Stay ahead of learning milestones! Enroll in a class over the summer!

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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.

Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Iran TST P8
TheBarioBario   7
N 5 minutes ago by bin_sherlo
Source: Iranian TST 2022 problem 8
In triangle $ABC$, with $AB<AC$, $I$ is the incenter, $E$ is the intersection of $A$-excircle and $BC$. Point $F$ lies on the external angle bisector of $BAC$ such that $E$ and $F$ lieas on the same side of the line $AI$ and $\angle AIF=\angle AEB$. Point $Q$ lies on $BC$ such that $\angle AIQ=90$. Circle $\omega_b$ is tangent to $FQ$ and $AB$ at $B$, circle $\omega_c$ is tangent to $FQ$ and $AC$ at $C$ and both circles pass through the inside of triangle $ABC$. if $M$ is the Midpoint od the arc $BC$, which does not contain $A$, prove that $M$ lies on the radical axis of $\omega_b$ and $\omega_c$.

Proposed by Amirmahdi Mohseni
7 replies
TheBarioBario
Apr 2, 2022
bin_sherlo
5 minutes ago
Parallel lines with incircle
buratinogigle   1
N 25 minutes ago by luutrongphuc
Source: Own, test for the preliminary team of HSGS 2025
Let $ABC$ be a triangle with incircle $(I)$, which touches sides $CA$ and $AB$ at points $E$ and $F$, respectively. Choose points $M$ and $N$ on the line $EF$ such that $BM = BF$ and $CN = CE$. Let $P$ be the intersection of lines $CM$ and $BN$. Define $Q$ and $R$ as the intersections of $PN$ and $PM$ with lines $IC$ and $IB$, respectively. Assume that $J$ is the intersection of $QR$ and $BC$. Prove that $PJ \parallel MN$.
1 reply
buratinogigle
Yesterday at 11:23 AM
luutrongphuc
25 minutes ago
Inequality with x,y
GeoMorocco   1
N 35 minutes ago by Mathzeus1024
Let $x,y\ge 0$ such that $ 5(x^3+y^3) \leq 16(1+xy)$. Prove that:
$$8+xy\geq 3(x+y) $$
1 reply
GeoMorocco
Apr 20, 2025
Mathzeus1024
35 minutes ago
Function from the plane to the real numbers
AndreiVila   3
N 39 minutes ago by ItzsleepyXD
Source: Balkan MO Shortlist 2024 G7
Let $f:\pi\rightarrow\mathbb{R}$ be a function from the Euclidean plane to the real numbers such that $$f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$$for any acute triangle $ABC$ with circumcenter $O$, centroid $G$ and orthocenter $H$. Prove that $f$ is constant.
3 replies
AndreiVila
4 hours ago
ItzsleepyXD
39 minutes ago
No more topics!
19th kmo #5
lightrhee   3
N Sep 18, 2024 by hanulyeongsam
Source: KMO round 2, problem 5
Let $P$ be a point that lies outside of circle $O$. A line passes through $P$ and meets the circle at $A$ and $B$, and another line passes through $P$ and meets the circle at $C$ and $D$. The point $A$ is between $P$ and $B$, $C$ is between $P$ and $D$. Let the intersection of segment $AD$ and $BC$ be $L$ and construct $E$ on ray $(PA$ so that $BL \cdot PE = DL \cdot PD$.

Show that $M$ is the midpoint of the segment $DE$, where $M$ is the intersection of lines $PL$ and $DE$.
3 replies
lightrhee
Feb 3, 2006
hanulyeongsam
Sep 18, 2024
19th kmo #5
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G H BBookmark kLocked kLocked NReply
Source: KMO round 2, problem 5
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lightrhee
557 posts
#1 • 2 Y
Y by Adventure10 and 1 other user
Let $P$ be a point that lies outside of circle $O$. A line passes through $P$ and meets the circle at $A$ and $B$, and another line passes through $P$ and meets the circle at $C$ and $D$. The point $A$ is between $P$ and $B$, $C$ is between $P$ and $D$. Let the intersection of segment $AD$ and $BC$ be $L$ and construct $E$ on ray $(PA$ so that $BL \cdot PE = DL \cdot PD$.

Show that $M$ is the midpoint of the segment $DE$, where $M$ is the intersection of lines $PL$ and $DE$.
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GHAVR
68 posts
#2 • 3 Y
Y by Adventure10, Mango247, and 1 other user
lightrhee wrote:
This is the fifth problem of the second round of 19th Korean Mathematics Olympiad.
5.
P is a point that is outside of circle O. A line passes through P and meets the circle at A and B, and another line passes through P and meets the circle at C and D. A is between P and B, C is between P and D. Let the intersection of segment AD and BC be L and construct E on ray PA so that $BL \cdot PE = DL \cdot PD$. Show that M is the midpoint of the segment DE, where M is the intersection of line PL and DE.


We have $\frac{BL}{DL}=\frac{PD}{PE} \Rightarrow$
Under the sin theorem we have $\frac{EM}{MD}=\frac{EP sin \angle BPL}{DP sin \angle DPL}=\frac{DL sin \angle BPL}{BL sin \angle DPL}$

$\frac{DL}{sin \angle DPL} * \frac{sin \angle BPL}{BL} = \frac{LP}{sin \angle PDL} * \frac{sin \angle PBL}{LP} = 1$ (as $\angle LDP=\angle PBL$)

Nice problem :)
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Sailor
256 posts
#3 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Since $\frac{BL}{DL}=\frac{PD}{PE}=\frac{AL}{LC}$ $\Longleftrightarrow$ $S_{\triangle{PEL}}=S_{\triangle{PLD}}$ and the conclusion follows.
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hanulyeongsam
170 posts
#4
Y by
BL:DL=triangle LAB:triangle LCD=x:y
Let H_1,H_4 vertically from M CD,AB
H_2,H_3 vertically from the meeting point of AD and BC to CD,AB

BL:DL=LH_3:LH_2=MH_4:MH_1 MH_4=bx, MH_1=by
BL:DL=PD:PE=> PD=ax,PE=ay
|triangle MEP|=1/2*ay*bx=abxy/2
|triangle MDP|=1/2*ax*by=abxy/2
|MEP|=|MDP| so, MD=ME
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