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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
2023 Japan Mathematical Olympiad Preliminary
parkjungmin   0
20 minutes ago
Please help me if I can solve the Chinese question
0 replies
parkjungmin
20 minutes ago
0 replies
Polynomial divisible by x^2+1
Miquel-point   1
N an hour ago by luutrongphuc
Source: Romanian IMO TST 1981, P1 Day 1
Consider the polynomial $P(X)=X^{p-1}+X^{p-2}+\ldots+X+1$, where $p>2$ is a prime number. Show that if $n$ is an even number, then the polynomial \[-1+\prod_{k=0}^{n-1} P\left(X^{p^k}\right)\]is divisible by $X^2+1$.

Mircea Becheanu
1 reply
Miquel-point
Apr 6, 2025
luutrongphuc
an hour ago
the same prime factors
andria   5
N an hour ago by bin_sherlo
Source: Iranian third round number theory P4
$a,b,c,d,k,l$ are positive integers such that for every natural number $n$ the set of prime factors of $n^k+a^n+c,n^l+b^n+d$ are same. prove that $k=l,a=b,c=d$.
5 replies
andria
Sep 6, 2015
bin_sherlo
an hour ago
A sharp estimation of the product
mihaig   0
an hour ago
Source: VL
Let $n\geq4$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$a_1+\cdots+a_n+2^{n-1}\geq n+2^{n-1}\cdot\prod_{i=1}^{n}{a_i}.$$
0 replies
mihaig
an hour ago
0 replies
No more topics!
Geometry
youochange   8
N Apr 7, 2025 by RANDOM__USER
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 $\triangle ANK$.
8 replies
youochange
Apr 6, 2025
RANDOM__USER
Apr 7, 2025
Geometry
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youochange
174 posts
#1 • 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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youochange
174 posts
#2 • 1 Y
Y by PikaPika999
Bump :first:
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youochange
174 posts
#3 • 1 Y
Y by PikaPika999
Helpmmmmmmme
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Double07
85 posts
#4 • 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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RANDOM__USER
7 posts
#5 • 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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lolsamo
12 posts
#6 • 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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Captainscrubz
67 posts
#7 • 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 :P

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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SimplisticFormulas
116 posts
#8 • 2 Y
Y by RANDOM__USER, youochange
unless im seriously mistaken, the simplest solution is using projective
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RANDOM__USER
7 posts
#9 • 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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