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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
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Geometry from Iranian TST 2017
bgn   17
N 8 minutes ago by SimplisticFormulas
Source: Iranian TST 2017, first exam, day1, problem 3
In triangle $ABC$ let $I_a$ be the $A$-excenter. Let $\omega$ be an arbitrary circle that passes through $A,I_a$ and intersects the extensions of sides $AB,AC$ (extended from $B,C$) at $X,Y$ respectively. Let $S,T$ be points on segments $I_aB,I_aC$ respectively such that $\angle AXI_a=\angle BTI_a$ and $\angle AYI_a=\angle CSI_a$.Lines $BT,CS$ intersect at $K$. Lines $KI_a,TS$ intersect at $Z$.
Prove that $X,Y,Z$ are collinear.

Proposed by Hooman Fattahi
17 replies
bgn
Apr 5, 2017
SimplisticFormulas
8 minutes ago
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Fixed point in a small configuration
Assassino9931   1
N 31 minutes ago by mathuz
Source: Balkan MO Shortlist 2024 G3
Let $A, B, C, D$ be fixed points on this order on a line. Let $\omega$ be a variable circle through $C$ and $D$ and suppose it meets the perpendicular bisector of $CD$ at the points $X$ and $Y$. Let $Z$ and $T$ be the other points of intersection of $AX$ and $BY$ with $\omega$. Prove that $ZT$ passes through a fixed point independent of $\omega$.
1 reply
Assassino9931
Yesterday at 10:23 PM
mathuz
31 minutes ago
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JBMO Shortlist 2021 G4
Lukaluce   1
N an hour ago by s27_SaparbekovUmar
Source: JBMO Shortlist 2021
Let $ABCD$ be a convex quadrilateral with $\angle B = \angle D = 90^{\circ}$. Let $E$ be the point of intersection of $BC$ with $AD$ and let $M$ be the midpoint of $AE$. On the extension of $CD$, beyond the point $D$, we pick a point $Z$ such that $MZ = \frac{AE}{2}$. Let $U$ and $V$ be the projections of $A$ and $E$ respectively on $BZ$. The circumcircle of the triangle $DUV$ meets again $AE$ at the point $L$. If $I$ is the point of intersection of $BZ$ with $AE$, prove that the lines $BL$ and $CI$ intersect on the line $AZ$.
1 reply
Lukaluce
Jul 2, 2022
s27_SaparbekovUmar
an hour ago
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incircle with center I of triangle ABC touches the side BC
orl   40
N 2 hours ago by Ilikeminecraft
Source: Vietnam TST 2003 for the 44th IMO, problem 2
Given a triangle $ABC$. Let $O$ be the circumcenter of this triangle $ABC$. Let $H$, $K$, $L$ be the feet of the altitudes of triangle $ABC$ from the vertices $A$, $B$, $C$, respectively. Denote by $A_{0}$, $B_{0}$, $C_{0}$ the midpoints of these altitudes $AH$, $BK$, $CL$, respectively. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$, $CA$, $AB$ at the points $D$, $E$, $F$, respectively. Prove that the four lines $A_{0}D$, $B_{0}E$, $C_{0}F$ and $OI$ are concurrent. (When the point $O$ concides with $I$, we consider the line $OI$ as an arbitrary line passing through $O$.)
40 replies
orl
Jun 26, 2005
Ilikeminecraft
2 hours ago
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Coin Probability
girishpimoli   1
N 3 hours ago by lpieleanu
A fair coin is repeatedly tossed. If the probability that the first time head is tossed , twice in a row , is on the $9$ th and $10$ th toss, is
1 reply
girishpimoli
3 hours ago
lpieleanu
3 hours ago
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BrUMO 2025 Team Round Problem 7
lpieleanu   1
N 4 hours ago by enaschair
Digits $1$ through $9$ are placed on a $3 \times 3$ square such that all rows and columns sum to the same value. Please note that diagonals do not need to sum to the same value. How many ways can this be done?
1 reply
lpieleanu
Yesterday at 11:09 PM
enaschair
4 hours ago
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A Hard Function Problem
Saucepan_man02   1
N 5 hours ago by Saucepan_man02
If $f(x)=\frac{(\sqrt{5}+1)x-1}{(10-2 \sqrt{5})x+(\sqrt{5}+1)}$, find the value of$\underbrace{f(f(....f(\sqrt{5})..)}_{61 \text{times}}$.
1 reply
Saucepan_man02
5 hours ago
Saucepan_man02
5 hours ago
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Geometry Basic
AlexCenteno2007   2
N 5 hours ago by AlexCenteno2007
Let $ABC$ be an isosceles triangle such that $AC=BC$. Let $P$ be a dot on the $AC$ side.
The tangent to the circumcircle of $ABP$ at point $P$ intersects the circumcircle of $BCP$ at $D$. Prove that CD$ \parallel$AB
2 replies
AlexCenteno2007
Today at 12:11 AM
AlexCenteno2007
5 hours ago
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Basic geometry
AlexCenteno2007   1
N 5 hours ago by imbadatmath1233
Let $ABC$ be an equilateral triangle.M and N are the midpoints of $AB$ and $BC$ respectively. Externally to the triangle $ABC$ an isosceles right triangle APC is constructed, with the angle $APC =90°$.If point $I$ is the intersection of $AN$ and $MP$, show that $CI$ is a bisector of the angle $ACM$
1 reply
AlexCenteno2007
Today at 12:21 AM
imbadatmath1233
5 hours ago
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BrUMO 2025 Team Round Problem 6
lpieleanu   1
N 5 hours ago by lpieleanu
$4$ bears — Aruno, Bruno, Cruno and Druno — are each given a card with a positive integer and are told that the sum of their $4$ numbers is $17.$ They cannot show each other their cards, but discuss a series of observations in the following order:

Aruno: "I think it is possible that the other three bears all have the same card."
Bruno: "At first, I thought it was possible for the other three bears to have the same card. Now I know it is impossible for them to have the same card."
Cruno: "I think it is still possible that the other three bears have the same card."
Druno: "I now know what card everyone has."

What is the product of their four card values?
1 reply
lpieleanu
Yesterday at 11:08 PM
lpieleanu
5 hours ago
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BrUMO 2025 Team Round Problem 8
lpieleanu   1
N 5 hours ago by lpieleanu
Define the operation $\oplus$ by $$x \oplus y = xy-2x-2y+6.$$Compute all complex numbers $a$ such that $$a \oplus(a\oplus(a\oplus a))=a.$$
1 reply
lpieleanu
Yesterday at 11:10 PM
lpieleanu
5 hours ago
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BrUMO 2025 Team Round Problem 1
lpieleanu   1
N 6 hours ago by AlexCenteno2007
Find the smallest positive integer $n$ such that $n$ is divisible by exactly $25$ different positive integers.
1 reply
lpieleanu
Yesterday at 11:01 PM
AlexCenteno2007
6 hours ago
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BrUMO 2025 Team Round Problem 2
lpieleanu   1
N Today at 12:09 AM by MathCosine
Two squares, $ABCD$ and $AEFG,$ have equal side length $x.$ They intersect at $A$ and $O.$ Given that $CO=2$ and $OA=2\sqrt{2},$ what is $x$?
1 reply
lpieleanu
Yesterday at 11:04 PM
MathCosine
Today at 12:09 AM
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\phi (n) related 2024 TMC AIME Mock #7
parmenides51   1
N Today at 12:09 AM by maromex
Let $\phi (n)$ denote the number of positive integers less than$ n$ that are relatively prime to $n$. Let $S$ denote the set of positive integers $n$ such that when $\frac{n}{\phi (n)}$ is expressed as a simplified fraction, the denominator is a power of $2$. Find the smallest prime number $p$ that satisfies the following:
$\bullet$ $p -1$ is not divisible by any square of a positive integer greater than $1$,
$\bullet$ No element in $S$ is divisible by $p$.
1 reply
parmenides51
Saturday at 8:08 PM
maromex
Today at 12:09 AM
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Two circumscribed circles
BogG   3
N Feb 24, 2012 by horizon
Source: Swiss Imo Selection 2006
Let $D$ be inside $\triangle ABC$ and $E$ on $AD$ different to $D$. Let $\omega_1$ and $\omega_2$ be the circumscribed circles of $\triangle BDE$ and $\triangle CDE$ respectively. $\omega_1$ and $\omega_2$ intersect $BC$ in the interior points $F$ and $G$ respectively. Let $X$ be the intersection between $DG$ and $AB$ and $Y$ the intersection between $DF$ and $AC$. Show that $XY$ is $\|$ to $BC$.
3 replies
BogG
May 25, 2006
horizon
Feb 24, 2012
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Two circumscribed circles
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Reveal topic tags
geometry
circumcircle
power of a point
radical axis
geometry proposed
L
G H BBookmark kLocked kLocked NReply
Source: Swiss Imo Selection 2006
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BogG
60 posts
BogG
#1 h  May 25, 2006, 9:44 PM • 2 Y
Y by Adventure10, Mango247
Let $D$ be inside $\triangle ABC$ and $E$ on $AD$ different to $D$. Let $\omega_1$ and $\omega_2$ be the circumscribed circles of $\triangle BDE$ and $\triangle CDE$ respectively. $\omega_1$ and $\omega_2$ intersect $BC$ in the interior points $F$ and $G$ respectively. Let $X$ be the intersection between $DG$ and $AB$ and $Y$ the intersection between $DF$ and $AC$. Show that $XY$ is $\|$ to $BC$.
This post has been edited 3 times. Last edited by BogG, May 26, 2006, 2:15 PM
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grobber
7849 posts
grobber
#2 h  May 26, 2006, 7:22 AM • 2 Y
Y by Adventure10 and 1 other user
Let $P=DE\cap BC$. By Menelaus in $ABP,ACP$ with transversals $DG$ and $DF$ respectively, we see that $\frac{XB}{XA}=\frac{DP}{DA}\cdot\frac{GB}{GP}$, and $\frac{YC}{YA}=\frac{DP}{DA}\cdot\frac{FC}{FP}$. Proving that the two LHS's in the expressions above are equal is then equivalent to proving that $\frac{GB}{GP}=\frac{FC}{FP}$, which, in turn, is equivalent to $PF\cdot PB=PG\cdot PC$. This is just another way of saying that the powers of $P$ wrt $\omega_1,\omega_2$ are equal, which is clear because $P$ belongs to the radical axis $DE$ of the two circles.


P.S.

You said $\omega_1$ is the circumcircle of $BDC$. I'm pretty sure, however, that you meant $BDE$, right?
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BogG
60 posts
BogG
#3 h  May 26, 2006, 8:47 AM • 2 Y
Y by Adventure10, Mango247
I'm sorry you are right it's $\triangle BDE$ :D

And it was also my solution with menelaos :roll:
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horizon
404 posts
horizon
#4 h  Feb 24, 2012, 3:56 PM • 2 Y
Y by Adventure10, Mango247
see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79779
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