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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
Number theory
MuradSafarli   0
16 minutes ago
Prove that for any natural number \( n \) :

\[
1 \cdot 3 \cdot 5 \cdot \ldots \cdot (2n + 1) \mid (4n + 3)(4n + 5) \cdot \ldots \cdot (8n + 3).
\]
0 replies
MuradSafarli
16 minutes ago
0 replies
The number of integers
Fang-jh   16
N 43 minutes ago by ihategeo_1969
Source: ChInese TST 2009 P3
Prove that for any odd prime number $ p,$ the number of positive integer $ n$ satisfying $ p|n! + 1$ is less than or equal to $ cp^\frac{2}{3}.$ where $ c$ is a constant independent of $ p.$
16 replies
Fang-jh
Apr 4, 2009
ihategeo_1969
43 minutes ago
functional equation interesting
skellyrah   12
N an hour ago by jasperE3
find all functions IR->IR such that $$xf(x+yf(xy)) + f(f(x)) = f(xf(y))^2  + (x+1)f(x)$$
12 replies
skellyrah
Apr 24, 2025
jasperE3
an hour ago
Easy Combinatorics
MuradSafarli   1
N an hour ago by Nuran2010
A student firstly wrote $x=3$ on the board. For each procces, the stutent deletes the number x and replaces it with either $(2x+4)$ or $(3x+8)$ or $(x^2+5x)$. Is this possible to make the number $(20^{25}+2024)$ on the board?
1 reply
MuradSafarli
an hour ago
Nuran2010
an hour ago
No more topics!
quadrilateral ....
sam-n   3
N Mar 21, 2004 by amfulger
Source: Iran(2003)
assume ABCD a convex quadrilatral. P and Q are on BC and DC respectively such that angle BAP= angle DAQ .prove that [ADQ]=[ABP]
([ABC] means its area ) iff the line which crosses through the orthocenters of these traingles , is perpendicular to AC.
3 replies
sam-n
Mar 13, 2004
amfulger
Mar 21, 2004
quadrilateral ....
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G H BBookmark kLocked kLocked NReply
Source: Iran(2003)
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sam-n
793 posts
#1 • 2 Y
Y by Adventure10, Mango247
assume ABCD a convex quadrilatral. P and Q are on BC and DC respectively such that angle BAP= angle DAQ .prove that [ADQ]=[ABP]
([ABC] means its area ) iff the line which crosses through the orthocenters of these traingles , is perpendicular to AC.
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grobber
7849 posts
#2 • 2 Y
Y by Adventure10, Mango247
It's a bit boring to write it all. Let H<sub>b</sub> and H<sub>d</sub> be the orthocenters of ABP and ADQ respectively. Let B' and P' be the feet of the perpendiculars from B and P respectively in ABP and let D' and Q' be defined similarly for triangle ADQ. It's easy to show that if <BAP=<DAQ and their areas are equal then AP*AB'=AQ*AD'.

Now consider the inversion of pole A and power AP*AB'=AQ*AD'=AH<sub>b</sub>*AP'=AH<sub>d</sub>*AQ'. This inversion turns H<sub>b</sub>H<sub>d</sub> into the circle which passes through A, P' and Q', and since CP' perpendicular to AP' and CQ' perpendicular to AQ', it means that the circle AP'Q' has AC as its diameter, and a line is always turned into a circle which passes through the pole s.t. the diameter of the circle which passes through the pole is perpendicular to the initial line, so, in our case, AC perpendicular to H<sub>b</sub>H<sub>d</sub>.

This is only one of the implications, but the other one isn't hard to derive from this one:

We know that the lines are perpendicular if the areas are equal, and if we move one of P and Q a bit, say P, the orthocenter H<sub>b</sub> moves along AP', which is fixed, H<sub>d</sub> is fixed, AC is fixed, so AC and H<sub>b</sub>H<sub>d</sub> can't be perpendicular anymore.
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amfulger
993 posts
#3 • 2 Y
Y by Adventure10, Mango247
I have an inversion free solution.

Let H1, H2 be the orthocenters of DAQ, BAP. Let R1, R2 be the circumradi of DAQ and BAP.
Let DAQ=BAP=a.
Let C1 be the angle ACD and C2 be the angle ACB.
Let F be a point such as CF and BC are perpendicular and AF is paralel to DC.
Let E be a point such as CE and CD are perpendicular and AE and BC are paralel.
It is easy to see that C is the orthocenter of AEF and CFE=C2 and CEF=C1. We get CE/CF=sinC2/sinC1

It is well known that AH1=2R1cosa and AH2=2R2cosa.

AH1H2 and CEF have AH1||CE and AH2||CF. C is the orthocenter of AEF, so AC and EF are perpendicular.

It is clear that H1H2 and AC are perpendicular iff H1H2||EF iff AH1H2 and CEF are similar iff CE/CF=AH1/AH2 iff sinC2/sinC1=R1/R2 iff R1sinC1=R2sinC2.

Now Area[DAQ]=Area[BAP] iff R1*AD*sina*sinD=R2*AB*sina*sinB iff
R1*(AC*sinC1/sinD)*sina*sinD=R2*(AC*sinC2/sinB)*sina*sinB iff R1*sinC1=R2*sinC2.

Problem solved. I'll do my best to offer you a picture.
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amfulger
993 posts
#4 • 2 Y
Y by Adventure10, Mango247
If C=90<sup>0</sup> then the construction of E and F doesn't work.

In this case, take M to be AH1 \cap CD and N=AH2 \cap BC. Then AMCN is a ractangle and H1AC=C2 and H2AC=C1.

As before, we have [DAQ]=[BAP] iff R1sinC1=R2sinC2.

H1H2 is perpendicular to AC iff AC is an altitude in AH1H2, iff AH1H2=C1 and AH2H1=C2, iff R1/R2=AH1/AH2=sinC2/sinC1 iff R1sinC1=R2sinC2.
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