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)!

Be sure to mark your calendars for the following events:
[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
[*]April 8th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS State Discussion
April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]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
Proving bisection of an angle
Kyj9981   1
N 23 minutes ago by Kyj9981
Source: A lemma used to prove 11th IGO advanced level P3
Given $\triangle{ABC}$ with Incenter $I$ and A-Excenter $I_{A}$, let $D$ be the foot of perpendicular from $A$ to $BC$. Prove that $BC$ bisects $\angle{IDI_{A}}$
1 reply
Kyj9981
24 minutes ago
Kyj9981
23 minutes ago
Indonesian Geometry Olympiad
somebodyyouusedtoknow   14
N an hour ago by Ihatecombin
Source: Indonesian National Mathematical Olympiad 2024, Problem 3
The triangle $ABC$ has $O$ as its circumcenter, and $H$ as its orthocenter. The line $AH$ and $BH$ intersect the circumcircle of $ABC$ for the second time at points $D$ and $E$, respectively. Let $A'$ and $B'$ be the circumcenters of triangle $AHE$ and $BHD$ respectively. If $A', B', O, H$ are not collinear, prove that $OH$ intersects the midpoint of segment $A'B'$.
14 replies
1 viewing
somebodyyouusedtoknow
Aug 28, 2024
Ihatecombin
an hour ago
Parallel lines with incircle
buratinogigle   2
N an hour ago by buratinogigle
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 $PJ \parallel MN$.
2 replies
buratinogigle
Sunday at 11:23 AM
buratinogigle
an hour ago
Greek gods flood the world
a1267ab   16
N an hour ago by awesomeming327.
Source: USA Winter TST for IMO 2020, Problem 3, by Nikolai Beluhov
Let $\alpha \geq 1$ be a real number. Hephaestus and Poseidon play a turn-based game on an infinite grid of unit squares. Before the game starts, Poseidon chooses a finite number of cells to be flooded. Hephaestus is building a levee, which is a subset of unit edges of the grid (called walls) forming a connected, non-self-intersecting path or loop*.

The game then begins with Hephaestus moving first. On each of Hephaestus’s turns, he adds one or more walls to the levee, as long as the total length of the levee is at most $\alpha n$ after his $n$th turn. On each of Poseidon’s turns, every cell which is adjacent to an already flooded cell and with no wall between them becomes flooded as well. Hephaestus wins if the levee forms a closed loop such that all flooded cells are contained in the interior of the loop — hence stopping the flood and saving the world. For which $\alpha$ can Hephaestus guarantee victory in a finite number of turns no matter how Poseidon chooses the initial cells to flood?
-----
*More formally, there must exist lattice points $\mbox{\footnotesize \(A_0, A_1, \dotsc, A_k\)}$, pairwise distinct except possibly $\mbox{\footnotesize \(A_0 = A_k\)}$, such that the set of walls is exactly $\mbox{\footnotesize \(\{A_0A_1, A_1A_2, \dotsc , A_{k-1}A_k\}\)}$. Once a wall is built it cannot be destroyed; in particular, if the levee is a closed loop (i.e. $\mbox{\footnotesize \(A_0 = A_k\)}$) then Hephaestus cannot add more walls. Since each wall has length $\mbox{\footnotesize \(1\)}$, the length of the levee is $\mbox{\footnotesize \(k\)}$.

Nikolai Beluhov
16 replies
a1267ab
Dec 16, 2019
awesomeming327.
an hour ago
No more topics!
isosceles
k2c901_1   3
N Aug 13, 2005 by Skyward_Sea
Source: Taiwan 2nd TST 2005, 2nd independent study, problem 2
It is known that $\triangle ABC$ is an acute triangle. Let $C'$ be the foott of the perpendicular from $C$ to $AB$, and $D$, $E$ two distinct points on $CC'$. The feet of the perpendiculars from $D$ to $AC$ and $BC$ are $F$ and $G$, respectively. Show that if $DGEF$ is a parallelogram then $ABC$ is isosceles.
3 replies
k2c901_1
Aug 12, 2005
Skyward_Sea
Aug 13, 2005
isosceles
G H J
Source: Taiwan 2nd TST 2005, 2nd independent study, problem 2
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k2c901_1
146 posts
#1 • 2 Y
Y by Adventure10, Mango247
It is known that $\triangle ABC$ is an acute triangle. Let $C'$ be the foott of the perpendicular from $C$ to $AB$, and $D$, $E$ two distinct points on $CC'$. The feet of the perpendiculars from $D$ to $AC$ and $BC$ are $F$ and $G$, respectively. Show that if $DGEF$ is a parallelogram then $ABC$ is isosceles.
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Skyward_Sea
129 posts
#2 • 2 Y
Y by Adventure10, Mango247
From $DF \perp AC$ we get $EG \perp AC$, similarly $EF \perp AB$, so $CE \perp FG$, that is $DE \perp FG$, so DFEG is a diamond.
Then we soon get CC' is also a bisector of $\angle BCA$, that completes the proof.□
Z K Y
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k2c901_1
146 posts
#3 • 2 Y
Y by Adventure10, Mango247
Assuming you meant "rhombus" when you said "diamond", I think you're pretty much correct.
Z K Y
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Skyward_Sea
129 posts
#4 • 2 Y
Y by Adventure10, Mango247
k2c901_1 wrote:
Assuming you meant "rhombus" when you said "diamond", I think you're pretty much correct.
Thanks for teaching me an English word of maths.
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