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#1 h Apr 6, 2025, 11:27 AM • 2 Y

Y by PikaPika999, Rounak_iitr

m:}
Let $\triangle ABC$ be a triangle inscribed in a circle, where the tangents to the circle at points $B$ and $C$ intersect at the point $P$ . Let $M$ be a point on the arc $AC$ (not containing $B$ ) such that $M \neq A$ and $M \neq C$ . Let the lines $BC$ and $AM$ intersect at point $K$ . Let $P'$ be the reflection of $P$ with respect to the line $AM$ . The lines $AP'$ and $PM$ intersect at point $Q$ , and $PM$ intersects the circumcircle of $\triangle ABC$ again at point $N$ .

Prove that the point $Q$ lies on the circumcircle of $\triangle ANK$ .

This post has been edited 1 time. Last edited by youochange, Apr 6, 2025, 11:28 AM
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#2 h Apr 6, 2025, 12:40 PM • 1 Y

Y by PikaPika999

Bump

:first:

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#3 h Apr 6, 2025, 2:10 PM • 1 Y

Y by PikaPika999

Helpmmmmmmme

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#4 h Apr 6, 2025, 5:24 PM • 3 Y

Y by PikaPika999, RANDOM__USER, youochange

Try complex bashing:

Take $(ABC)$ to be the unit circle and WLOG suppose $m=1$ .

Denote by $S=AP\cap(ABC), S\neq A$ and $R=AP'\cap (ABC), R\neq A$ .

Since $P'$ is the reflection of $P$ over $AM\implies \widehat{SAM}=\widehat{MAR}\implies M$ is the midpoint of arc $\widehat{RS}$ , so $r\cdot s=m^2=1\implies r=\frac{1}{s}=\overline{s}$ .

Compute $p=\frac{2bc}{b+c}$ and $s=\frac{ab+ac-2bc}{2a-b-c}$ , so $r=\frac{b+c-2a}{2bc-ab-ac}$ .

Since $M, N, P$ are collinear and $|m|=|n|=1\implies -mn=\frac{p-m}{\overline{p}-\overline{m}}\implies n=\frac{2bc-b-c}{b+c-2}$ .

$Q=AR\cap MN\implies q=\frac{ar(m+n)-mn(a+r)}{ar-mn}=\frac{a(b+c-2a)(2bc-2)-(2bc-b-c)(2abc-a^2b-a^2c+b+c-2a)}{a(b+c-2a)(b+c-2)-(2bc-b-c)(2bc-ab-ac)}=$
$=\frac{(2a+2bc-ab-ac-b-c)(ab+ac+2a-2abc-b-c)}{2(a-bc)(2a+2bc-ab-ac-b-c)}=\frac{ab+ac+2a-2abc-b-c}{2(a-bc)}$ .

Compute $K=AM\cap BC\implies k=\frac{am(b+c)-bc(a+m)}{an-bc}=\frac{ab+ac-bc-abc}{a-bc}$ .

Now, to prove that $A, N, Q, K$ are concyclic, we need to prove that $\frac{a-n}{a-q}\cdot\frac{k-q}{k-n}\in\mathbb{R}$ .

$a-n=\frac{ab+ac+b+c-2a-2bc}{b+c-2}$

$a-q=\frac{2a^2-ab-ac-2a+b+c}{2(a-bc)}=\frac{(a-1)(2a-b-c)}{2(a-bc)}$

$k-q=\frac{ab+ac+b+c-2a-2bc}{2(a-bc)}$

$k-n=\frac{(ab+ac-bc-abc)(b+c-2)-(2bc-b-c)(a-bc)}{(a-bc)(b+c-2)}=\frac{(b-1)(c-1)(2bc-ab-ac)}{(a-bc)(b+c-2)}$

So $\frac{a-n}{a-q}\cdot\frac{k-q}{k-n}=\frac{(a-bc)(ab+ac+b+c-2a-2bc)^2}{(a-1)(b-1)(c-1)(2a-b-c)(2bc-ab-ac)}$ , which is real by conjugating, so we're done.

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#5 h Apr 6, 2025, 9:08 PM • 2 Y

Y by PikaPika999, youochange

Hmm, very interesting problem, sadly I only have minor results that might be useful.

Claim 1: If $D$ is the midpoint of $BC$ , then $ADNK$ is cyclic.
Proof: We intersect $AN$ with $BC$ at a point $F$ . Then because $BCMN$ is harmonic (the tangents from $B$ and $C$ intersect on $NM$ ) it must be that if we project this harmonic quad from $A$ onto $BC$ that $(B,C;F,K)=-1$ . Now using a very well known property of harmonic sets, we know that $BF \cdot FC = FD \cdot FK$ . However, due to PoP we know that $BF \cdot FC = AF \cdot FN$ , thus $AF \cdot FN = FD \cdot FK$ , meaning that, indeed $ADNK$ is cyclic. $\square$

Now for another cool observation,

Claim 2: The problem is equivelent to showing that $PDQP'$ is cyclic.
Proof: Assume $PDQP'$ is cyclic, then $\angle{DQA} = \angle{P'PD}$ . If $X = PP' \cup AK$ , then $\angle{PXK} = \frac{\pi}{2}$ and $\angle{PDK} = \frac{\pi}{2}$ . Thus $\angle{DPP'} = \angle{DKA}$ and thus $\angle{DKA} = \angle{DQA}$ which means that $AQKD$ is cyclic. Taking into account the result that $ADNK$ is cyclic, we obtain that $ANKQ$ is cyclic. $\square$

And finally the last observation I think is note worthy is the following,

Claim 3: If $E$ is the intersection of $AP$ and $(ABC)$ , then $E, N, K$ and $E,F,M$ are colinear.
Proof: Quite trivial through harmonics and projective ideas. $\square$

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#6 h Apr 6, 2025, 10:54 PM • 2 Y

Y by RANDOM__USER, youochange

Person above is just done, $\angle QAK=\angle PAK=\angle KNM=\angle KNQ$ , as desired

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#7 h Apr 7, 2025, 4:45 AM • 2 Y

Y by RANDOM__USER, youochange

ig the simplest solution take $\sqrt bc$ inversion
Suppose $X$ is a point in the plane let after inversion it be $X'$
forgive me cuz I used $P'$ as a point after inversion of $P$ while there was $P'$ in the problem

So $P \rightarrow P'$ where $P'$ will be the $A-$ Humpty point
$M'$ will be a random point on $\overrightarrow{CB}$
$N'=(M'AP')\cap BC$ , $P''=$ reflection of $P'$ in $AM'$ , $Q'=AP''\cap (M'AP')$ and $K'=AM'\cap (ABC)$
We need to prove that $Q'-K'-N'$
Let $E$ be the reflection of $P'$ in $BC$ see that $E$ will lie on $(ABC)$

We will use phantom points here
Let $K^*=AM'\cap EN'$
and Let $D$ be the midpoint of $BC$
So-
$\angle EN'D=\angle DN'P'=\angle M'AP'$ $\implies (AK^*N'D)$ $\implies \angle AK^*E =\angle AK^*N'=\angle ADC=\angle ABC+ \angle BAD=\angle EBC$ $\therefore K^*\equiv K'$ $\therefore \angle M'N'K'=\angle K'AP'=\angle M'AQ'$ $\blacksquare$

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#8 h Apr 7, 2025, 7:00 AM • 2 Y

Y by RANDOM__USER, youochange

unless im seriously mistaken, the simplest solution is using projective

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#9 h Apr 7, 2025, 7:38 AM • 1 Y

Y by youochange

Yea, that seems to be the correct solution! That is essentially my solution in addition to the comment that I somehow didn't notice to finish of my solution

This post has been edited 1 time. Last edited by RANDOM__USER, Apr 7, 2025, 7:39 AM

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