$ (Note that 3 is sufficient, say by inversion at $A$ ). Taking $P=B$ gives $E=Q=C$ , taking $P=C\infty_{AD} \cap \ell$ gives $E=A$ , and finally $P=\infty_{\ell}$ makes $Q=A$ . Thus, all three cases degenerate to given result. Hence, done. $\blacksquare$

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#11 h Dec 18, 2019, 5:17 PM • 1 Y

Y by Adventure10

anantmudgal09 wrote:

a1267ab wrote:

Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$ , $BC$ , or $CA$ . Distinct points $D$ , $E$ , and $F$ lie on lines $BC$ , $AC$ , and $AB$ , respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$ . Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$ .

Andrew Gu

Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point. So, we show for $D=B$ and $D=C$ that $Q$ lies on this circle. Both cases are analogous, so effectively, we just check $D=B$ . Then $E=B$ and say $F=F'$ with $\overline{BF'} \parallel \overline{CP}$ . Then $\measuredangle BQA=\measuredangle PCA=\measuredangle BF'A$ so $Q$ lies on $\odot(ABF')$ . The result follows. $\blacksquare$

How do you localized the point $Q$ in this step:
$\measuredangle BQA=\measuredangle PCA=\measuredangle BF'A$ .

Also you have a typo using $F$ like $E$ .

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#12 h Dec 27, 2019, 6:12 PM • 2 Y

Y by SomeUser221104, Adventure10

MarkBcc168 wrote:

Lemma (Generalized Reim's): Let $\mathcal{C}$ be a conic and $A,B$ be points on it. Let $\omega_1$ and $\omega_2$ be circles through $A,B$ which intersect $\mathcal{C}$ again at $\{C,D\}$ and $\{E,F\}$ respectively. Then $CD\parallel EF$ .

Does the converse hold? I mean the following:

Quote:

Let $\mathcal{C}$ be a conic and $A$ $\in$ $\mathcal{C}$ . Let $C,D,E,F$ $\in$ $\mathcal{C}$ , such that $CD||EF$ . Then, $\odot (ACD)$ $\cap$ $\odot (AEF)$ $\in$ $\mathcal{C}$

If this holds, then we can use it here

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#13 h Dec 27, 2019, 8:26 PM • 3 Y

Y by AlastorMoody, lachimolala, Adventure10

A possible proof

Quote:

Let $\mathcal{C}$ be a conic and $A$ $\in$ $\mathcal{C}$ . Let $C,D,E,F$ $\in$ $\mathcal{C}$ , such that $CD||EF$ . Then, $\odot (ACD)$ $\cap$ $\odot (AEF)$ $\in$ $\mathcal{C}$

Let $T=\odot{AEF}\cap \odot{ACD}$ .It suffices to show $(TC,TD,TE,TF)=(AC,AD,AE,AF)$ to show that $T\in \mathcal{C}$ .Now $\begin{align*}M=AD\cap\odot{AEF} \\ N=AC\cap \odot{AEF} \\ K=TC\cap \odot{AEF} \\ L=TD\cap \odot{AEF} \end{align*}$ .Now by Reim's Theorm we have $NL\parallel CD\parallel EF\parallel KM$ .Thus we have $NMEF$ and $LKFE$ are just reflections of each other in the perpendicular bisector of $EF$ .In particular we have $(N,M,E,F)=(K,L,E,F)$ .Thus we have $(AC,AD,AE,AF)\overset{A}{=}(N,M,E,F)=(K,L,E,F)\overset{T}{=}(TC,TD,TE,TF)$ so we conlcude. $T\in \mathcal{C} \blacksquare$

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#15 h Jan 8, 2020, 4:12 AM • 1 Y

Y by Adventure10

This geo is giving away a big hint. $D$ could be any point on $BC$ , hence we know that $\odot(AEF)$ has to pass through a fixed point. The problem is much easier when $D$ is either $B$ or $C$ , so we wish to prove that $\odot(AEF)$ passes through a fixed point. Let $O$ be the circumcenter of $\triangle AEF$ . $\triangle OEF$ is always similar to a fixed triangle. It's isosceles, and $\angle EOF=2\angle A$ . Notice that, as $D$ moves on $BC$ with velocity $v$ , $E$ and $F$ both move with constant velocity as well, and using a little complex numbers it's easy to see that $O$ also moves with constant velocity $\Longrightarrow$ $\odot(OEF)$ s are coaxial. We're done.

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#18 h Feb 22, 2020, 3:02 PM • 5 Y

Y by mueller.25, GeoMetrix, Hexagrammum16, Bumblebee60, Adventure10

Different sketch from all others, this solution mainly focusses on Isogonal Lines.

USATST for EGMO 2020 P2 wrote:

Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$ , $BC$ , or $CA$ . Distinct points $D$ , $E$ , and $F$ lie on lines $BC$ , $AC$ , and $AB$ , respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$ . Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$ .

Andrew Gu

Let $DE\cap \odot(ABC)=X$ and $DF\cap\odot(ABC)=Y$ . We claim that the point $Q=XB\cap YC\in\odot(ABC)$ . $XB\cap YC\in\odot(ABC)$ by Pascal's Theorem on $AFYQXE$ . Now $\angle ACP=\angle AEX=\angle AQB$ . So we just need to prove that $\{AP,AQ\}$ are isogonal WRT $\triangle AEF$ . Let $XQ\cap PC=T$ and $PB\cap CQ=J$ .

Claim:- $APBT$ and $APCJ$ are cyclic quadrilaterals.
Let $TC\cap YK=K$ . So, $\angle ACT=\angle AEX=\angle AQT\implies ACQT$ is a cyclic quad. So, $\angle ATC=\angle AQY=\angle AFY$ and as $PB\cap FY$ . So, by Reim's Theorem we get that $APBT$ is a cyclic quadrilateral. Similarly we get that $APCJ$ is a cyclic quadrilateral.

So, $\angle TAB=\angle TPB=\angle JPC=\angle JAC\implies \{AT,AJ\}$ are isogonal WRT $\triangle AFE$ so as $\{AB,AC\}$ . So, by Isogonal Line Lemma we get that $\{AP,AQ\}$ are also Isogonal WRT $\triangle AEF$ . Hence, $\triangle BAQ\sim\triangle CAP$ . So, $Q=CY\cap BC$ is the desired point. $\blacksquare$

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#19 h May 9, 2020, 5:04 AM

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alifenix- wrote:

anantmudgal09 wrote:

Fix $ABCP$ and animate $D$ on $BC$ with constant velocity. Since $D \mapsto E$ and $D \mapsto F$ are linear, $\odot(AEF)$ passes through a fixed point.

You can also use barycentric coordinates to show this, which is what I did.

Claim: Let the reflection of $AP$ over the angle bisector of $\angle BAC$ meet $(AEF)$ at a point $Q'.$ If we fix $P$ and vary $D$ along $BC,$ then $Q'$ is fixed too.

Proof: We use barycentric coordinates with reference triangle $\triangle ABC.$ Let $P=(t_1, t_2, t_3)$ and $D=(0,d,1-d).$ Note that the point at infinity along $PC$ must have coordinates $(t_1:t_2:-t_2-t_1),$ so $E=(x_1,0,1-x_1)$ must satisfy
$\begin{align*} 0&=\begin{vmatrix} 0 & d & 1-d \\ x_1 & 0 & 1-x_1 \\ t_1 & t_2 & -t_2 - t_1 \\ \end{vmatrix} \\ 0&=d((1-x_1)t_1+x_1(t_2+t_1))+(1-d)(x_1t_2) \\ 0&=dt_1+x_1t_2 \\ x_1&=-\frac{dt_1}{t_2}, \\ \end{align*}$ so $E=(-dt_1:0:t_2+dt_1).$ Similarly, $F=(x_2,1-x_2,0)$ satisfies
$\begin{align*} 0&=\begin{vmatrix} 0 & d & 1-d \\ x_2 & 1-x_2 & 0 \\ t_1 & -t_1 - t_3 & t_3 \\ \end{vmatrix} \\ 0&=dx_2t_3+(1-d)(x_2(t_1+t_3)+t_1(1-x_2))\\ 0&=x_2t_3+(1-d)t_1 \\ x_2&=-\frac{(1-d)t_1}{t_3}, \\ \end{align*}$ so $F=(-(1-d)t_1:t_3+(1-d)t_1:0).$ $(AEF)$ is parametrized by $a^2yz+b^2zx+c^2xy=(x+y+z)(ux+vy+wz);$ plugging in $A$ gives us $u=0,$ and plugging in $E$ and $F,$ we get
$\begin{align*} -b^2dt_1(t_2+dt_1)&=t_2w(t_2+dt_1) \\ w&=-\frac{b^2dt_1}{t_2} \\ \end{align*}$ and
$\begin{align*} -c^2(1-d)t_1(t_3+(1-d)t_1)&=t_3v(t_3+(1-d)t_1) \\ v&=-\frac{c^2(1-d)t_1}{t_3}, \\ \end{align*}$ respectively. Finally, plugging in $Q'=(t:\tfrac{b^2}{t_2}:\tfrac{c^2}{t_3})$ gives us
$b^2c^2\left(\frac{a^2}{t_2t_3}+\frac{t}{t_3}+\frac{t}{t_2}\right)=-\frac{b^2c^2}{t_2t_3}\left(t+\frac{b^2}{t_2}+\frac{c^2}{t_3}\right)((1-d)t_1+dt_1),$ which in particular implies that $t$ is fixed because the terms with $d$ cancel out. $\Box$

Back to the main problem: Let $BP\cap CQ' = G$ and $BQ'\cap CP = H.$ We could probably do this next step with some easy angle chasing, but instead we'll use DDIT because it's a crime not to use it when you can. By Dual of Desargues' Involution about $A$ with respect to $BQ'CP,$ we find that $\{AB,AC\},$ $\{AP,AQ'\},$ and $\{AG,AH\}$ are swapped under some involution about the pencil of lines through $A,$ which clearly must be reflection across the angle bisector of $\angle BAC.$ This implies that $\measuredangle GAB=\measuredangle GQ'B=\measuredangle CQ'H=\measuredangle CAH,$ so $A$ is the Miquel Point of $BQ'CP.$ Thus, $\measuredangle Q'GA=\measuredangle HBA=\measuredangle HPA,$ whence $\triangle BAQ'\stackrel{+}{\sim}\triangle PAC$ by AA, so $Q'$ is the point we want. $\blacksquare$

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NMN12

#20 h May 25, 2020, 8:55 AM

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If we construct the point $Q$ one can notice that $\angle AQB = \angle ACP$ by definition and $\angle AQC = \angle ABP$ by $\triangle ABP\sim \triangle AQC$ which follows from the spiral similarity with center $A$ . Denote these two angle relations by $(1)$ .

Let $\Gamma = (AEF)$ , $R = \Gamma \cap ED$ , $S = \Gamma \cap FD$ , $Q' = RB \cap SC$ .
We have $B = RQ' \cap FA$ , $D = RE\cap FS$ , $C = SQ'\cap AE$ are collinear.
From Pascal's theorem $\implies$ $A,R,F,Q',E,S$ lie on a conic but 5 of them lie on $\Gamma \implies Q'\in \Gamma$ .

Then $\angle AQ'B = \angle AQ'R = \angle AER = \angle AED = \angle ACP$ , $\angle AQ'C = \angle AQ'S = \angle AFS = \angle AFD = \angle ABP$ which means that $Q'$ satisfies $(1)$ . But there is only one such point (It is intersection of two circles through $AB$ and $AC$ and it is $\neq A$ ) $\implies Q\equiv Q' \implies Q\in \Gamma$

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#21 h Sep 19, 2020, 4:13 PM

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Basically the first solution in post #2.

Assume all angles are directed.

Let $G$ be the point on line $\overline{AC}$ such that $\overline{BG}\parallel\overline{CP},$ let $H$ be the point on line $\overline{AB}$ such that $\overline{CH}\parallel\overline{BP},$ and redefine $Q$ as the point such that $\triangle ABQ\sim\triangle APC.$

$\textbf{Claim: }$ $(ABG)$ and $(ACH)$ intersect at $Q.$

$\emph{Proof: }$ Let $X$ be the second intersection of $(ABG)$ and $(ACH).$ Note that $\angle BXA=\angle BGA=\angle BGC=\angle PCA.$ Similarly, $\angle CXA=\angle CHA=\angle CHB=\angle PBA.$ Therefore, $A$ is the center of the spiral similarity sending $\overline{XB}$ to $\overline{CP},$ so $X=Q,$ as desired. $\blacksquare$

Therefore, $Q$ is the center of the spiral similarity sending $\overline{GB}$ to $\overline{CH}.$ Hence, to show that $Q$ lies on $(AEF),$ it suffices to show that $\frac{BF}{FH}=\frac{GA}{AC}.$

Since both $\overline{DF}$ and $\overline{CH}$ are parallel to $\overline{BP},$ we have $\overline{DF}\parallel\overline{CH},$ so $\frac{BF}{FH}=\frac{BD}{DC}.$ Similarly, we can show that $\frac{GA}{AC}=\frac{BD}{DC},$ so we are done.

This post has been edited 1 time. Last edited by amuthup, Sep 19, 2020, 4:14 PM

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GeronimoStilton

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GeronimoStilton

#22 h Oct 18, 2020, 8:44 PM

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Amazing problem.

Let $M$ denote the second intersection of $PB$ and $(PAC)$ , define $N$ similarly. Let $BN\cap CM=Q$ . Note that $A$ is the Miquel point of $PBQCNM$ , so there is a spiral similarity about $A$ taking $BQ$ to $PC$ . It remains to demonstrate that $Q$ lies on $(AEF)$ .

Take $D=C$ , then $C=E$ and $\measuredangle DQA=\measuredangle MQA=\measuredangle MBA=\measuredangle DFA$ , so $Q$ lies on $(AEF)$ . Similarly, $Q$ lies on $(AEF)$ when $D=B$ . Let the value of $F$ in the case $D=C$ be $F^*$ and the value of $E$ in the case $D=B$ be $E^*$ . Now, consider some intermediate value of $D$ , and note that by the Mean Geometry Theorem (as $E$ and $F$ move linearly with respect to $D$ ), $\triangle QEF\sim\triangle QCF^*\sim\triangle QE^*B$ , hence $Q$ is the center of spiral similarity between $EF$ and $CF^*$ , and so lies on $(AEF)$ , as desired.

This post has been edited 1 time. Last edited by GeronimoStilton, Oct 18, 2020, 8:44 PM

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#23 h Nov 22, 2020, 5:11 PM

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Let the lines parallel to $BP$ and $CP$ through $C$ and $B$ hit at $C'$ and $B'$ . Let $Q$ be the intersections of the circumcircles of $BAB'$ and $CAC'$ . Note that since $AQ$ is a radical axis of the two such circles, then $\triangle QB'C \sim \triangle QBC'$ . Now, since $\frac{C'F}{FB} = \frac{CD}{DB} = \frac{CE}{EB}$ , then $\triangle QB'E \sim \triangle QBF$ and $\triangle QCE \sim \triangle QC'F$ and $\angle EQF = \angle CQC' = 180^{\circ} - \angle A$ so $AEQF$ is cyclic. Letting $B''$ and $C''$ be where $CP$ hits $BQ$ and where $BP$ hits $CQ$ respectively, we see $\angle QB''C = \angle QBB' = \angle QAB' = \angle QAC$ so $QB''AC$ is cyclic and so is $QC''AB$ . Now, $\angle B''AB = \angle B''AC' = \angle B''CC' = \angle B''PB$ so $B''BPA$ is cyclic. Finally, $\angle BAP = \angle BB''P = \angle BB''C = \angle QB''C = \angle QAC$ which means since $\angle ABP = \angle AC'C = \angle AQC$ , then $\triangle BAP \sim \triangle QAC$ and due to similarity, $\triangle BAQ = \triangle QAC$ and we are done.

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MP8148

#24 h Jan 17, 2021, 9:04 AM • 2 Y

Y by Mango247, Mango247

$[asy] size(6cm); defaultpen(fontsize(10pt)+linewidth(0.4)); dotfactor *= 1.5; pair A = dir(110), B = dir(210), C = dir(330), D = B+dir(B--C)*abs(B-C)*0.4, P = B+dir(B--C)*abs(B-C)*0.55+dir(90)*0.4, P1 = A+dir(A--P)*abs(A-B)*abs(A-C)/abs(A-P), Q = 2*foot(P1,A,dir(270))-P1, E = extension(D,D+dir(C--P),A,C), F = extension(D,D+dir(B--P),A,B), X = extension(B,Q,D,E); draw(D--F--B--C--E--X--B--P, heavyblue); draw(A--B--Q--A--P--C--A, heavyred); draw(circumcircle(A,E,F), heavygreen+dashed); dot("$A$", A, dir(120)); dot("$B$", B, dir(180)); dot("$C$", C, dir(30)); dot("$D$", D, dir(270)); dot("$E$", E, dir(315)); dot("$F$", F, dir(200)); dot("$X$", X, dir(150)); dot("$P$", P, dir(270)); dot("$Q$", Q, dir(225)); [/asy]$
Let $Q$ be the point such that $\triangle ABQ \sim \triangle APC$ and $X = \overline{BQ} \cap \overline{DE}$ . Since $\measuredangle XQA = \measuredangle BQA = \measuredangle PCA = \measuredangle XEA,$ $AXQE$ is cyclic, so it suffices to show that $AXFQ$ is cyclic, which can be done by length chasing: $\begin{align*} \frac{BF}{BX} &= \frac{\sin \angle BDF}{\sin \angle BDX} \cdot \frac{\sin \angle BXD}{\sin \angle BFD} \quad (\text{law of sines}) \\ &= \frac{\sin \angle PBC}{\sin \angle PCB} \cdot \frac{\sin \angle PAB}{\sin \angle PBA} \quad (\text{angle chase}) \\ &= \frac{PC}{PA} \quad (\text{trig ceva}) \\ &= \frac{BA}{BQ} \quad (\text{similarity})\end{align*}$ $\blacksquare$

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#25 h Feb 9, 2021, 11:37 PM • 1 Y

Y by Pranav1056

bary bash

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MathForesterCycle1

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MathForesterCycle1

#26 h May 27, 2021, 8:39 PM

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dame dame

This post has been edited 2 times. Last edited by MathForesterCycle1, Oct 17, 2021, 5:16 PM

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#29 h Sep 15, 2021, 5:49 PM

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Let $Q$ be the point such that $\triangle BAQ\sim\triangle PAC$ . This also means that $\triangle QAC\sim\triangle BAP$ . Now, let $G=DE\cap BQ$ , then we have $\measuredangle AEG=\measuredangle AED=\measuredangle ACP=\measuredangle AQB=\measuredangle AQG$ , which means that $AEQG$ is cyclic. Redefine $F,H$ so that $F=AB\cap (AEQG)$ and $H=QC\cap (AEQG)$ . By Pascal's theorem on hexagon $FAEGQH$ , we get that $D$ lies on $FH$ . Therefore, $\measuredangle PBA=\measuredangle CQA=\measuredangle HQA=\measuredangle HFA=\measuredangle DFA$ , thus $DF\parallel PB$ . We are done.

$[asy]import olympiad; size(10cm);defaultpen(fontsize(10pt)); pair O,A,B,C,P,Q,D,E,F,G,H; O=(0,0);A=dir(115);B=dir(195);C=dir(345);P=dir(255)*0.74;Q=-(C-P)/(P-A)*(A-B)+B;D=0.6B+0.4C;F=extension(A,B,D,B+D-P);E=extension(A,C,D,C+D-P);G=extension(E,D,B,Q);H=extension(F,D,Q,C); draw(A--B--C--cycle,red);draw(circumcircle(A,E,F),royalblue);draw(B--Q--C,orange);draw(F--H,orange);draw(G--E,orange);draw(B--P--C,red); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$F$",F,dir(F)); dot("$E$",E,dir(E)); dot("$D$",D,dir(D)); dot("$G$",G,dir(G)); dot("$H$",H,dir(H)); [/asy]$

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8868 posts

#30 h Sep 20, 2021, 7:26 PM

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The key (and difficulty) of this seems to be messing around with the symmetry and getting it to work.

Denote $X \neq D = (DEF) \cap \overline{BC}$ . We claim that the Miquel Point of $EFX$ with respect to $ABC$ is our desired point $Q$ . Observe that
$\begin{align*} \measuredangle BQA &= \measuredangle BQF + \measuredangle FQA \\ &= \measuredangle BXF + \measuredangle FEA \\ &= \measuredangle DEF+ \measuredangle FEA \\ &= \measuredangle DEA \\ &= \measuredangle PCA, \end{align*}$ and for similar reasons $\measuredangle CQA = \measuredangle PBA$ . However, the point $Q'$ such that $\triangle BAQ' \sim \triangle PAC$ also satisfies $\triangle CAQ \sim \triangle PAB$ by spiral similarity lemma, and thus fulfills the above angle conditions. In fact, $Q \neq A = (AQ'C) \cap (AQ'B)$ by our angle conditions, so $Q = Q'$ must hold. The conclusion follows naturally because $Q$ lies on $(AEF)$ by definition.

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#31 h Jul 16, 2023, 8:04 PM • 1 Y

Y by centslordm

bary/mmp solution by someone who does not actually know bary nor mmp

Fix $P$ and note that the choice of $Q$ is uniquely determined. Now vary $D:=(0,t,1-t)$ linearly along $\overline{BC}$ , so the (normalized) components of the coordinates of $E$ and $F$ vary linearly as well because $\overline{DE},\overline{DF}$ are parallel to fixed lines. Now consider the "normalized" formula for the circle $(AEF)$ , which is of the form $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)=0$ . Since $A$ always lies on this circle, we have $u=0$ , and then by plugging in $E$ and $F$ we find that both $v$ and $w$ vary linearly in terms of $t$ .

We wish to prove that $Q$ always lies on this circle. Since it is fixed, this is equivalent to some linear function in $t$ being identically zero, so we only have to check it at two points. At $D=B$ , we have $E=B$ , and because $\measuredangle AQB=\measuredangle ACP=\measuredangle AFB$ , $ABQF=AEFQ$ is cyclic. Similarly, when $D=C$ , $ACQE=AEFQ$ is cyclic, so we are done. $\blacksquare$

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#32 h Oct 7, 2023, 4:27 PM

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Ok @v_Enhance forgive me for what i'm about to show u.
Let $Q$ the inverse of $P$ down $\sqrt{bc}$ inversion, now let $CQ \cap FD=K$ and $BQ \cap DE=L$ . (Note that $\triangle APB \sim \triangle ACQ$ and $\triangle APC \sim \triangle ABQ$ )
Claim 1: $ALQE, AFQK$ are cyclic.
Proof: By simple angle u just need:
$\angle AFD=\angle ABP=\angle AQK \; \text{and} \; \angle AED=\angle ACP=\angle AQL$ Claim 2: $ALFQEK$ lies in a conic $\mathcal C$
Proof: Follows from the converse of Pascal's theorem.
Finishing: Assume FTSOC that $\mathcal C$ isn't a circle, then by Degenerate 3 conics theorem over $\mathcal C, (ALQE), (AFQK)$ we get that $LE \parallel FK$ , contradiction!. Hence $\mathcal C$ is a circle, so $Q$ lies in $(AEF)$ as desired, thus we are done

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This post has been edited 3 times. Last edited by MathLuis, Oct 7, 2023, 4:30 PM

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#33 h Dec 27, 2023, 7:54 PM

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Construct $X$ on $AB$ with $BX \parallel DE$ , and $Y$ analogously.

Claim 1: $(ABC)$ , $(AEF)$ , and $(AXY)$ are coaxial.

Simply note there exists a spiral similarity sending $BX \mapsto FE \mapsto YC$ , as
$\frac{BF}{FY} = \frac{BD}{DC} = \frac{XE}{EC}. \quad {\color{blue} \Box}$
Claim 2: The second concurrency point is the desired point $Q$ .

Let $Q'$ be the desired point such that $\triangle BAQ' \sim \triangle PAC$ . Then
$\measuredangle BQA = \measuredangle PCA = \measuredangle BXA, \quad \measuredangle AQC = \measuredangle ABP = \measuredangle AYC,$
so $Q' \ne A$ lies on $(ABX)$ and $(AYC)$ , and thus $Q' = Q$ . $\blacksquare$

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AN1729

#34 h Oct 23, 2024, 10:15 AM

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Solved with Janhavi Kulkarni

Let $F' \in \overline{AB}$ be such that $CF' \parallel \overline{BP}$
Define $E'$ analogously.
Thus, $\frac{BF}{FF'} = \frac{BD}{DC} = \frac{E'E}{EC}$
Consider $Q$ such that $A$ is spiral center taking $\overline{BQ}$ to $\overline{PC}$
$\implies \triangle AQB \sim \triangle APC$
$\implies \angle AQB = \angle ACP = \angle AE'B \implies (AQE'B)$ and similarly $(AQF'C)$
Now, consider spiral similarity centered at $Q$ taking $\overline{BF'}$ to $\overline{E'C}$
Note that the same spiral similarity takes $F$ to $E$
$\implies \measuredangle QFA = \measuredangle QFF' = \measuredangle QEC = \measuredangle QEA \implies (AQEF)$

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#35 h Nov 13, 2024, 3:16 PM

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So basically, we want $Q$ , the image of $P$ under $\sqrt{bc}$ inversion, to be on $(AEF)$ .

First, check this for $D=B$ . Then call this $E$ $E_1$ . Then $DE_1\parallel PC$ , so the image of $(ADE_1)$ will be the line through $C$ and $CP\cap AB$ , that is, line $CP$ . Similarly, this also works for $D=C$ .

Now note that $E$ and $F$ are moving in the way of the ratio thing of the Miquel point config. So basically, indeed $Q$ will be the Miquel point of self-intersecting quadrilateral $BF_1CE_1$ , and the ratio thing gives that it lies on all such circles $(AEF)$ . $\blacksquare$

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#36 h Dec 13, 2024, 6:12 PM

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Subjective Rating (MOHs) $\text{[math]}$

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#37 h May 11, 2025, 11:14 PM

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what the 5 minute solve
(Typed on phone so might have error)

Let $K$ denote the intersection of $CE$ with $(AEF)$ . Define $L$ similarly.
Define $Q$ the intersection of $BK$ with $(AEF)$ . We prove $Q$ is the desired point.
Pascals on $KQLFAE$ implies $QCL$ are collinear.
Define $X$ to be intersection of $BQ$ and $CP$ . Define $Y$ similarly. I.e. we completed the quad $BPCQ$ .
We have $\angle QAC=\angle QKC=\angle QXC$ implying that $XACQ$ is cyclic. Similarly, $AYBQ$ is cyclic. Therefore, $A$ is the Miquel point of quad $BPCQ$ implying the desired statement.

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