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Source: 4th Iranian Geometry Olympiad (Intermediate) P2

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178 posts

bgn

#1 h Sep 15, 2017, 6:20 AM • 4 Y

Y by GammaBetaAlpha, chessgocube, Adventure10, Rounak_iitr

Two circles $\omega_1,\omega_2$ intersect at $A,B$ . An arbitrary line through $B$ meets $\omega_1,\omega_2$ at $C,D$ respectively. The points $E,F$ are chosen on $\omega_1,\omega_2$ respectively so that $CE=CB,\ BD=DF$ . Suppose that $BF$ meets $\omega_1$ at $P$ , and $BE$ meets $\omega_2$ at $Q$ . Prove that $A,P,Q$ are collinear.

Proposed by Iman Maghsoudi

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436 posts

#2 h Sep 15, 2017, 7:20 AM • 5 Y

Y by chessgocube, Adventure10, Mango247, ohiorizzler1434, Rounak_iitr

Simple angle chasing.

Firstly, we have $\angle EBC = \angle BEC = \angle BPC$ , so $\angle PCB = 180^{\circ} - 2\angle EBC - \angle ABE$ . Thus, we have $\angle PAB = 180^{\circ} - \angle PEB = 180^{\circ} - \angle PCB = 2\angle EBC + \angle ABE$ .

We note that $\angle PBC = \angle DBF$ and $\angle EBC = \angle DBQ$ . Therefore, $\angle DFQ = \angle DBQ = \angle EBC$ , and $\angle DFB = \angle DBF = \angle EBC + \angle PBE$ . Therefore, $\angle QAB = 180^{\circ} - \angle QFB = 180^{\circ} - (\angle DFQ + \angle QFB) = 180^{\circ} - (2 \angle EBC + \angle PBC) = 180^{\circ} - \angle PAB$ .

Therefore, $\angle PAQ = 180^{\circ}$ , so $A$ , $P$ and $Q$ are collinear.

This post has been edited 2 times. Last edited by SHARKYKESA, Sep 16, 2019, 10:14 AM

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125 posts

#4 h Sep 15, 2019, 4:59 AM • 3 Y

Y by chessgocube, Adventure10, Rounak_iitr

Can anynone please provide a diagram for this problem.

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495 posts

NJOY

#5 h Sep 15, 2019, 5:26 AM • 7 Y

Y by T-E-A-S-IND, John-Wick, chessgocube, Adventure10, Rounak_iitr, ehuseyinyigit, Ismayil07

Here is the attached diagram.

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125 posts

#6 h Sep 15, 2019, 5:29 AM • 4 Y

Y by chessgocube, Adventure10, Mango247, Rounak_iitr

SHARKYKESA wrote:

Simple angle chasing.

Firstly, we have $\angle EBC = \angle BEC = \angle BPC$ , so $\angle PCB = 180^{\circ} - 2\angle EBC - \angle ABE$ . Thus, we have $\angle PAB = 180^{\circ} - \angle PEB = 180^{\circ} \angle PCB = 2\angle EBC + \angle ABE$ .

We note that $\angle PBC = \angle DBF$ and $\angle EBC = \angle DBQ$ . Therefore, $\angle DFQ = \angle DBQ = \angle EBC$ , and $\angle DFB = \angle DBF = \angle EBC + \angle PBE$ . Therefore, $\angle QAB = 180^{\circ} - \angle QFB = 180^{\circ} - (\angle DFQ + \angle QFB) = 180^{\circ} - (2 \angle EBC + \angle PBC) = 180^{\circ} - \angle PAB$ .

Therefore, $\angle PAQ = 180^{\circ}$ , so $A$ , $P$ and $Q$ are collinear.

Is there any other method for proving this result?

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867 posts

#7 h Sep 15, 2019, 5:34 AM • 3 Y

Y by chessgocube, Adventure10, Mango247

John-Wick wrote:

SHARKYKESA wrote:

Simple angle chasing.

Firstly, we have $\angle EBC = \angle BEC = \angle BPC$ , so $\angle PCB = 180^{\circ} - 2\angle EBC - \angle ABE$ . Thus, we have $\angle PAB = 180^{\circ} - \angle PEB = 180^{\circ} \angle PCB = 2\angle EBC + \angle ABE$ .

We note that $\angle PBC = \angle DBF$ and $\angle EBC = \angle DBQ$ . Therefore, $\angle DFQ = \angle DBQ = \angle EBC$ , and $\angle DFB = \angle DBF = \angle EBC + \angle PBE$ . Therefore, $\angle QAB = 180^{\circ} - \angle QFB = 180^{\circ} - (\angle DFQ + \angle QFB) = 180^{\circ} - (2 \angle EBC + \angle PBC) = 180^{\circ} - \angle PAB$ .

Therefore, $\angle PAQ = 180^{\circ}$ , so $A$ , $P$ and $Q$ are collinear.

Is there any other method for proving this result?

Bashing

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495 posts

NJOY

#8 h Sep 15, 2019, 5:36 AM • 4 Y

Y by T-E-A-S-IND, chessgocube, Adventure10, Mango247

@above I guess bashing shouldn't be done for such an easy problem!!

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54 posts

#10 h Aug 11, 2021, 10:04 AM • 1 Y

Y by chessgocube

Solution:
Construction :Join $AE,AC,PE,PC,AF,BQ$
Here we can see that
$\angle CBE=\angle CEB \text{ (Since given that BC=CE)}$ we can see that
$\angle CBE=\angle CPE=\angle CAE=\alpha\text{ (Since we know that CBPE and CBAE is cyclic)}$ $\angle CEB=\angle CAB=\alpha \text{ (since CBAE is cyclic)}$ now we can see in another circle that
$\angle FBD=\angle BFD=\beta \text{ (Since BD=DF)}$ and we can see here that line QE and line CD intersect at point B so we can say that
$\angle CBE=\angle QBF=\alpha$ $\angle QBD=\angle QFD=\beta \text{ (Since QBFD is cyclic)}$ and here we can see that in $\triangle BPE$ have exterior angle $\angle PBQ$ so we can say that $\angle BEP+\angle BPE=\angle PBQ\implies \angle BEP+2\alpha=\alpha+\beta\implies \angle BEP=\beta-\alpha$ so we can see here that $BEAP$ is cyclic quadrilateral so we can say that
$\angle BEP=\angle BAP=\beta-\alpha$ and we can see that $\angle DFB-\angle DFQ=\beta-\alpha$ and since we know that $Q,B,A,F$ lie on circle $\omega_2$ so we can say that $\angle BAQ$ is $\beta-\alpha$ since we can see here that $\angle BAP=\angle BAQ$ and we are done ! $\blacksquare$

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322 posts

BVKRB-

#11 h Aug 11, 2021, 3:22 PM • 1 Y

Y by chessgocube

One-liner

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41 posts

#12 h Dec 16, 2021, 5:35 PM • 2 Y

Y by chessgocube, SPHS1234

does this work?

Storage

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Mahdi_Mashayekhi

690 posts

Mahdi_Mashayekhi

#13 h Dec 27, 2021, 11:09 AM • 1 Y

Y by Rounak_iitr

∠CEB = ∠CBE = ∠QBD = ∠QFD and ∠DFB = ∠DBF = ∠CEP
∠PAB = ∠PEB = ∠PEC - ∠BEC = ∠DFB - ∠DFQ = ∠QFB = ∠QAB so Q,P,A are collinear.

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lian_the_noob12

173 posts

lian_the_noob12

#14 h Apr 25, 2023, 10:36 AM

Y by

$\color{blue} \boxed {\textbf{SOLUTION}}$

$AP \cap \omega_2 \equiv Q'$
We want to show $Q \equiv Q'$

$\textbf{Step 1 : }$
$\angle CEP=\angle PBD=\angle FBD=\angle BFD$
$\angle CEB=\angle CBE=\angle QBD=\angle QFD$

$\textbf{Step 2 : }$
$\angle BFQ=\angle BFD - \angle QFD=\angle CEP - \angle CEB =\angle BEP=\angle BAP=\angle BAQ'=\angle BFQ'$

So, $FQ \equiv FQ'$
As, $Q$ and $Q'$ both lies on $\omega_2$ and $F$ is also on $\omega_2$ ,

$\implies Q \equiv Q'$
$\blacksquare$

This post has been edited 3 times. Last edited by lian_the_noob12, Apr 25, 2023, 10:37 AM

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22 posts

#15 h Aug 8, 2023, 11:37 AM

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To avoid configuration issues let $\measuredangle$ be a directed angle modulo $\pi$ .
Also, let $\measuredangle DBF = \measuredangle BFD = \alpha$ and $\measuredangle EBC = \measuredangle CEB = \beta$ .

Note that $\begin{align*} \measuredangle PCD = \measuredangle PCE - \measuredangle BCE = \measuredangle CEB + \measuredangle EBC - \measuredangle EBP = \beta - \alpha \\ \measuredangle QDC = \measuredangle DQB + \measuredangle QBD = \measuredangle EBC - \measuredangle BFD = \beta - \alpha \end{align*}$ So $CP \parallel QD$ and thus, by the converse of Reim's theorem (or simple angle chasing), we conclude that $Q - A - P$ .

This post has been edited 1 time. Last edited by Alfombra12, Aug 8, 2023, 11:38 AM

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59 posts

#16 h Jul 2, 2024, 9:27 AM

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Angle chase

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455 posts

#17 h Aug 28, 2024, 7:50 AM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -15.28, xmax = 15.36, ymin = -8.88, ymax = 8.96; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); /* draw figures */ draw(circle((-6,-1), 5.168558793319468), linewidth(1) + blue); draw(circle((4,0), 7.65341638476964), linewidth(1) + ccqqqq); draw((-9.783183088799102,-4.521580002871505)--(10.205488237608218,-4.479586821523434), linewidth(1)); draw((-9.783183088799102,-4.521580002871505)--(-10.557747413852905,1.4374040517561189), linewidth(1)); draw((-10.557747413852905,1.4374040517561189)--(1.603952459053537,-7.268682035980443), linewidth(1)); draw((-2.96,3.18)--(-2.1934123141811837,-4.496268037518368), linewidth(1)); draw((-2.1934123141811837,-4.496268037518368)--(3.3249944877483792,7.623591667782995), linewidth(1)); draw((3.3249944877483792,7.623591667782995)--(10.205488237608218,-4.479586821523434), linewidth(1)); draw((-2.96,3.18)--(1.603952459053537,-7.268682035980443), linewidth(1) + linetype("4 4")); draw((3.3249944877483792,7.623591667782995)--(1.603952459053537,-7.268682035980443), linewidth(1)); draw((-10.557747413852905,1.4374040517561189)--(-0.8635976763904211,-1.5756484778214144), linewidth(1)); draw((-10.5,-1.64)--(-9.82,-1.64), linewidth(1)); draw((-10.5,-1.86)--(-9.82,-1.86), linewidth(1)); draw((-6.475575499931073,-4.28013728777642)--(-6.475575499931073,-5.000117339767044), linewidth(1)); draw((-6.210010795541072,-4.320554086134865)--(-6.210010795541072,-5.040357440247402), linewidth(1)); draw((6.615779589588047,0.48722713274971746)--(7.5001993819101065,1.5798386093215109), linewidth(1)); draw((4.757719405262763,-4.001111239961445)--(4.757719405262763,-5.258639480686026), linewidth(1)); /* dots and labels */ dot((-6,-1),dotstyle); label("$O_{1}$", (-5.92,-0.8), NE * labelscalefactor); dot((4,0),dotstyle); label("$O_{2}$", (4.08,0.2), NE * labelscalefactor); dot((-2.96,3.18),dotstyle); label("$A$", (-3.12,3.44), NE * labelscalefactor); dot((-2.1934123141811837,-4.496268037518368),dotstyle); label("$B$", (-2.3,-5.04), NE * labelscalefactor); dot((-9.783183088799102,-4.521580002871505),dotstyle); label("$C$", (-9.7,-4.32), NE * labelscalefactor); dot((10.205488237608218,-4.479586821523434),dotstyle); label("$D$", (10.28,-4.28), NE * labelscalefactor); dot((-10.557747413852905,1.4374040517561189),dotstyle); label("$E$", (-10.48,1.64), NE * labelscalefactor); dot((1.603952459053537,-7.268682035980443),dotstyle); label("$Q$", (1.68,-7.06), NE * labelscalefactor); dot((3.3249944877483792,7.623591667782995),dotstyle); label("$G$", (3.4,7.82), NE * labelscalefactor); dot((-0.8635976763904211,-1.5756484778214144),linewidth(4pt) + dotstyle); label("$P$", (-0.64,-1.76), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
USING ABOVE DIAGRAM.

$\color{red}\textbf{Claim:-}$ $A,P,Q$ are collinear.
$\color{blue}\textbf{Proof:-}$ First we observe that the points $E,A,P,B,C$ and $G,A,B,Q,D$ are cyclic points. Therefore we get some angle condition,
$\angle CBE=\angle CEB=\angle QBD=\angle QGD=2\angle EPB=2\angle EAB.$ Also we get, $\angle EBP+\angle BEP=\angle ECB$ Now we get, $\angle PEB=\angle PEC-\angle BEC\implies \angle BGQ=\angle BGD-\angle QGD$ $\boxed{\angle QAB=\angle QGB.}$ Therefore the points $A,P,Q$ are collinear.

This post has been edited 1 time. Last edited by Rounak_iitr, Aug 28, 2024, 7:52 AM

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