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 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
Orthocenter config once again
Assassino9931   4
N 11 minutes ago by cj13609517288
Source: Bulgaria National Olympiad 2025, Day 2, Problem 4
Let \( ABC \) be an acute triangle with \( AB < AC \), midpoint $M$ of side $BC$, altitude \( AD \) (\( D \in BC \)), and orthocenter \( H \). A circle passes through points \( B \) and \( D \), is tangent to line \( AB \), and intersects the circumcircle of triangle \( ABC \) at a second point \( Q \). The circumcircle of triangle \( QDH \) intersects line \( BC \) at a second point \( P \). Prove that the lines \( MH \) and \( AP \) are perpendicular.
4 replies
+1 w
Assassino9931
5 hours ago
cj13609517288
11 minutes ago
a functional equation on positive reals
littletush   10
N 14 minutes ago by Frd_19_Hsnzde
Source: Czech and Slovak third round,2004,p6
Find all functions $f:\mathbb R^+ \rightarrow \mathbb R^+$ such that for all positive real numbers $x,y$,
\[x^2[f(x)+f(y)]=(x+y)f(yf(x)).\]
10 replies
littletush
Mar 3, 2012
Frd_19_Hsnzde
14 minutes ago
Set with a property
socrates   4
N 24 minutes ago by sadat465
Let $n\in \Bbb{N}, n \geq 4.$ Determine all sets $ A = \{a_1, a_2, . . . , a_n\} \subset \Bbb{N}$ containing $2015$ and having the property that $ |a_i - a_j|$ is prime, for all distinct $i, j\in \{1, 2, . . . , n\}.$
4 replies
socrates
May 29, 2015
sadat465
24 minutes ago
Number Theory Chain!
JetFire008   21
N 44 minutes ago by Primeniyazidayi
I will post a question and someone has to answer it. Then they have to post a question and someone else will answer it and so on. We can only post questions related to Number Theory and each problem should be more difficult than the previous. Let's start!

Question 1
21 replies
JetFire008
Yesterday at 7:14 AM
Primeniyazidayi
44 minutes ago
No more topics!
Quadrilateral formed by these four segments is cyclic
Amir Hossein   4
N Apr 12, 2017 by sturdyoak2012
Source: All Russian Olympiads 1993 - Grade 9 - Day 2 - Problem 2
A convex quadrilateral intersects a circle at points $A_1,A_2,B_1,B_2,C_1,C_2,D_1,$ and $D_2$. (Note that for some letter $N$, points $N_1$ and $N_2$ are on one side of the quadrilateral. Also, the points lie in that specific order on the circle.) Prove that if $A_1B_2=B_1C_2=C_1D_2= D_1A_2$, then quadrilateral formed by these four segments is cyclic.
4 replies
Amir Hossein
Jul 13, 2011
sturdyoak2012
Apr 12, 2017
Quadrilateral formed by these four segments is cyclic
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G H BBookmark kLocked kLocked NReply
Source: All Russian Olympiads 1993 - Grade 9 - Day 2 - Problem 2
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Amir Hossein
5452 posts
#1 • 2 Y
Y by Adventure10, Mango247
A convex quadrilateral intersects a circle at points $A_1,A_2,B_1,B_2,C_1,C_2,D_1,$ and $D_2$. (Note that for some letter $N$, points $N_1$ and $N_2$ are on one side of the quadrilateral. Also, the points lie in that specific order on the circle.) Prove that if $A_1B_2=B_1C_2=C_1D_2= D_1A_2$, then quadrilateral formed by these four segments is cyclic.
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Vo Duc Dien
341 posts
#2 • 2 Y
Y by Adventure10, Mango247
Problem is very easy but not always true. If each pair of the four equal segments parallel to each other, the resulting quadrilateral is a parallelogram (not a square) and is not cyclic.
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Merlinaeus
162 posts
#3 • 2 Y
Y by Adventure10, Mango247
What exactly does the 'quadrilateral formed by these four segments' mean?

I took it to mean the quadrilateral whose vertices are at the intersections of the segments $ A_1B_2$, $ B_1C_2$, $ C_1D_2$, $D_1A_2$ (the four intersections which are internal to the original circle) - but then concluded that what we are supposed to prove is not generally true.
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sturdyoak2012
864 posts
#4 • 2 Y
Y by Adventure10, Mango247
I believe the problem should state:
Prove that if $A_1B_2 = B_1C_2 = C_1D_2 = D_1A_2,$ then the quadrilateral formed by the segments $A_1A_2,B_1B_2,C_1C_2,D_1D_2$ is cyclic.
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sturdyoak2012
864 posts
#5 • 2 Y
Y by Adventure10, Mango247
Let $A=D_1D_2\cap A_1A_2$ and so forth such that the points on the quadrilateral are $A,A_1,A_2,B,B_1,B_2,\ldots$ in order.
ok i'm getting latex error for overarc D:<, henceforth assume there's an arc over everything
Then $\angle D = 180^{\circ} - \angle C_2C_1D_2 - \angle C_1D_2D_1  = \dfrac{360^{\circ} - C_2D_2 - C_1D_1}{2}.$ Similarly, $\angle B = \dfrac{360^{\circ} - A_2B_2 - A_1B_1}{2}.$ Then it suffices to show that $A_2B_2+A_1B_1 + C_2D_2+C_1D_1 = 360^{\circ}.$ Because arcs subtended by chords of the same length have equal measure, we find that $A_2B_2 = A_1D_1$ and $C_2D_2 = B_1C_1.$ Then it suffices to show that $A_1D_1+A_1B_1 + B_1C_1+C_1D_1 = 360^{\circ},$ but this is clearly true because $B_1A_1D_1+B_1C_1D_1 = 360^{\circ}.$
This post has been edited 1 time. Last edited by sturdyoak2012, Apr 12, 2017, 7:59 PM
Reason: bad latex D:<
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