IMO Shortlist 2010 - Problem G1

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5452 posts

#1 h Jul 17, 2011, 2:10 AM • 16 Y

Y by -[]-, sandeepreddy, siavosh, DIGGER1, jam10307, Davi-8191, mathlearner2357, donotoven, megarnie, Adventure10, Mango247, ItsBesi, buddyram, Rounak_iitr, Sadigly, namanrobin08

Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom

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Jutaro

388 posts

Jutaro

#2 h Jul 17, 2011, 3:22 AM • 7 Y

Y by DIGGER1, jam10307, Amir Hossein, karitoshi, Adventure10, Mango247, Rounak_iitr

By construction $ABCP$ is cyclic so that $\angle APB=\angle BCA$ . On the other hand since $AD$ and $CF$ are altitudes $ACDF$ is also cyclic, whence $\angle BFD=\angle BCA$ . Putting this together one gets that $\angle APQ=\angle BFQ$ , that is, $APQF$ is cyclic.

Now by angle chasing $\angle AFE = \angle BFD = \angle BCA$ , and using the previous fact gives $\angle AQP = \angle AFP =\angle BCA$ . Thus $APQ$ is isosceles with $AP=AQ$ .

This was my argument for the case when $Q$ lies inside the triangle. When it is outside you do basically the same.

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Jutaro

388 posts

Jutaro

#3 h Jul 17, 2011, 3:55 AM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Sorry to double post, I just realized the cases can be avoided by using directed angles from the beginning: $\angle AFQ = \angle AFD = \angle ACD = \angle ACB \pmod \pi$ and $\angle APQ=\angle APB = \angle ACB \pmod \pi$ , so $\angle APQ = \angle AFQ \pmod \pi$ . But $\angle AQP = \angle AFP = \angle EFB= \angle ECB = \angle ACB \pmod \pi$ . Therefore $\angle APQ = \angle AQP \pmod \pi$ .

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dragon96

3212 posts

dragon96

#4 h Jul 17, 2011, 4:54 AM • 11 Y

Y by meowme, siavosh, Amir Hossein, KamalDoni, AoPS_0829, ike.chen, Adventure10, Mango247, Math_.only., and 2 other users

Hmm, this looks familiar...

$[asy] import olympiad; defaultpen(linewidth(0.65)); defaultpen(fontsize(10)); size(250); pair C=origin, A=(5,12), B=(14,0), D=foot(A,B,C), E=foot(B,A,C), F=foot(C,A,B), H=orthocenter(A,B,C), om=extension(E,F,C,B), nom=extension(E,F,D,(9,3)); path circ = circumcircle(A,B,C); pair P1=intersectionpoint(F--om, circ), P2=intersectionpoint(E--nom, circ), Q1=intersectionpoint(D--F, B--P1), Q2=extension(B,P2,D,F), Ep=reflect(A,B)*E, Cp=reflect(A,B)*C, Hp=reflect(A,B)*H, Dp=reflect(A,B)*D; draw(A--B--C--A--D--F--C); draw(B--E); draw(P1--P2); draw(F--Q2); draw(P1--B--Q2, dotted); dot(A^^B^^C^^D^^E^^F^^Q1^^Q2^^H^^P1^^P2); draw(circ); markscalefactor=0.05; draw(rightanglemark(C,F,A)); draw(rightanglemark(B,E,C)); draw(rightanglemark(A,D,B)); label("$A$", A, N); label("$B$", B, SE); label("$C$", C, SW); label("$D$", D, S); label("$E$", E, NW); label("$F$", F, dir(340)); label("$H$", H, E); label("$P_1$", P1, W); label("$P_2$", P2, NE); label("$Q_1$", Q1, dir(290)); label("$Q_2$", Q2, N); [/asy]$
Solution

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arshakus

769 posts

arshakus

#5 h Jul 27, 2011, 12:50 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

$\angle{C}=\angle{APB}=\angle{BFC}=a=\angle{AEC}=>APQF$ is cyclic.
$\angle{QAP}=\angle{DFP}=180-2a=>\angle{AQP}=a=>AP=AQ$

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mavropnevma

15142 posts

mavropnevma

#6 h Jul 27, 2011, 1:21 PM • 19 Y

Y by jatin, biomathematics, Not_a_Username, Amir Hossein, AlastorMoody, OlympusHero, Adventure10, and 12 other users

Let's correct the above, since it is indeed a short proof (although basically the proof at #2).

$\angle{C}=\angle{APB}=\angle{BFD}=a=>APQF$ is cyclic.
$\angle{QAP}=\angle{DFP}=180-2a=>\angle{AQP}=a=>AP=AQ$

EDIT. It seems some of our readers think this was not only unneeded, but spammish. Indeed, why take the trouble to correct some patently wrong mathematical statements, or mention the similarity with some already posted solution? Some of our users like it muddy, the same their own thinking processes are ...

This post has been edited 1 time. Last edited by mavropnevma, Aug 9, 2011, 2:46 PM

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Pascal96

124 posts

Pascal96

#7 h Aug 9, 2011, 2:25 PM • 9 Y

Y by Adventure10, Mango247, Math_DM, ehuseyinyigit, AlexRuiz, and 4 other users

Quite a simple angle chase. Can't believe this made an IMO shortlist!

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Bertus

37 posts

Bertus

#8 h Aug 15, 2011, 6:04 AM • 3 Y

Y by leru007, Adventure10, Mango247

Since : $\angle{PQF}=\angle{BQD}=\pi-\angle{PBC}-\angle{FBD}=\pi-\angle{PAC}-\angle{BAC}=\pi-\angle{PAF}$ . Hence $APQF$ is cyclic, and then $\angle{AQP}=\angle{AFE}=\angle{ACB}=\angle{APB}=\angle{APQ}$ , and so we are done !

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SKhan

297 posts

SKhan

#9 h Aug 18, 2011, 12:26 PM • 5 Y

Y by Amir Hossein, Adventure10, and 3 other users

My Solution referring to dragon96's diagram

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dgreenb801

1896 posts

dgreenb801

#10 h Sep 22, 2011, 2:50 AM • 2 Y

Y by Adventure10, Mango247

Here is my solution:
Let $O$ be the circumcenter of $\triangle ABC$ .
Let the other intersection of $EF$ with the circumcircle be $R$ .
Then it is well known that $AO \perp EF$ , so $AP=AR$ . Thus $\angle PBA= \angle ABR$ .
Also $\angle AFR= \angle BFD=\angle AFQ$ , so $\angle BFR= \angle BFQ$ .
So by $ASA$ congruency, $\triangle BFQ \cong \triangle BFR$ . Then $\triangle BAQ \cong \triangle BAR$ .
So $\triangle BAQ$ and $\triangle BAR$ have congruent circumcircles.
Since $AP$ and $AQ$ are both intercepted by $\angle ABQ$ in these circumcircles, $AP=AQ$ .

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exmath89

2572 posts

exmath89

#11 h Jun 28, 2013, 11:17 PM • 2 Y

Y by Adventure10, Mango247

Solution

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IDMasterz

1412 posts

IDMasterz

#12 h Jun 29, 2013, 9:57 AM • 4 Y

Y by Amir Hossein, mikestro, Adventure10, Mango247

Looks projective...

Lemma 1: $PAFQ$ is cyclic, with circle $w$ .

$\angle APQ = \angle APB = \angle ACB = \angle BFD$ where the last follows from anti-parallel $DF$ wrt $\angle ABC$ .

Lemma 2: Let $FC \cap BP = K$ , then $(P, Q; K, B)$ is harmonic.

Let $EF \cap BC = X$ , so $(X, D; C, B) = -1 \implies F(X, D; C, B) = -1 \implies (P, Q; K, B) = -1$ .

Proof: Take the pencil $F(P, Q; K, B)$ and intersect it with $w$ to get $PQA'A$ is a harmonic quadrilateral, where $A' = FK \cap w$ . Since $\angle KFB = 90 \implies KF$ bisects $\angle PFQ$ , so $A'$ is the midpoint of arc $PQ$ , and thus $A$ is the midpoint of the supplementary arc $PQ$ , so $AP = AQ$ .

Truth be told, Jutaro's is the one I saw first

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bonciocatciprian

41 posts

bonciocatciprian

#13 h Jun 17, 2014, 8:22 AM • 1 Y

Y by Adventure10

Take $P'$ the second point of intersection of $EF$ with the circle. We will prove that $Q$ is the mirror image of $P'$ about $AB$ . To do this, we take $Q'$ the mirror image of $P'$ about $AB$ , and try to prove that is equivalent to $Q$ . Firstly, we show that $Q' \in BP \Leftrightarrow \angle ABQ' \equiv \angle ABP \Leftrightarrow \angle P'BA \equiv \angle ABP \Leftrightarrow \angle P'A \equiv \angle PA$ (as arcs of circles). But this is equivalent to $EF || t_A$ (the tangent through $A$ at the circumcircle), which is obvious ( $\angle AFE \equiv \angle ACB$ ). Now, to prove that $Q' \in FD$ : $Q' \in FD \Leftrightarrow \angle BFQ' \equiv \angle BFD \Leftrightarrow \angle BFP' \equiv \angle BFD \equiv \angle EFA$ , which is obvious. So, $Q \equiv Q'$ . Since $Q$ and $P'$ are symmetric, we have $AP' = AQ$ . Since $\angle AP' \equiv \angle AP$ , we also have $AP' = AP$ . Hence, the conclusion follows.

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AnonymousBunny

339 posts

AnonymousBunny

#14 h Jun 19, 2014, 5:31 AM • 2 Y

Y by Adventure10, Rounak_iitr

This is just angle chasing. Let $\angle PCB = \angle PAC = \theta.$ We have that $\angle QFB = C$ and $\angle FBP = B - \theta,$ so $\angle FQB = A + \theta = \angle BAC + \angle CAP = \angle BAP,$ so $APQF$ is cyclic. It follows that $\angle AQP = C.$ From cyclic $ABCP,$ we have that $\angle APB = C,$ so $\angle APQ = \angle AQP,$ which implies $\triangle APQ$ is isoceles with $AP=AQ.$ $\blacksquare$

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sayantanchakraborty

505 posts

sayantanchakraborty

#15 h Jun 19, 2014, 4:25 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

This problem,I should say,is a really easy geometry problem.First of all,we have $\angle{APQ}=\angle{APB}=\angle{ACB}=\angle{BFD}=\angle{BFQ} \Rightarrow APQF$ is cyclic.Finally $\angle{AQP}=\angle{AFP}=\angle{AFE}=\angle{ACB}=\angle{APQ} \Rightarrow AP=AQ %Error. "Blackbox" is a bad command.$ .

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Jishnu4414l

154 posts

Jishnu4414l

#124 h Apr 26, 2024, 12:10 PM

Y by

Very easy even for G1

@above, Ig you need to use directed angles or prove for all possible configurations...

This post has been edited 2 times. Last edited by Jishnu4414l, Apr 26, 2024, 12:11 PM

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P2nisic

406 posts

P2nisic

#125 h Jun 23, 2024, 8:28 AM

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Amir Hossein wrote:

Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom

$\angle PQF=\angle BQD=180-\angle QBD-\angle BDQ=180-\angle B-\angle A-\angle PBA=\angle C-\angle PBA=\angle PAF$
So $P,Q,F,A$ are cyclic.

$\angle QPA=\angle QFA=180-\angle AFD=180-(180-\angle C)=\angle C=\angle AFE=\angle PQA$
Which gives $AQ=AP$

Attachments:

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Mathandski

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Mathandski

#126 h Aug 23, 2024, 10:03 PM

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

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ezpotd

1260 posts

ezpotd

#127 h Aug 31, 2024, 6:15 PM

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We claim that $(AFPQ)$ is cyclic, this is trivial by observing $\angle QPA = 180 - \angle BPA = \angle ACB = \angle BFD = \angle AFQ$ . Now we see that $\angle AQP = 180 - \angle AFP = 180 - \angle BFD = \angle ACB = 180 - \angle BPA = \angle QPA$ , so we are done.

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little-fermat

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little-fermat

#128 h Oct 11, 2024, 7:45 PM

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I have discussed this problem in my EGMO YouTube tutorial ch1 angle chasing practice part

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ehuseyinyigit

810 posts

ehuseyinyigit

#129 h Oct 30, 2024, 3:04 PM

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I will show the case where point $Q$ is inside of the triangle $ABC$ , other situation can be shown similarly.

Since $\angle APQ=\angle C$ and $\angle AFQ=90^{\circ}+\angle CFD=90^{\circ}+\angle CAD=180^{\circ}-\angle C$ , we have that the points $A$ , $P$ , $F$ and $Q$ are concyclic.

Now, observe that we wish to show $\angle APQ=\angle AFP$ which is equilavent to $\angle C=\angle AFP=90^{\circ}-\angle CAD=\angle C$ as desired.

This post has been edited 1 time. Last edited by ehuseyinyigit, Oct 30, 2024, 3:05 PM

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Vedoral

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Vedoral

#130 h Dec 2, 2024, 2:55 AM

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Click to reveal hidden text

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g0USinsane777

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g0USinsane777

#131 h Dec 30, 2024, 6:11 PM

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Since $APBC$ , $BFEC$ and $AFDC$ are cyclic, we have that $\angle AFQ = \angle BFD = \angle ACD = \angle ACB = \angle APQ$ , which means that $AFPQ$ is cyclic. Then, because of the above cyclic quadrilaterals, we have that $\angle AQP = \angle AFE = \angle ACB = \angle AFQ = \angle APQ$ , giving that triangle $APQ$ is isosceles.
The above proof is for $Q$ outside triangle $ABC$ . An analogous proof follows for $Q$ inside $ABC$ .

This post has been edited 1 time. Last edited by g0USinsane777, Dec 30, 2024, 6:13 PM

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eg4334

633 posts

eg4334

#132 h Dec 31, 2024, 10:38 PM

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Assume $P$ is closer to $E$ , the other config is the same solution. $AFQP$ is cyclis because $\angle QPA = 180 - \angle QFA$ . Now $\angle AQP = \angle AFP = \angle AFE = \angle C$ but $\angle APQ = \angle C$ as well.

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Maximilian113

560 posts

Maximilian113

#133 h Jan 7, 2025, 5:48 AM

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Suppose that $APQF$ forms a non-self-intersecting quadrilateral. Then $\angle APQ = \angle ACB = \angle AFE$ so it suffices to show that $APQF$ is cyclic. But this is easy since $\angle APQ = \angle ACB = 180^\circ - \angle AFD,$ completing the proof. Note that the case when $AQPF$ is cyclic can be dealt with similarly. QED

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thdwlgh1229

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thdwlgh1229

#135 h Feb 19, 2025, 5:50 AM

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Its too easy for IMO Shortlist!

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Ilikeminecraft

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Ilikeminecraft

#137 h Feb 27, 2025, 7:51 AM

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Note that $\angle APQ = \angle C = \angle DFB,$ so $AFQP$ is cyclic. Hence, $\angle AQP = \angle AFP = \angle C = \angle APQ$ so we are done.

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blueprimes

343 posts

blueprimes

#138 h Mar 3, 2025, 11:21 PM

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We use directed angles throughout this proof. Note that
$\angle AFQ = 180^\circ - \angle BFD = 180^\circ - \angle C = 180^\circ - \angle APQ$ so $APQF$ is cyclic. Thus
$\angle AQP = \angle ACE = \angle C = \angle APQ$ which finishes.

This post has been edited 1 time. Last edited by blueprimes, Mar 3, 2025, 11:21 PM

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LeYohan

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LeYohan

#139 h Mar 29, 2025, 6:30 PM

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We handle the two cases of the location of $P$ :
Let $P_{1}$ be the intersection of $EF$ and $(ABC)$ so that $F$ is between $E$ and $P_{1}$ , and $P_{2}$ be the intersection of $EF$ and $(ABC)$ so that $E$ is between $F$ and $P_{2}$ .

Case 1: $P \equiv P_{1}$ .

Claim: $AFPQ$ is cyclic.

Proof:
We notice that $AFDC$ is cyclic because $\angle AFC = \angle ADC = 90$ implying that $\angle AFQ = \angle C$ , and similarly $\angle APQ = \angle C$ so we're done. $\square$

$BFEC$ is cyclic as $\angle BFC = \angle BEC = 90$ meaning that $\angle AFE = \angle C \implies \angle PFA = 180 - \angle C$ , and remembering that $AFPQ$ is cyclic we obtain that $\angle AQP = \angle APQ = \angle C$ , so $\triangle APQ$ is isosceles with $AP = AQ$ , as desired. $\square$

Case 2: $P \equiv P_{2}$ .

Claim: $AFPQ$ is cyclic.

Proof:
From the previous case we get that $AFDC$ is cyclic meaning that $\angle BFD = \angle C = \angle QPA$ and we're done. $\square$

From the previous case we also get that $BFEC$ is cyclic so $\angle C = \angle AFP = \angle AQP = \angle QPA$ , so $\triangle AQP$ is isosceles with $AQ = AP$ and after covering both configurations we're finally done. $\square$

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Avron

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Avron

#140 h Apr 11, 2025, 1:00 PM

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$\textbf{Case 1}$ : $P$ lies on the arc $AC$ not containing $B$ .

Notice that $\angle QDC=180 - \angle A = 180 - \angle QPC$ so $QPCD$ is cylic, thus $BQ\cdot BP = BD \cdot BC = BF \cdot BA$ so $AFQP$ is also cyclic. Now $\angle AQP = \angle AFP = \angle C = \angle APQ$ as desired.

$\textbf{Case 2}$ : $P$ lies on the arc $AC$ containing $B$ .

Similarly, note that $\angle QDC = 180 - \angle A = 180 - \angle BPC = \angle QPC$ so $QPDC$ is cylic and $BP\cdot BQ = BD \cdot BC = BF \cdot BA$ , thus $QPFA$ is also cyclic so $\angle QPA = \angle QFA = \angle BFD = \angle C = \angle AFE = \angle PQA$ and we're done.

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