Easy for a #6 -- Prove AP and AQ are isogonal

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Easy for a #6 -- Prove AP and AQ are isogonal

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Source: Taiwan 2014 TST2, Problem 6

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#1 h Jul 18, 2014, 7:15 PM • 3 Y

Y by HamstPan38825, Adventure10, Mango247

Let $P$ be a point inside triangle $ABC$ , and suppose lines $AP$ , $BP$ , $CP$ meet the circumcircle again at $T$ , $S$ , $R$ (here $T \neq A$ , $S \neq B$ , $R \neq C$ ). Let $U$ be any point in the interior of $PT$ . A line through $U$ parallel to $AB$ meets $CR$ at $W$ , and the line through $U$ parallel to $AC$ meets $BS$ again at $V$ . Finally, the line through $B$ parallel to $CP$ and the line through $C$ parallel to $BP$ intersect at point $Q$ . Given that $RS$ and $VW$ are parallel, prove that $\angle CAP = \angle BAQ$ .

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Radar

#2 h Jul 18, 2014, 11:32 PM • 3 Y

Y by mhq, Adventure10, Mango247

Let $X$ and $Y$ be intersections of $PB$ and $PC$ with $AC$ and $AB$ . First, we have $RS$ is antiparallel to $BC$ in $\angle BPC$ , therefore so is $VW$ . Then $V,W,C$ and $B$ are concyclic, so $PV\cdot PB=PW\cdot PC$ .

Then from similar triangles $\triangle PXA$ and $\triangle PVU$ we have $\frac{PV}{PU}=\frac{PX}{PA}$ and analogically $\frac{PW}{PU}=\frac{PY}{PA}$ . Dividing these we get $\frac{PV}{PW}=\frac{PX}{PY}$ , so $PX\cdot PB=PY\cdot PC$ . Therefore $XYBC$ is cyclic, so $\angle ABP=\angle YBX=\angle YCX=\angle PCA$ .

Finally, let $A'$ be image of $A$ in translation, which moves $P$ to $B$ (and $C$ to $Q$ , because $BPCQ$ is parallelogram). Then $\angle A'QB=\angle ACP=\angle PBA=\angle A'AB$ . Therefore, $AA'BQ$ is cyclic. Therefore $\angle CAP=\angle QA'B=\angle BAQ$ , which is what we wanted.

Q.E.D.

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#3 h Mar 29, 2016, 1:27 PM • 8 Y

Y by Ankoganit, rkm0959, PRO2000, BBeast, CeuAzul, Muradjl, AlastorMoody, Adventure10

$[asy] unitsize(3.2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10pt); /* Positions of pairs X,U */ real x=0.54, u=2/3; pair sipc(pair A=(0,0), pair P, path c=unitcircle,real ext=10) { return IP(L(A,P,-1/ext,ext),c); } pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), T=sipc(A,P), S=sipc(B,P), R=sipc(C,P), U=WP(P--T,u), V=extension(B,P,U,U+C-A), W=extension(C,P,U,U+B-A), Q=B+C-P, Ap=A+B-P; filldraw(U--V--W--cycle,palegrey,pathpen); filldraw(A--X--Y--cycle,palegrey,invisible); D(unitcircle,heavygreen); D(circumcircle(B,C,X),red); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--T,purple); DPA(B--Q--C^^A--Ap--B^^Ap--Q^^X--Y^^R--S); DPA(B--S^^C--R,pathpen+linetype("4 4")+linewidth(1)); /* Angle marks */ DPA(anglemark(Ap,A,B,5)^^anglemark(X,B,A,5)^^anglemark(X,C,Y,5)^^anglemark(Ap,Q,B,5),orange); /* Dots and labels */ D("A",A,dir(A)); D("B",B,dir(B)); D("C",C,dir(C)); D("X",X,dir(75)); D("Y",Y,dir(100)); D("P",P,dir(120)); D("T",T,dir(T)); D("S",S,dir(S)); D("R",R,dir(R)); D("U",U,dir(U)); D("V",V,N); D("W",W,SSE); D("Q",Q,dir(Q)); D("A'",Ap,dir(Ap)); [/asy]$
Let $X=BP\cap AC,Y=CP\cap AB$ . Then $\triangle AXY,\triangle UVW$ are in perspective, so it follows by Desargue's theorem that $XY\cap VW$ is at infinity: i.e. $XY\parallel VW\parallel RS$ . Hence $BCXY$ is cyclic by Reim's theorem.

Let $A'$ be such that $APBA',ACQA'$ are parallelograms. Then $\angle A'AB=\angle XBA=\angle XCY=\angle A'QB$ , so $AA'BQ$ is cyclic. But then $\angle BAQ=\angle BA'Q=\angle PAC$ , as required.

Comment. Note that the position of $U$ on $AT$ is irrelevant, as the direction of $VW$ is independent of $U$ . If we take $U=A$ , then removing some more extraneous fluff gives the following problem:

The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$ . The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$ .
$[asy] unitsize(2.8cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10pt); real x=0.54; pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), Q=B+C-P; D(unitcircle,heavygreen); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--P,purple); D(B--P--C--Q--cycle); /* Angle marks */ DPA(anglemark(P,B,A,5)^^anglemark(A,C,P,5),orange); /* Dots and labels */ D("A",A,dir(A)); D("B",B,dir(B)); D("C",C,dir(C)); D("P",P,dir(130)); D("Q",Q,dir(Q)); [/asy]$
Which is precisely BrMO2 2013/2.

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#4 h Mar 29, 2016, 3:00 PM • 1 Y

Y by Adventure10

Solution with Ceva's Theorem:
Let $X\equiv RS\cap AB, Y\equiv RS\cap AC, D\equiv BS\cap AC, E\equiv CR\cap AB$ .

Note that $\overline{RS}$ is anti-parallel to $\overline{BC},$ so $B,X,Y,C$ are concyclic $\implies \widehat{AS}+\widehat{RB}=$ $\angle AXY=\angle ACB=\widehat{AR}+\widehat{RB}$ $\implies \widehat{AS}=\widehat{AR}\implies\angle BEC=\angle BDC\implies B,E,D,C$ are concyclic.

Apply Ceva's Theorem for $P$ WRT $\triangle ABC$ we get
$1=\frac{\sin{\angle BAP}}{\sin{\angle PAC}}\cdot\frac{\sin{\angle ACP}}{\sin{\angle PCB}}\cdot\frac{\sin{\angle PBC}}{\sin{\angle PBA}}=\frac{\sin{\angle BAP}}{\sin{\angle PAC}}\cdot\frac{\sin{\angle PBC}}{\sin{\angle PCB}}.\qquad (1)$ where $\angle ACP=\angle PBA$ follows by $B,E,D,C$ are concyclic.

On the other hand, apply Ceva's Theorem for $Q$ WRT $\triangle ABC$ and use the fact that $\angle ABQ=\angle AEC=\angle ADB=\angle ACQ$ we get
$1=\frac{\sin{\angle CAQ}}{\sin{\angle QAB}}\cdot\frac{\sin{\angle ABQ}}{\sin{\angle QBC}}\cdot\frac{\sin{\angle QCB}}{\sin{\angle QCA}}=\frac{\sin{\angle CAQ}}{\sin{\angle QAB}}\cdot\frac{\sin{\angle QCB}}{\sin{\angle QBC}}.\qquad (2)$
Combine $(1)$ and $(2)$ and remember the obvious condition that $BQCP$ is a parallelogram, from $\angle BAQ+\angle PAC=\angle CAQ+\angle QAB=\angle BAC<180^\circ$ we deduce that $AP,AQ$ are isogonal with respect to $\angle BAC,$ as desired. $\blacksquare$
$[asy] size(9cm); defaultpen(fontsize(10pt)); pair A,B,C,R,S,M,T,X,Y,E,D,V,W,Q,P,X,Y; draw(unitcircle); A=dir(110); B=dir(-155); C=dir(-25); P=dir(-185)/3; D=4*A/7+3*C/7; E=IP(A--B,circumcircle(B,C,D)); P=IP(B--D,C--E); T=IP(P--dir(P-A)*2+P,unitcircle); S=IP(P--dir(P-B)*2+P,unitcircle); R=IP(P--dir(P-C)*2+P,unitcircle); M=5*T/7+2*P/7; W=extension(C,R,M,dir(A-B)+M); V=extension(B,S,M,dir(A-C)+M); Q=extension(B,B+dir(C-P),C,C+dir(B-P)); X=IP(A--B,R--S); Y=IP(A--C,R--S); draw(A--B--C--cycle); draw(A--T); draw(B--S); draw(C--R); draw(R--S,blue); draw(V--W,blue); draw(D--E,blue+dashed); draw(M--T); draw(M--W); draw(B--Q); draw(C--Q); draw(M--V); draw(circumcircle(B,D,E),red+linetype("4 4")); draw(A--Q); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$T$",T,dir(T)); dot("$S$",S,dir(S)); dot("$R$",R,dir(R)); dot("$P$",P,dir(120)*1.5); dot("$W$",W,dir(90)); dot("$V$",V,dir(90)); dot("$D$",D,dir(-90)*1.5); dot("$E$",E,dir(-240)); dot("$U$",M,dir(-135)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Q$",Q,dir(Q)); [/asy]$

Remark: It seems quite easy and similar to All Rusia Olympiad 2011/8.

Edit: All right, by a bit of angle-chasing we get $\angle ABP=\angle ACP,$ and then we just use same method in ARO 2011/8 to finish the rest:

liberator wrote:

The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$ . The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$ .

This post has been edited 1 time. Last edited by Complex2Liu, Mar 29, 2016, 3:12 PM
Reason: typo

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Kayak

#5 h Nov 21, 2018, 1:28 PM • 2 Y

Y by Adventure10, Mango247

Isn't this way too easy even for a P1 or I am missing something ? (Took me <10 minutes to solve using geogebra ...) (also I don't know if the proof of the last claim is correct or not)

Let $V_U$ , $W_U$ denote the points $V, W$ as a function of $U$ for some $U \in \overline{PT}$ .

Claim For any two $U_1, U_2 \in \overline{AT}$ , $V_{U_1}W_{U_1}||V_{U_2}W_{U_2}$
Proof : Observe that $\angle V_{U_1}U_1P = \angle PAE = \angle V_{U_2}U_2P$ , and $\angle U_1PV_{U_1} = \angle V_{U_1}PT = \angle U_2PV_{U_2}$ , so $\Delta U_1PV_{U_1} \sim U_2PV_{U_2}$ . Similarly $\Delta U_1PW_{U_1} \sim \Delta U_2PW_{U_2}$ , and so $\Delta PV_{U_1}W_{U_1} \sim \Delta PV_{U_2}W_{U_2}$ . The conclusion follows. $\blacksquare$

So we can WLOG set $U = A$ . Let $BP, CP$ hit $AC, AB$ at $E, D$ respectively. Then $\ell_{VW} = \ell_{DE}$

Claim: $RS || DE \Leftrightarrow (B,C,D,E) \text{ is cyclic}$
Proof: $\angle CRS = \angle SBC = \angle EBC$ , so $PS || DE \Leftrightarrow \angle CRS = \angle CDE \Leftrightarrow \angle CBE = \angle CDE \Leftrightarrow (B,C,D,E) \text{cyclic}$ $\blacksquare$

Claim: $(B,C,D,E) \text{is cyclic} \Rightarrow \angle PAC = \angle BAQ$ .
Proof Let $\ell_1$ be the line through $A$ parallel to $DC$ , and $\ell_2$ be the line through $A$ parallel to $BE$ . Let $P^{\infty}_1, P^{\infty}_2$ be the points at infinity on $\ell_1, \ell_2$ respectively. Note that $P = BP^{\infty}_1 \cap CP^{\infty}_2$ and $Q = BP^{\infty}_2 \cap CP^{\infty}_1$ .

Then since $\angle P^{\infty}_2AB = \angle DBE = \angle DCE = \angle P^{\infty}_1AC$ , so the two pair of lines $(AP^{\infty}_1, AP^{\infty}_2), (AB, AC)$ are isogonal w.r.t $(AB, AC)$ . By the forgotten isogonality lemma, thus $AP, AQ$ are isogonal, as desired (?) $\blacksquare$

This post has been edited 1 time. Last edited by Kayak, Nov 21, 2018, 1:30 PM

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#6 h Nov 29, 2018, 8:06 AM • 4 Y

Y by potentialenergy, DonaldJ.Trump, Adventure10, Mango247

liberator wrote:

$[asy] unitsize(3.2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10pt); /* Positions of pairs X,U */ real x=0.54, u=2/3; pair sipc(pair A=(0,0), pair P, path c=unitcircle,real ext=10) { return IP(L(A,P,-1/ext,ext),c); } pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), T=sipc(A,P), S=sipc(B,P), R=sipc(C,P), U=WP(P--T,u), V=extension(B,P,U,U+C-A), W=extension(C,P,U,U+B-A), Q=B+C-P, Ap=A+B-P; filldraw(U--V--W--cycle,palegrey,pathpen); filldraw(A--X--Y--cycle,palegrey,invisible); D(unitcircle,heavygreen); D(circumcircle(B,C,X),red); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--T,purple); DPA(B--Q--C^^A--Ap--B^^Ap--Q^^X--Y^^R--S); DPA(B--S^^C--R,pathpen+linetype("4 4")+linewidth(1)); /* Angle marks */ DPA(anglemark(Ap,A,B,5)^^anglemark(X,B,A,5)^^anglemark(X,C,Y,5)^^anglemark(Ap,Q,B,5),orange); /* Dots and labels */ D("A",A,dir(A)); D("B",B,dir(B)); D("C",C,dir(C)); D("X",X,dir(75)); D("Y",Y,dir(100)); D("P",P,dir(120)); D("T",T,dir(T)); D("S",S,dir(S)); D("R",R,dir(R)); D("U",U,dir(U)); D("V",V,N); D("W",W,SSE); D("Q",Q,dir(Q)); D("A'",Ap,dir(Ap)); [/asy]$
Let $X=BP\cap AC,Y=CP\cap AB$ . Then $\triangle AXY,\triangle UVW$ are in perspective, so it follows by Desargue's theorem that $XY\cap VW$ is at infinity: i.e. $XY\parallel VW\parallel RS$ . Hence $BCXY$ is cyclic by Reim's theorem.

Let $A'$ be such that $APBA',ACQA'$ are parallelograms. Then $\angle A'AB=\angle XBA=\angle XCY=\angle A'QB$ , so $AA'BQ$ is cyclic. But then $\angle BAQ=\angle BA'Q=\angle PAC$ , as required.

Comment. Note that the position of $U$ on $AT$ is irrelevant, as the direction of $VW$ is independent of $U$ . If we take $U=A$ , then removing some more extraneous fluff gives the following problem:

The point $P$ lies inside triangle $ABC$ so that $\angle ABP = \angle PCA$ . The point $Q$ is such that $PBQC$ is a parallelogram. Prove that $\angle QAB = \angle CAP$ .
$[asy] unitsize(2.8cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10pt); real x=0.54; pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), Q=B+C-P; D(unitcircle,heavygreen); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--P,purple); D(B--P--C--Q--cycle); /* Angle marks */ DPA(anglemark(P,B,A,5)^^anglemark(A,C,P,5),orange); /* Dots and labels */ D("A",A,dir(A)); D("B",B,dir(B)); D("C",C,dir(C)); D("P",P,dir(130)); D("Q",Q,dir(Q)); [/asy]$
Which is precisely BrMO2 2013/2.

Just nice.... :coolspeak:

:coolspeak:

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yayups

#7 h Apr 9, 2019, 8:10 PM • 3 Y

Y by Gaussian_cyber, Adventure10, Mango247

https://lh3.googleusercontent.com/-FpZ_PUoDdJs/XKz7_fKEyBI/AAAAAAAAE_0/FUt6HpM3QjAUtvZ1Frmby1KuzGy2-SdFQCK8BGAs/s0/2019-04-09.png

It's clear that the choice of $U$ is irrelevant, so we may choose $U=A$ , so $V=BS\cap AC$ and $W=CR\cap AB$ .

Claim: The condition of the problem implies $\angle ABP=\angle ACP$ .

Proof of Claim: We see that $RS$ is anti-parallel to $BC$ and since $RS\parallel WV$ , we have $WV$ antiparallel to $BC$ , so $WVCB$ cyclic. Thus, $\angle WBV=\angle WCV$ , or $\angle ABP=\angle ACP$ . $\blacksquare$

Let $D,E\in(ABC)$ so that $AD\parallel BP$ and $AE\parallel CP$ . By DDIT on $PCQB$ with $A$ , we see that there is an involution swapping $(AB,AC)$ , $(AP,AQ)$ , and $(AD,AE)$ . We see that
$\angle DAB=\angle ABP=\angle ACP=\angle EAC,$ so $AD$ and $AE$ are isogonal. Clearly $AB$ and $AC$ are isogonal, so this involution is isogonality, so $AP$ and $AQ$ are isogonal, as desired. $\blacksquare$

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#8 h Apr 9, 2019, 9:55 PM • 1 Y

Y by Adventure10

Note that, by Reim's theorem, $BVWC$ is cyclic. Now, let $BP$ and $CP$ hit $AC$ , $AB$ at $E$ , $F$ respectively. $PVU$ and $PEA$ are similar, so $PE/PV=PU/PA$ , and $PFA,PWU$ are also similar, implying that $PF/PW=PV/PE$ . Then, by Power of a Point, $BFEC$ is cyclic. So, naturally redefine point $P$ to be $BE\cap CF$ , where $BFEC$ is cyclic.

Now, we will prove that $AP$ and $AQ$ are isogonal with barycentric coordinates. They are reflections about $M$ , the midpoint of $BC$ , so if the homogenized coordinates of $P$ are $(x_P,y_P,z_P)$ , then $Q$ is $(-x_P,1-y_P,1-z_P)$ . So, we wish to show that $\frac{y_P(1-y_P)}{z_P(1-z_P)}=\frac{b^2}{c^2}$

Letting $E=(e,0,1-e)$ and $F=(f,1-f,0)$ , the circumcircle of $(BCE)$ has equation $-a^2yz-b^2xz-c^2xy+(x+y+z)b^2(1-e)x=0$ , which means $f=1-\frac{b^2}{c^2}(1-e)$ . $BE$ and $CF$ are cevians, which makes it easy to find $P$ . Ater homogenization, it is $\left(\frac{ef}{e+f-ef},\frac{e-ef}{e+f-ef},\frac{f-ef}{e+f-ef}\right)$ . So, $\frac{y_P(1-y_P)}{z_P(1-z_P)}=\frac{ef(1-f)}{ef(1-e)}=\frac{1-f}{1-e}=\frac{b^2}{c^2}$ as desired.

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#9 h Jul 23, 2020, 7:12 AM

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Let $E, F$ be the infinite point lie on $AC, AB$ . Note that $W=SP \cap UE$ , $V=RP \cap UF$ , and $\infty_{VW} = EF\cap RS$ , by Desargues's theorem, we hence that $\triangle PSR$ and $\triangle UEF$ are perspective. Let $UP, ES, FR$ concurrence at point $T$ . Denote $K=AC \cap PS$ , $L= AB \cap PR$ . While $PSTR \sim PKAL$ , conclude that $KL \parallel RS$ . By Reim's theorem, $BCKL$ is cyclic. Now $\measuredangle ACP = \measuredangle PBA$ , hence that $A\infty_{CP}$ and $A\infty_{BP}$ are isogonal conjugate WRT $\angle BAC$ . By Desargues's Involution with quadrilateral $\{ BP, PC, CQ, QB \}$ and point $A$ , hence that $AP$ and $AQ$ are isogonal conjugate WRT $\angle BAC$ , so we're done.

This post has been edited 2 times. Last edited by Cindy.tw, Jul 23, 2020, 7:16 AM
Reason: first time latex

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#10 h Jul 27, 2021, 6:31 PM

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Taiwan 2014 TST2, Problem 6 wrote:

Let $P$ be a point inside triangle $ABC$ , and suppose lines $AP$ , $BP$ , $CP$ meet the circumcircle again at $T$ , $S$ , $R$ (here $T \neq A$ , $S \neq B$ , $R \neq C$ ). Let $U$ be any point in the interior of $PT$ . A line through $U$ parallel to $AB$ meets $CR$ at $W$ , and the line through $U$ parallel to $AC$ meets $BS$ again at $V$ . Finally, the line through $B$ parallel to $CP$ and the line through $C$ parallel to $BP$ intersect at point $Q$ . Given that $RS$ and $VW$ are parallel, prove that $\angle CAP = \angle BAQ$ .

Solution. Let $X=AC\cap BS$ , and $Y=AB\cap CR$ .
We need to prove that $AP$ , and $AQ$ are isogonal w.r.t $\angle BAC$ , but since $BPCQ$ is a parallelogram by the Parallelogram Isogonality Lemma it is enough to show that $\angle ABP=\angle ACP$ , or $BYXC$ cyclic. But by Desargue's Theorem on $\triangle AXY,\triangle UVW\implies XY\parallel VW\parallel RS$ . Now By the converse of Reim's Theorem $BYXC$ cyclic. $\blacksquare$

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#11 h Feb 18, 2024, 3:20 PM

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It suffices to show $\angle ABP = \angle ACP$ , as then the result follows by the Parallelogram Isogonality Lemma. (In particular, we may ignore the point $Q$ from the diagram.)

Let $X=BP\cap AC$ and $Y=CP\cap AB$ . Since $AX \parallel WU$ and $AY \parallel VU$ , by Thales' theorem (or the homothety centered at $P$ which maps $A$ to $U$ ) we obtain $XY \parallel VW$ . Now the problem condition $RS \parallel VW$ implies $XY \parallel RS$ . Thus $\angle BYX = \angle BSR = \angle CBR = \angle BCX$ , i.e. $BCXY$ is cyclic and so $\angle ABP = \angle XBY = \angle XCY = \angle ACP$ , as desired.

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#12 h Feb 18, 2024, 8:21 PM

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By the First Isogonality Lemma it suffices to show $\angle ABP = \angle ACP$ . Let $W' = \overline{AB} \cap \overline{CP}$ and $V' = \overline{AC} \cap \overline{BP}$ . Then we wish to show $BW'V'C$ is cyclic.

Note that from Reim's we find that $BRSC$ cyclic implies $BVWC$ cyclic. Thus it suffices to show that $\overline{W'V'} \parallel \overline{WV}$ and we may finish by Reim's. To see this consider the negative homothety $\mathcal{H}$ at $P$ mapping $U \mapsto A$ . Consider the image of $\triangle UVW$ under $\mathcal{H}$ . Clearly $V \mapsto V'$ and $W \mapsto W'$ due to the condition $\overline{UV} \parallel \overline{AV'}$ and $\overline{UW} \parallel \overline{AW'}$ . However then we must have $\overline{VW} \parallel \overline{V'W'}$ so we are done. $\square$

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