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The familiar right angle from the orthocenter
buratinogigle   3
N 12 minutes ago by luutrongphuc
Source: Own, HSGSO P6
Let $ABC$ be a triangle inscribed in a circle $\omega$ with orthocenter $H$ and altitude $BE$. Let $M$ be the midpoint of $AH$. Line $BM$ meets $\omega$ again at $P$. Line $PE$ meets $\omega$ again at $Q$. Let $K$ be the orthogonal projection of $E$ on the line $BC$. Line $QK$ meets $\omega$ again at $G$. Prove that $GA\perp GH$.
3 replies
buratinogigle
4 hours ago
luutrongphuc
12 minutes ago
Calculus
youochange   13
N 13 minutes ago by wh0nix
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
13 replies
youochange
Yesterday at 2:38 PM
wh0nix
13 minutes ago
No more topics!
Strange angle condition and concyclic points
lminsl   127
N May 3, 2025 by reni_wee
Source: IMO 2019 Problem 2
In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle PP_1C=\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle CQ_1Q=\angle CBA$.

Prove that points $P,Q,P_1$, and $Q_1$ are concyclic.

Proposed by Anton Trygub, Ukraine
127 replies
lminsl
Jul 16, 2019
reni_wee
May 3, 2025
Strange angle condition and concyclic points
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G H BBookmark kLocked kLocked NReply
Source: IMO 2019 Problem 2
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lminsl
544 posts
#1 • 31 Y
Y by e_plus_pi, Hexagrammum16, abdelkrim, cauchyguess, char2539, Mr.Chagol, Seicchi28, a_simple_guy, Carpemath, FadingMoonlight, AlastorMoody, samuel, RAMUGAUSS, Davrbek, Vietjung, JustKeepRunning, poplintos, sabrinamath, OlympusHero, k12byda5h, bariboaa, math31415926535, HWenslawski, megarnie, TFIRSTMGMEDALIST, ImSh95, SADAT, GeoKing, Adventure10, deplasmanyollari, Rounak_iitr
In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle PP_1C=\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle CQ_1Q=\angle CBA$.

Prove that points $P,Q,P_1$, and $Q_1$ are concyclic.

Proposed by Anton Trygub, Ukraine
This post has been edited 1 time. Last edited by djmathman, Jul 18, 2019, 4:40 AM
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lminsl
544 posts
#4 • 20 Y
Y by TUGUMEANDREW, Assassino9931, jeteagle, Wizard_32, pavel kozlov, JustKeepRunning, HolyMath, Gumnaami_1945, jnzm02, myh2910, Kagebaka, k.ata.tkm, AllanTian, Illuzion, ImSh95, rayfish, Adventure10, Mango247, ihatemath123, Rounak_iitr
Let $R$ be the intersection of $AA_1$, $BB_1$. Let $\Omega$ denote the circumcircle of triangle $ABC$, and let $A_0=AA_1 \cap \Omega, B_0=BB_1 \cap \Omega$.

Step 1. The points $P,Q, A_0, B_0$ are concyclic.
Since $PQ \parallel BC$, $\angle PQR=\angle ABB_0=\angle PA_0B_0$. The conclusion follows.

Step 2. The points $P,Q, B_0, P_1$ are concyclic, hence $P_1$ and $Q_1$ lies in $\odot(PQA_0B_0)$.

Note that $\angle B_1B_0C=\angle BAC=\angle B_1P_1C$, or $(B_0B_1P_1C)$ are concyclic. Hence $$\angle B_0P_1P=\angle B_1CB_0=\angle ABB_0=\angle PQB_0,$$or $P, Q, B_0, P_1$ are cyclic. This concludes the proof. $\square$
This post has been edited 3 times. Last edited by lminsl, Jul 16, 2019, 3:51 PM
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mofumofu
179 posts
#5 • 14 Y
Y by Flash_Sloth, zhangruichong, jeteagle, Wizard_32, JustKeepRunning, HolyMath, Not_real_name, ImSh95, ike.chen, Quidditch, CyclicISLscelesTrapezoid, SADAT, Adventure10, Rounak_iitr
[asy]
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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 = -7.63900015685, xmax = 14.779588377, ymin = -2.38665690592, ymax = 8.57946862605;  /* image dimensions */

 /* draw figures */
draw((1.,4.96)--(-0.798898966671,-0.708008219337)); 
draw((1.,4.96)--(7.68847429913,-0.833514663193)); 
draw((7.68847429913,-0.833514663193)--(-0.798898966671,-0.708008219337)); 
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Let $AA_1,BB_1$ intersect the circumcircle of $\triangle ABC$ again at $D,E$ respectively, and $PB_1,QA_1$ meet $BC$ again at $R,S$ respectively. It is clear that $A,R,C,P_1$ and $B,S,C,Q_1$ are cyclic. By Reim's theorem, we know that $P,Q,D,E$ are cyclic. Taking power of a point at $B_1$, we get $B_1B\cdot B_1E=B_1A\cdot B_1C=B_1R\cdot B_1P_1$. Also, since $\triangle B_1PQ\sim \triangle B_1RB$, we get $B_1B\cdot B_1P=B_1R\cdot B_1Q$. Combining the two gives us $B_1P_1\cdot B_1P= B_1Q\cdot B_1E$, whence $P_1\in (PQDE)$. Similarly $Q_1\in (PQDE)$ are we are done.
This post has been edited 4 times. Last edited by mofumofu, Jul 17, 2019, 12:39 PM
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nguyenhaan2209
111 posts
#6 • 5 Y
Y by top1csp2020, jeteagle, ImSh95, Pranav1056, Adventure10
$PB_1$ cuts $CB$,$AB$ at $E$,$D$,$QA_1$ cuts $CA$,$AB$ at $G$,$F$. Apply Pappus for ${B_1GA}$ and ${A_1BE}$ we have $B_1B$-$A_1G$=$Q$, $B_1E$-$A_1A$=$P$, $GE$-$AB$ collinear or parallel but $PQ$//$AB$ thus $GE//AB$. It's obvious that $P_1$,$Q_1$ also lies on $(AGE)$ thus by Reim we have the conclusion!
This post has been edited 1 time. Last edited by nguyenhaan2209, Jul 16, 2019, 12:48 PM
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Seicchi28
252 posts
#7 • 6 Y
Y by a_simple_guy, jeteagle, Wizard_32, CrazyMathMan, ImSh95, Adventure10
Nice problem :)

Let $A_1Q \cap AC = X$, $B_1P \cap BC = Y$, and extension of $PQ$ meets $AC,BC$ at $M,N$ respectively.
From the condition of the problem, $MN||AB$.

Lemma $XY||AB||MN$.
Proof

Construct the circumcircle $\omega$ of $\bigtriangleup CXY$. Denote $P'$ as the second intersection of $YP$ and $\omega$. Since \[ \angle PP'C = \angle YP'C = \angle YXC = \angle BAC, \]therefore $P'=P_1$. Similarly, $Q'=Q_1$. So, $C,X,Y,P_1,Q_1$ are concyclic and
\[  \angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP.  \]
Done. $\blacksquare$
This post has been edited 2 times. Last edited by Seicchi28, Jul 16, 2019, 1:00 PM
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JustKeepRunning
2958 posts
#9 • 2 Y
Y by ImSh95, Adventure10
mofumofu wrote:
[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(15cm); 
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 = -7.63900015685, xmax = 14.779588377, ymin = -2.38665690592, ymax = 8.57946862605;  /* image dimensions */

 /* draw figures */
draw((1.,4.96)--(-0.798898966671,-0.708008219337)); 
draw((1.,4.96)--(7.68847429913,-0.833514663193)); 
draw((7.68847429913,-0.833514663193)--(-0.798898966671,-0.708008219337)); 
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draw(circle((0.279066234699,2.06933899636), 2.97920568808), green); 
draw(circle((1.75406921626,4.18630229154), 4.07948863865), blue); 
draw(circle((3.47180131167,1.0560363319), 4.62068545199), red); 
draw((3.88543194857,7.66473878747)--(2.88060830443,-0.762418678059)); 
draw((-2.28184506812,3.59163983096)--(1.27443840444,-0.738667551996)); 
draw((-1.14173984034,1.31288664799)--(7.68847429913,-0.833514663193)); 
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draw((0.386107411678,0.343008384736)--(3.00779285736,0.304240392932)); 
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dot((1.27443840444,-0.738667551996),dotstyle); 
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dot((2.88060830443,-0.762418678059),dotstyle); 
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dot((-2.28184506812,3.59163983096),dotstyle); 
label("$P_1$", (-2.22653476554,3.68471731564), NE * labelscalefactor); 
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label("$Q_1$", (3.9546575944,7.76367674098), NE * labelscalefactor); 
dot((-1.14173984034,1.31288664799),dotstyle); 
label("$E$", (-1.08128846535,1.40991302074), NE * labelscalefactor); 
dot((5.73321976419,5.08551776613),dotstyle); 
label("$D$", (5.7901893358,5.17510633644), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Let $AA_1,BB_1$ intersect the circumcircle of $\triangle ABC$ again at $D,E$ respectively, and $PB_1,QA_1$ meet $BC$ again at $R,S$ respectively. It is clear that $A,R,C,P_1$ and $B,S,C,Q_1$ are cyclic. By Reim's theorem, we know that $P,Q,D,E$ are cyclic. Taking power of a point at $B_1$, we get $B_1B\cdot B_1E=B_1A\cdot B_1C=B_1R\cdot B_1P_1$. Also, since $\triangle B_1PQ\sim \triangle B_1RB$, we get $B_1B\cdot B_1P=B_1R\cdot B_1Q$. Combining the two gives us $B_1P_1\cdot B_1P= B_1Q\cdot B_1E$, whence $P_1\in (PQDE)$. Similarly $Q_1\in (BQDE)$ are we are done.

How do you import a graph form geogebra?
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Wictro
119 posts
#10 • 3 Y
Y by ImSh95, Adventure10, Mango247
Seicchi28 wrote:
Nice problem :)
Let $A_1Q$ meet $AC$ at $X$, $B_1P$ meet $BC$ at $Y$, extension of $PQ$ meets $AC,BC$ at $M,N$ respectively.
From the question, $MN||AB$.

Lemma

Now by the lemma, create the circumcircle $\omega$ of $\bigtriangleup CXY$. Denote $P'$ as the second intersection of $YP$ and $\omega$. Since \[ \angle PP'C = \angle YP'C = \angle YXC = \angle BAC, \]therefore $P'=P_1$. Similarly, $Q'=Q_1$. So, $C,X,Y,P_1,Q_1$ are concyclic and
\[  \angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP.\]Done.

Exactly how I did it. Though I proved the lemma with Pappus and then Desargues on perspective triangles.
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TUGUMEANDREW
5 posts
#13 • 2 Y
Y by ImSh95, Adventure10
Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry?
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richrow12
411 posts
#14 • 2 Y
Y by ImSh95, Adventure10
Seicchi28 wrote:
Nice problem :)

Let $A_1Q \cap AC = X$, $B_1P \cap BC = Y$, and extension of $PQ$ meets $AC,BC$ at $M,N$ respectively.
From the condition of the problem, $MN||AB$.

Lemma $XY||AB||MN$.
Proof

Construct the circumcircle $\omega$ of $\bigtriangleup CXY$. Denote $P'$ as the second intersection of $YP$ and $\omega$. Since \[ \angle PP'C = \angle YP'C = \angle YXC = \angle BAC, \]therefore $P'=P_1$. Similarly, $Q'=Q_1$. So, $C,X,Y,P_1,Q_1$ are concyclic and
\[  \angle Q_1P_1P = \angle Q_1P_1Y = \angle Q_1XY = \angle Q_1QP.  \]
Done. $\blacksquare$

Just the same but my proof of the lemma as follows:

Consider two lines $(P, B_1, Y)$ and $(Q, A_1, X)$. By Pappus's theorem points $S=PA_1\cap QB_1$, $T=PX\cap QY$ and $C=B_1Y\cap A_1X$ are collinear. Now, consider trinagles $APX$ and $BQY$. We claim that they are perspective. Indeed, $AP\cap BQ = S$, $AX\cap BY=C$ and $PX\cap QY=T$ are colinear. Hence, by Desargues's theorem lines $AB$, $PQ$ and $XY$ are concurrent. Since $AB\parallel PQ$ we have $AB\parallel PQ\parallel XY$.
TUGUMEANDREW wrote:
Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry?

Usually (at least in recent years) problems 1,2,4,5 have different topics (ACGN).
This post has been edited 1 time. Last edited by richrow12, Jul 16, 2019, 1:56 PM
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-[]-
97 posts
#15 • 5 Y
Y by richrow12, ImSh95, carefully, Adventure10, Mango247
Finally, the first medium geometry problem since the introduction of the new protocol! So surprised that they didn't vote for G1 this time.
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Math-wiz
6107 posts
#16 • 3 Y
Y by ImSh95, Adventure10, Mango247
TUGUMEANDREW wrote:
Since Problem 2 was geometry and problem 2 is medium level, is there a chance that Problem 4 won't be geometry?

P6 can be a Geo. P4,5 would be NT and Combi
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MS_Kekas
275 posts
#17 • 65 Y
Y by a_simple_guy, Wictro, richrow12, MathPassionForever, Mr.Chagol, rmtf1111, RudraRockstar, SHREYAS333, FadingMoonlight, AlastorMoody, UlanKZ, 62861, MNJ2357, GourmetSalad, Illuzion, Systematicworker, anantmudgal09, juckter, niyu, samuel, Anzoteh, Kayak, AlbertEinstein1905, jeteagle, Euler1728, Wizard_32, myh2910, Gumnaami_1945, Bassiskicking, Geronimo_1501, Abbas11235, aa1024, MarkBcc168, Aryan-23, amar_04, Atpar, k12byda5h, CHLORG1, CrazyMathMan, blackbluecar, 554183, tigerzhang, Bubu-Droid, Seicchi28, Wizard0001, rayfish, ImSh95, ike.chen, Pranav1056, Quidditch, parmenides51, IMUKAT, sabkx, wenwenma, CyclicISLscelesTrapezoid, kamatadu, Adventure10, Mango247, LoloChen, EpicBird08, Kingsbane2139, khina, ihatemath123, farhad.fritl, Sedro
Lol, this is my problem! Didn't expect to see this at P2
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AngleChasingXD
109 posts
#18 • 5 Y
Y by jeteagle, ImSh95, DashTheSup, Adventure10, Mango247
Beautiful problem!!! :)
My solution is kinda long, especially because I used Moving Points during it.

Let $B_1P_1$ cut $AB$ at $U$ and $Q_1A_1$ cut $AB$ at $V$. Moreover, let $L$ be the intersection point of lines $B_1P_1$ and $BC$ and $K$ the intersection point of lines $Q_1A_1$ and $AC$.

Since $\angle PP_1C\equiv \angle BAC$, we conclude that $AP_1CU$ is cyclic. Similarly, since $\angle QQ_1C\equiv \angle ABC$, $BQ_1CV$ is cyclic, too. Keep these in mind as (1).

The conclusion is equivalent to $\angle QPP_1\equiv \angle QQ_1P_1$. However, since $PQ\parallel AB$, we have $\angle QPP_1\equiv \angle AUP_1$, so it suffices to prove $\angle AUP_1\equiv \angle QQ_1P_1$, which is equivalent to $UVP_1Q_1$ being cyclic.

$\textbf{Claim.}$ We have that $KL\parallel AB$.

$\textbf{Proof:}$ We show this by the Moving Points method. Fix $\Delta ABC$, points $A_1$, $B_1$ and animate $P$ on $AA_1$. Consider $L'$ on $AB$ such that $KL'\parallel AB$. We prove that the map $$L\to P\to Q\to K\to L'~~by~~\overline{BC}\to \overline{AA_1}\to \overline{BB_1}\to \overline{AC}\to \overline{BC}$$is projective, thus it would ve enough to check $L=L'$ for $3$ distinct positions of $P$.
$\textbf{(1)}$ $L\to P$ by $\overline{BC}\to \overline{AA_1}$ is projective since it is a perspectivity through $B_1$ from $BC$ to $AA_1$.
$\textbf{(2)}$ $P\to Q$ by $\overline{AA_1}\to \overline{BB_1}$ is projective since it preserves linear motion.
$\textbf{(3)}$ $Q\to K$ by $\overline{BB_1}\to \overline{AC}$ is projective since it is a perspectivity through $A_1$ from $BB_1$ to $AC$.
$\textbf{(4)}$ $K\to L'$ by $ \overline{AC}\to \overline{BC}$ is projective since it preserves linear motion.

Now it suffices to check for $3$ distinct positions of $P$ the desired parallelism.
$\textbf{(1)}$ When $P\in AA_1\cap BB_1$, we actually have $Q=P$, $L=B$ and $K=A$ so there is nothing to prove.
$\textbf{(2)}$ When $P=A$, $Q=B$ so $C=K=L$, again nothing to prove.
$\textbf{(3)}$ When $P$ is the point at infinity on $\overline{AA_1}$, $Q$ is the point at infinity on $\overline{BB_1}$ so $B_1L\parallel AA_1$ and $A_1K\parallel BB_1$. Then apply Thales' Lemma to obtain $CB_1\cdot CA_1=CK\cdot CB$ and $CA_1\cdot CB_1=CL\cdot CA$, hence $CK\cdot CB=CL\cdot CA$ so indeed $KL\parallel BC$.$\blacksquare$

Now we are ready to finish. From (1) and Power of a Point Theorem we get $PB_1\cdot B_1U=AB_1\cdot CB_1$. By our lemma and Thales' Lemma, $B_1K\cdot B_1U=B_1L\cdot B_1A$. Divide these two last relations in order to get $PB_1\cdot B_1L=CB_1\cdot B_1K$, which again by Power of a Point leads to $CP_1KL$ being cyclic. In a similar manner one proves the cyclicity of $CQ_1KL$, so the pentagon $CP_1KLQ_1$ is cyclic. But then $\angle KQ_1P_1\equiv \angle KP_1C$, and since by (1) we have $\angle KCP_1\equiv \angle AUP_1$, we arrive at $\angle AUP_1\equiv \angle VQ_1P_1$, hence $UVQ_1P_1$ is cyclic and done.
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prague123
230 posts
#19 • 2 Y
Y by ImSh95, Adventure10
-[]- wrote:
Finally, the first medium geometry problem since the introduction of the new protocol! So surprised that they didn't vote for G1 this time.

This probably just means that G1 and G2 got eliminated as known problems, and that G3 contained traces of combinatorial geometry.
This post has been edited 1 time. Last edited by prague123, Jul 16, 2019, 2:13 PM
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Math-wiz
6107 posts
#20 • 2 Y
Y by ImSh95, Adventure10
MS_Kekas wrote:
Lol, this is my problem! Didn't expect to see this at P2

Wowww. Where are you from?
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