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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
chat gpt
fuv870   0
5 minutes ago
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
0 replies
fuv870
5 minutes ago
0 replies
IZHO 2017 Functional equations
user01   50
N 9 minutes ago by HamstPan38825
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
50 replies
user01
Jan 14, 2017
HamstPan38825
9 minutes ago
Waiting for a dm saying me again "old geometry"
drmzjoseph   0
22 minutes ago
Source: Idk easy
Given $ABCD$ a tangencial quadrilateral that is not a rhombus, let $a,b,c,d$ be lengths of tangents from $A,B,C,D$ to the incircle of the quadrilateral which center is $I$. Let $M,N$ be the midpoints of $AC,BD$ resp. Prove that
\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]
0 replies
drmzjoseph
22 minutes ago
0 replies
Finally hard NT on UKR MO from NT master
mshtand1   2
N 24 minutes ago by IAmTheHazard
Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 11.4
A pair of positive integer numbers \((a, b)\) is given. It turns out that for every positive integer number \(n\), for which the numbers \((n - a)(n + b)\) and \(n^2 - ab\) are positive, they have the same number of divisors. Is it necessarily true that \(a = b\)?

Proposed by Oleksii Masalitin
2 replies
mshtand1
Mar 13, 2025
IAmTheHazard
24 minutes ago
No more topics!
An extension of a property from mixtilinear incircle
buratinogigle   6
N May 10, 2015 by Luis González
Source: Own
Let $ABC$ be a triangle inscribed circle $(O)$ and $P,Q$ are two isogonal conjugate points. $AP$ cuts $(O)$ again at $D$. $DE$ is diameter of $(O)$. $EQ$ cuts $(O)$ again at $F$. Prove that $BC, DF$ and the line passing through $P$ and perpendicular to $AQ$ are concurrent.
6 replies
buratinogigle
Dec 27, 2014
Luis González
May 10, 2015
An extension of a property from mixtilinear incircle
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buratinogigle
2314 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be a triangle inscribed circle $(O)$ and $P,Q$ are two isogonal conjugate points. $AP$ cuts $(O)$ again at $D$. $DE$ is diameter of $(O)$. $EQ$ cuts $(O)$ again at $F$. Prove that $BC, DF$ and the line passing through $P$ and perpendicular to $AQ$ are concurrent.
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Luis González
4145 posts
#2 • 6 Y
Y by Arab, buratinogigle, Adventure10, Mango247, MS_asdfgzxcvb, and 1 other user
Let $AQ$ cut $(O)$ again at $D'$ and let $X \equiv DF \cap BC.$ If we fix $AD,AD',$ the relation between $P$ and the intersection of $BC$ with the perpendicular from $P$ to $AD'$ is obviously a perspectivity and $EP \ \overline{\wedge} \ EQ \equiv EF \ \overline{\wedge} \ DF \equiv DX.$ So it suffices to show that the desired concurrency holds for at least 3 positions of $P.$

When $P \equiv AD \cap BC,$ then $X \equiv P$ and $Q \equiv A \equiv F$ $\Longrightarrow$ the concurrency trivially holds. When $P$ is at infinity, then $Q \equiv F \equiv D'$ $\Longrightarrow$ $DF \parallel BC$ $\Longrightarrow$ $X$ is at infinity $\Longrightarrow$ the concurrency holds. Finally, when $P \equiv A,$ then $Q \equiv AD' \cap BC$ and we have $\angle FAD'=\angle FDD'=\angle FXQ$ $\Longrightarrow$ $AXFQ$ is cyclic $\Longrightarrow$ $\angle XAQ=\angle DFQ=90^{\circ},$ i.e. $AX \perp AQ$ $\Longrightarrow$ the concurrency holds.
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TelvCohl
2311 posts
#3 • 6 Y
Y by buratinogigle, Adventure10, Mango247, MS_asdfgzxcvb, and 2 other users
My solution:

Let $ X $ be the projection of $ P $ on $ AQ $ .
Let $ Y $ be the projection of $ Q $ on $ BC $ .
Let $ Z=AQ \cap \odot (ABC), T=YZ \cap \odot (ABC), U=AP \cap BC, V=DF \cap BC $ .

It's suffices to prove $ V \in PX $

Easy to see $ F, Q, V, Y $ lie on a circle with diameter $ VQ $ .

From Reim theorem ( for $ T-Y-Z $ and $ A-U-D $ ) we get $ A, T, U, Y $ are concyclic .
From my post at Concurrent with PQ line (lemma 3) we get $ A, T, P, X $ all lie on $ \odot (AP) $ .
From Reim theorem ( for $ F-D-V $ and $ T-Z-Y $ ) we get $ T \in \odot (VQ) $ . ... $ (\star) $

Since $ \angle AXT=90^{\circ}-\angle TAP=90^{\circ}-\angle TYU=\angle QYT $ ,
so we get $ T, Q, X, Y $ are concyclic ,
hence combine with $ (\star) $ we get $ F, T, Q, V, X, Y $ all lie on $ \odot (VQ) $ ,
so we get $ \angle AXV=90^{\circ} =\angle AXP $ . ie. $ X, P, V $ are collinear

Q.E.D
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buratinogigle
2314 posts
#4 • 3 Y
Y by shinichiman, Adventure10, Mango247
Lemma. Let $ABC$ be a triangle inscribed circle $(O)$. $P,Q$ are isogonal conjugate points. $AP$ cuts $(O)$ agian at $M$. $QM$ cuts $BC$ at $E$ then $PE\parallel AQ$.

Proof by my pupil Phan Anh Quan. $AQ$ cuts $(O)$ again at $N$ and cut $BC$ at $H$. From $P,Q$ are isogonal conjugate we have $\triangle CHN\sim\triangle ACM$ and $\triangle CPM\sim\triangle QCN$ (g.g) deduce $HN.AM=CM.CN=QN.PM$, therefore $\dfrac{MP}{MA}=\dfrac{NH}{NQ}=\dfrac{ME}{MQ}$, thus $PE\parallel AQ$. We are done.

Proof of the problem by my pupil Trinh Huy Vu. Let $FD$ cut $BC$ at $S$, we will prove that $PS\perp AQ$, ineed. Let $QD$ cuts $BC$ at $R$. $T$ is projection of $R$ on $FD$. From the lemma, $PR\parallel AQ$. From this, $\dfrac{DT}{DF}=\dfrac{DR}{DQ}=\dfrac{DP}{DA}$ deduce $PT\parallel AF$. We have angle charsing,

\[\angle TPR=\angle FAQ=\angle BAQ-\angle BAF=\angle PAC-\angle BDF=\angle DBC-\angle BDF=\angle TSR.\]

Therfore, $STRP$ is cyclic or $SP\perp PR\parallel AQ$. We are done.
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shinichiman
3212 posts
#5 • 3 Y
Y by buratinogigle, Adventure10, Mango247
buratinogigle wrote:
Let $ABC$ be a triangle inscribed circle $(O)$ and $P,Q$ are two isogonal conjugate points. $AP$ cuts $(O)$ again at $D$. $DE$ is diameter of $(O)$. $EQ$ cuts $(O)$ again at $F$. Prove that $BC, DF$ and the line passing through $P$ and perpendicular to $AQ$ are concurrent.

$AQ$ cuts $(O)$ again at $M$, $DF$ cuts $BC$ at $T$. Since $P,Q$ are isogonal conjugates wrt $\triangle ABC$ so $\triangle DBP \sim \triangle MQB \; ( \text{A.A})$. Hence $MQ \cdot DP=MB \cdot DB$. We can also prove that $\triangle MFB \sim \triangle DBT \; ( \text{A.A})$ so $MB \cdot DB= MF \cdot DT$. Thus, $MF \cdot DT=MQ \cdot DP$ or $\frac{MQ}{MF}= \frac{DT}{DP}$. From here we get $\triangle MQF \sim \triangle DTP \; ( \text{S.A.S})$. Therefore $\angle DPT= \angle MFQ= \angle MAE=90^{\circ}- \angle PAQ$. This follows $TP\perp AQ$.
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buratinogigle
2314 posts
#6 • 3 Y
Y by shinichiman, Adventure10, Mango247
A better extension

Let $ABC$ be a triangle inscribed circle $(O)$ and $P,Q$ are two isogonal conjugate point. $AP$ cuts $(O)$ again at $D$. $E$ is a point on $(O)$. $EQ$ cuts $(O)$ again at $F$. $DF$ cuts $BC$ at $M$. $AQ $ cuts $(FMQ)$ again at $N$. Prove that $M,N,P$ are colliear.

The solution is the same as in #4 or in #5 of shinichiman.
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Luis González
4145 posts
#7 • 2 Y
Y by buratinogigle, Adventure10
We can also solve the extension with the same method I gave in my previous post.

Fix $D$ and $E$ and animate $P.$ If we redefine $N \in AQ,$ such that $\angle (NA,NP)=\angle (FE,FD)=\text{const},$ then all lines $NP$ are parallel to each other $\Longrightarrow$ series $P,N$ are similar, but clearly $P \ \overline{\wedge} \ Q \ \overline{\wedge} \ F \ \overline{\wedge} \ M.$ So it suffices to prove that $M,N,P$ are collinear for at least 3 positions of $P.$

When $P$ is at infinity and $P \equiv D$ the collinearity trivially holds. When $P \equiv A,$ then $Q \in BC$ and we have $\angle AQB=\angle ACD=\angle AFD$ making $AQFM$ cyclic $\Longrightarrow$ $\angle (AQ,AM)=\angle (FE,FD)$ $\Longrightarrow$ the collinearity holds. Thus $M,N,P$ are collinear for any $P.$
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