Equal angles

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Source: All Russian Olympiad - Problem 11.7

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1270 posts

April

#1 h May 10, 2009, 3:34 PM • 3 Y

Y by Infinityfun, Adventure10, Mango247

Let be given a parallelogram $ABCD$ and two points $A_1$ , $C_1$ on its sides $AB$ , $BC$ , respectively. Lines $AC_1$ and $CA_1$ meet at $P$ . Assume that the circumcircles of triangles $AA_1P$ and $CC_1P$ intersect at the second point $Q$ inside triangle $ACD$ . Prove that $\angle PDA = \angle QBA$ .

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2643 posts

yetti

#2 h May 10, 2009, 10:59 PM • 4 Y

Y by doxuanlong15052000, Kayak, Adventure10, Mango247

Let $(X) \equiv \odot(AA_1P)$ cut $DA$ again at $A_2$ and let $(Y) \equiv \odot (CC_1P)$ cut $CD$ again at $C_2.$

$\angle CQP = \angle PC_1B = \angle PAA_2 = \pi - \angle A_2QP$

$\Longrightarrow$ $C, Q, A_2$ are collinear and similarly, $A, Q, C_2$ are collinear. From $(X)$ , $A_1Q$ is antiparallel of $AA_2 \parallel BC$ WRT angle $\angle (AB, A_2C)$ $\Longrightarrow$ $A_1BCQ$ is cyclic, and $A_2P$ is antiparallel of $AA_1 \parallel DC$ WRT angle $\angle (DA, CA_1)$ $\Longrightarrow$ $CDA_2P$ is also cyclic. From $\odot (A_1BCQ), \odot(CDA_2P),$

$\angle QBA = \angle QBA_1 = \angle QCA_1 = \angle QCP = \angle A_2CP = \angle A_2DP = \angle ADP.$

Attachments:

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374 posts

Ahiles

#3 h May 11, 2009, 3:21 PM • 2 Y

Y by Adventure10, Mango247

From Virgil Nicula's extension we have

$\frac {\sin \angle ABQ}{\sin \angle CBQ} = \frac {AA_1}{CC_1}$

Denote by $E$ the intersection of the lines $AD$ and $CP$ . Then

$\dfrac{\sin \angle ADP}{\sin \angle CDP} = \dfrac{EP}{PC} \cdot \dfrac{CD}{ED}$

But

$\dfrac{EP}{PC} = \dfrac{EA}{CC_1}$

and

$\dfrac{AD}{ED} = 1 - \dfrac{EA}{ED} = 1 - \dfrac{AA_1}{CD} = \dfrac{CD - AA_1}{CD}$

$ED = \dfrac{CD \cdot AD}{CD - AA_1} = \dfrac{CD \cdot AD}{AB - AA_1} = \dfrac{CD \cdot AD}{BA_1}$

So

$\dfrac{\sin \angle ADP}{\sin \angle CDP} = \dfrac{EA}{CC_1} \cdot \dfrac{CD}{\dfrac{CD \cdot AD}{BA_1}} = \dfrac{EA}{BC} \cdot \dfrac{BA_1}{CC_1} = \dfrac{AA_1}{BA_1} \cdot \dfrac{BA_1}{CC_1} = \frac {AA_1}{CC_1} = \frac {\sin \angle ABQ}{\sin \angle CBQ}$

As function $f(x) = \dfrac{\sin (\alpha - x)}{\sin x}$ is strictly decreasing on $(0,\pi)$ we get that $\angle PDA = \angle QBA$ .

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212 posts

#4 h Aug 26, 2010, 7:48 PM • 3 Y

Y by Swad, Adventure10, Mango247

Sorry to revive this old thread, But I have an alternative solution to this beautifull problem.
My solution

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2326 posts

yugrey

#5 h Sep 23, 2014, 11:49 PM • 2 Y

Y by Adventure10, Mango247

Use complex numbers. Let $Q=0$ . Note that in angle measure we have $QAP=QA_1P=QA_1C=QBC$ , $QPA=QA_1A=QCC_1=QCB$ since $Q$ takes $AC_1$ to $A_1C$ . So $Q$ takes $AP$ to $BC$ and $pb=ac$ . So $p(a+c-d)=ac$ , $-pd=-ap-cp+ac$ , and $p(p-d)=(p-a)(p-c)$ . Thus, $P$ takes $AD$ to $QC$ . So now $PDA=PCA=A_1CQ=A_1BQ=QBA$ , so we're done.

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1980 posts

#6 h Dec 4, 2016, 8:09 AM • 4 Y

Y by Anar24, TheCosineLaw, Rg230403, Adventure10

Nice angle chasing exercise!

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.182cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 15.96, ymin = -5.24, ymax = 6.3; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen bfffqq = rgb(0.7490196078431373,1.,0.); pen ffqqff = rgb(1.,0.,1.); pen xfqqff = rgb(0.4980392156862745,0.,1.); /* draw figures */ draw((4.32,3.2)--(5.36,0.), xdxdff); draw((2.66,0.)--(5.36,0.), xdxdff); draw((1.62,3.2)--(4.32,3.2), xdxdff); draw((1.62,3.2)--(2.66,0.), xdxdff); draw((4.32,3.2)--(4.1,0.)); draw((4.859378365410247,1.5403742602761632)--(2.66,0.)); draw(circle((3.38,0.5817777444167372), 0.9256702133582054), linetype("2 2") + ffxfqq); draw(circle((3.919179933099226,2.152271624016165), 1.1217804936225562), linetype("2 2") + ffxfqq); draw(circle((4.73,1.56425), 1.686350515907057), linetype("4 4") + bfffqq); draw((3.518359866198457,3.2)--(2.66,0.), linewidth(1.2) + linetype("4 4") + ffqqff); draw(circle((2.569179933099227,1.739483478257249), 1.7418527652134657), dotted + xfqqff); /* dots and labels */ dot((2.66,0.),dotstyle); label("$A$", (2.38,-0.26), NE * labelscalefactor); dot((5.36,0.),dotstyle); label("$B$", (5.46,-0.38), NE * labelscalefactor); dot((4.32,3.2),dotstyle); label("$C$", (4.38,3.36), NE * labelscalefactor); dot((1.62,3.2),linewidth(3.pt) + dotstyle); label("$D$", (1.36,3.04), NE * labelscalefactor); dot((4.1,0.),dotstyle); label("$A_1$", (4.14,-0.44), NE * labelscalefactor); dot((4.859378365410247,1.5403742602761632),dotstyle); label("$C_1$", (5.02,1.52), NE * labelscalefactor); dot((4.172843888664785,1.059547471487769),linewidth(3.pt) + dotstyle); label("$P$", (4.34,0.78), NE * labelscalefactor); dot((4.172843888664785,1.059547471487769),linewidth(3.pt) + dotstyle); dot((3.0478152622528927,1.4457908484299304),linewidth(3.pt) + dotstyle); label("$Q$", (3.12,1.56), NE * labelscalefactor); dot((4.32,3.2),linewidth(3.pt) + dotstyle); dot((3.518359866198457,3.2),linewidth(3.pt) + dotstyle); label("$A_2$", (3.6,3.32), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Lemma 1. Points $A_1, B, C, Q$ lie on a circle.

Proof. We have $\angle A_1QC=\angle A_1QP+\angle PQC=\angle A_1AP+\angle PC_1B=180^{\circ}-\angle A_1BC_1 \Longrightarrow A_1, B, C, Q \, \text{are concyclic}. \, \square$
Lemma 2. Points $A, Q, A_2$ are collinear where $A_2$ is the second intersection of $(CC_1P)$ with $CD$ .

Proof. Note that from $A_2C \parallel A_1B$ we have $\angle AQP+\angle A_2QP=\angle PA_1B+\left(180^{\circ}-\angle PCA_2\right)=180^{\circ} \Longrightarrow A, Q, A_2 \, \text{are collinear}. \, \square$
Lemma 3. Points $D, A_2, P, A$ are concyclic.

Proof. We have $\angle APA_2=\angle A_2CC_1=180^{\circ}-\angle ADA_1 \Longrightarrow A, D, A_2, P \, \text{are concyclic}. \, \square$
As $\angle PDA=\angle PA_2Q=\angle QCA_1=\angle QBA,$ we are done.

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947 posts

#7 h Apr 16, 2017, 3:16 AM • 2 Y

Y by Adventure10, Mango247

Let $APQ$ interrsect $AD$ at $X$ and let $CPQ$ intersect $CD$ at $Y$ .

Observe that $\angle XAP = \angle PC_1B = \angle PQC$ . Thus $X, Q, C$ are collinear. Similarly, $A, Q, Y$ are collinear.

This mean that $DAPY$ is cyclic and $BA_1QC$ is cyclic (properties of the Miquel Point). This gives us

$\angle PDA = \angle PYA = \angle PYQ = \angle PCQ = \angle A_1CQ = \angle ABQ$ as required.

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1298 posts

Kayak

#8 h Jan 4, 2019, 1:02 PM • 2 Y

Y by Adventure10, Mango247

Pick $A_2 \in \overline{AD}, C_2 \in \overline{CD}$ such that $AA_1A_2PQ$ , $CC_1C_2PQ$ are concylic. Observe that $Q$ is the Miquel point of $PA_1BC_1$ .

$\angle PQA_2 + \angle PQC = 180 - \angle PAA_2 + 180 - \angle PC_1C = 180$ , and hence $A_2, Q, C$ are colinear. Similarly $C_2, A, Q$ are colinear. Thus $P$ is the Miquel point of $QC_2DA_2$ .

Now we're done since $\angle QBA = \angle QC_1A = \angle PC_1Q = \angle PC_2Q = \angle AC_2Q = \angle ADP$

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Mahdi_Mashayekhi

694 posts

Mahdi_Mashayekhi

#9 h May 14, 2022, 5:15 AM

Y by

Let circle $AA_1P$ and $CC_1P$ meet $AD$ and $CD$ at $S,K$ .
Claim $: BA_1QC$ is cyclic.
Proof $:$ Note that $\angle CQA_1 = \angle CQP + \angle PQA_1 = \angle BC_1A + \angle PAB = \angle 180 - \angle A_1BC$ .
Claim $: A,Q,K$ and $C,Q,S$ are collinear.
Proof $:$ Note that $\angle PQA = \angle PA_1B = \angle PCK = \angle 180 - \angle PQK$ . we prove the other one with same approach.
Claim $: CPSD$ is cyclic.
Proof $:$ Note that $\angle PSA = \angle PQA = \angle PCK = \angle PCD$ .
Now Note that $\angle QBA = \angle QCA_1 = \angle QCP = \angle SCP = \angle SDP = \angle PDA$ .
we're Done.

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559 posts

#10 h May 14, 2022, 12:46 PM • 1 Y

Y by Infinityfun

Denote by $R$ reflection of $P$ wrt center of parallelogram. In $A_1BC_1P$ Miquel point $Q$ is the isogonal conjugate of $\infty_{BR},$ since $BR$ is homothetic to it's Gauss line with center $P$ and coefficient $2,$ therefore $\angle PDA\stackrel{\text{symmetry}}{=}\angle RBC=\angle QBA.$

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1603 posts

#11 h Aug 14, 2023, 9:27 AM

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solution

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