Plan ahead for the next school year. Schedule your class today!

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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
and train with the best! Please note that early bird pricing ends August 19th!
Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

Our full course list for upcoming classes is below:
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0 replies
jwelsh
Jul 1, 2025
0 replies
Easy Geometry
ys-lg   0
21 minutes ago
Source: 2024 Aug SZM MO Second Round-1
Let \(\Omega\) be the circumcircle of the isosceles trapezoid \(ABCD\), and let \(I, J\) be the incenters of \(\triangle ABC\) and \(\triangle ACD\), respectively. Two circles with diameters \(AI\) and \(CJ\) intersect \(\Omega\) again at \(P\) and \(Q\), respectively. Prove that $PQ \parallel AD.$

Proposed by Site Mu, Beijing 101 Middle School
0 replies
+1 w
ys-lg
21 minutes ago
0 replies
Peru IMO TST 2024
diegoca1   0
33 minutes ago
Source: Peru IMO TST 2024 D2 P3
Determine the minimum possible value of \( x^2 + y^2 + z^2 \), where \( x, y, z \) are positive real numbers satisfying:
\[
    \begin{cases} 
    x + y + z = 5, \\ 
    \frac{1} {x} + \frac{1} {y} + \frac{1} {z} = 2.
    \end{cases}
    \]
0 replies
diegoca1
33 minutes ago
0 replies
Peru IMO TST 2024
diegoca1   0
36 minutes ago
Source: Peru IMO TST 2024 D2 P2
Consider the system of equations:
\[
    \begin{cases} 
    b^2 + 1 = ac, \\ 
    c^2 + 1 = bd, 
    \end{cases}
    \qquad (1)
    \]where \( a, b, c, d \) are positive integers.
a) Prove that there are infinitely many positive integer solutions to system (1).
b) Prove that if \((a, b, c, d)\) is a solution of (1), then
\[
        a = 3b - c, \quad d = 3c - b.
        \]
0 replies
diegoca1
36 minutes ago
0 replies
Peru IMO TST 2024
diegoca1   0
39 minutes ago
Source: Peru IMO TST 2024 D2 P1
For \( n \geq 2 \), an \( n \)-omino is a figure without holes formed by \( n \) unit squares such that each unit square shares an edge with at least one other square. We say two \( n \)-ominoes are identical if they have the same shape when rotated or if they are the reflection of one of the other.
(In the image we can see the 5 different 4-ominoes)
Determine all integers \( n \geq 2 \) such that we can partitionate an \( n \times n \) board with pairwise distinct \( n \)-ominoes.
0 replies
diegoca1
39 minutes ago
0 replies
No more topics!
Old hard problem
ItzsleepyXD   3
N May 14, 2025 by Funcshun840
Source: IDK
Let $ABC$ be a triangle and let $O$ be its circumcenter and $I$ its incenter.
Let $P$ be the radical center of its three mixtilinears and let $Q$ be the isogonal conjugate of $P$.
Let $G$ be the Gergonne point of the triangle $ABC$.
Prove that line $QG$ is parallel with line $OI$ .
3 replies
ItzsleepyXD
Apr 25, 2025
Funcshun840
May 14, 2025
Old hard problem
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G H BBookmark kLocked kLocked NReply
Source: IDK
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ItzsleepyXD
166 posts
#1 • 1 Y
Y by Funcshun840
Let $ABC$ be a triangle and let $O$ be its circumcenter and $I$ its incenter.
Let $P$ be the radical center of its three mixtilinears and let $Q$ be the isogonal conjugate of $P$.
Let $G$ be the Gergonne point of the triangle $ABC$.
Prove that line $QG$ is parallel with line $OI$ .
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ItzsleepyXD
166 posts
#2
Y by
Bump Bump
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ItzsleepyXD
166 posts
#4
Y by
Bump again ...
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Funcshun840
63 posts
#5 • 1 Y
Y by Lufin
very cool!!!
Lemma 1: The point $P$ lies on ray $OI$ with $IP:OP = r:2R$, where $r$, $R$ are the inradius and circumradius respectively.
Proof:Take a homothety with ratio $0.5$ at $I$ of the intouch triangle. Then it suffices to prove that $P$ is the exsimilicenter of this homothetied triangle and the medial triangle of the excentral triangle. Consider the $B$- and $C$- mixtilinear incircles $\omega _B$ and $\omega _ C$. We have that the south pole lies on the radical axis of $\omega _B$ and $\omega _ C$ by the shooting lemma. Additionally, the midpoint of $ID$ lies on the radical axis by a simple application of linpop and Pythagoras, hence the radical axis of $\omega _B$ and $\omega _ C$ passes through the desired exsimilicenter. By symmetry, we find that $P$ is the desired point.
Lemma 2:Let T be the insimilicenter of the incircle and the circumcircle and $J$ the inverse of $J$ wrt the circumcircle, then we have $$(J, I; P, T) = -1$$.
Proof: This is a length chase, recall that $OI = \sqrt{R^2 - 2rR}$, and $OJ= \frac{R^2}{OI}$. We also have $\frac{OT}{TI} = \frac{R}{r}$, so $OT = \frac{R}{r+R} OI$. We also have $OP = \frac{2R}{2R-r} OI$. Hence
$$\frac{JP}{IP} = \frac{JO - PO}{PO - IO} = \frac{\frac{R^2}{R^2 - 2rR} - \frac{2R}{2R-r}}{\frac{2R}{2R-r}-1} = \frac{2R^3 - rR^2 - 2R^3 + 4r R^2}{rR(R-2r)} = \frac{3R}{(R-2r)}.$$and $$\frac{JT}{IT} = \frac{JO - TO}{IO - TO} = \frac{\frac{R^2}{R^2-2rR} - \frac{R}{r+R}}{1- \frac{R}{r+R}} = \frac{R^2r + R^3 - R^3 + 2rR^2 }{rR(R-2r)}=\frac{3R}{R-2r}.$$
Now we are ready for the problem. Consider an isogonal conjugation, which preserves cross ratios. Then $I$ is fixed and $J$ is sent to the antipode $K$ of $I$ on the Feuerbach hyperbola (proof: let $OI$ intersect the circumcircle at $X$, $Y$ Then $-1=(X,Y;I,J)=(X', Y';I, K)$ and $X'$, $Y'$ are the points at infinity of the hyperbola). The point $T$ is sent to the Gergonne point $G$ (proof: Monge + mixtilinear properties). Finally $P$ is sent to $Q$ by definition. Now since $J$, $I$, $P$, $T$ lie on $OI$ with $(J, I; P, T) = -1$, we have $K$, $G$, $Q$, $I$ lie on the Feuerbach hyperbola with $(K, I; Q, G) = -1$. However, as the tangents to the hyperbola are parallel to $OI$ (they are antipodes and $IO$ is tangent at $I$), we find that $QG$ is parallel to $OI$.
OwO
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