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In convex cyclic quadrilateral $ABCD$, we know that lines $AC$ and $BD$ intersect at $E$, lines $AB$ and $CD$ intersect at $F$, and lines $BC$ and $DA$ intersect at $G$. Suppose that the circumcircle of $\triangle ABE$ intersects line $CB$ at $B$ and $P$, and the circumcircle of $\triangle ADE$ intersects line $CD$ at $D$ and $Q$, where $C,B,P,G$ and $C,Q,D,F$ are collinear in that order. Prove that if lines $FP$ and $GQ$ intersect at $M$, then $\angle MAC = 90^\circ$.

My Progress:

enter image description here

Claim : $PBQD$ is cyclic

Proof: Note that $CQ\cdot CD=CE\cdot CA=CB\cdot CP \implies PQDB$ is cyclic.

Claim: $APQC$ is cyclic

Proof : angle chase! Note that for this to be true , it is enough to show that $\angle AEB=\angle AQC$ or it is enough to show that $\angle AEB=\angle AQC $ or it is enough to show that $\angle AED=\angle AQD$ which is true since $AEDQ$ is cyclic.

Claim: $E\in PQ$

Proof: So enough to show that $\angle AEQ+\angle AEP=180 $

or enough to show that $180- \angle ADC + \angle AEP=180 $

or enough to show that $\angle ADC= \angle ABC$ , which is true since $ABCD$ is cyclic.

after that I am stuck.

I observed that $FG , AM, PQ$ concur but was not able to prove. Can someone give hints?

Thanks in advance.

enter image description here

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    $\begingroup$ As a beginner in Olympiad math just seeing 5+ circles in one diagram makes me anxious $\endgroup$
    – Amadeus
    Commented Aug 21, 2020 at 17:19
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    $\begingroup$ @l1mbo lol, even I was like you , but now, the more the circles in the diagram , more I like the problem :) $\endgroup$ Commented Aug 22, 2020 at 0:21
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    $\begingroup$ Have you guys learned calculus, linear algebra and simillar topics? $\endgroup$
    – 1b3b
    Commented Aug 22, 2020 at 0:31
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    $\begingroup$ @1b3b no , not in school atleast and I know only basic calculus.. $\endgroup$ Commented Aug 22, 2020 at 1:18
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    $\begingroup$ @Shubhangi, yes. I am interested in what math competitors learn in other countries. For example, I qualified for the state (Croatia) competition this year, but I learned Calculus 1 and basics of Calculus 2 on my own this summer (when I said learned I meant on understanding all rules, theorems, etc.). Also a lot of other topics are not even mentioned before college. But I think olympiad approach is better because it develops problem solving skills which is ground for math research $\endgroup$
    – 1b3b
    Commented Aug 22, 2020 at 11:03

1 Answer 1

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Continuing from where you finished...

So, we have $PBDQ$ cyclic and $E\in PQ$. Now focus on quadrilateral $PBDQ$. From definition $A$ is the Miquel Point of the quadrilateral $PBDQ$. Now let $X:=PD\cap BQ$ and thus, by Miquel point properties, we get that $A$ is projection of $X$ on $CE$. Thus, its enough to show that $M,A,X$ are collinear but this is trivial. Just apply Pappus Theorem on $\{PGB,QFD\}$ completing the proof. $\blacksquare$

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    $\begingroup$ Thank you so much ! I saw your profile , OMG , congrats for INMO, sharygin and IGO . You are so Pro !!! Can you give me some Geo and NT advice , please ? Plus Can I pm you in AOPS if I get any doubt ? $\endgroup$ Commented Aug 22, 2020 at 13:21
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    $\begingroup$ @Shubhangi Thanks but I'm a noob! My AoPS id is Jupiter_is_BIG what's yours? $\endgroup$
    – Anand
    Commented Aug 22, 2020 at 13:38
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    $\begingroup$ mine is Jelena_ivanchic :) . But looks like due to some reasons I can't PM you .. Anyways Thanks for your solution . $\endgroup$ Commented Aug 22, 2020 at 13:51
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    $\begingroup$ @Shubhangi GOD, you won't believe but I got two major hints for that... 1) You put a space in front and after every comma or question mark or exclamation mark 2) You are the only AoPSer who know bout SOR (expect Muler) $\endgroup$
    – Anand
    Commented Aug 22, 2020 at 13:54

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