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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
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Geometry
Whatisthepurposeoflife   0
18 minutes ago
Source: Lithuania TST 2013
From the point of intersection of the bisectors of the acute triangle ABC (where AB > AC), the base of the perpendicular descending from the point of intersection of the bisectors of the acute triangle ABC to the side BC is the point D. Find the ratio BD : BA if AD is the bisector of the angle BAC.
0 replies
Whatisthepurposeoflife
18 minutes ago
0 replies
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Sets with ab+1-closure
pieater314159   28
N 22 minutes ago by john0512
Source: ELMO 2019 Problem 5, 2019 ELMO Shortlist N3
Let $S$ be a nonempty set of positive integers such that, for any (not necessarily distinct) integers $a$ and $b$ in $S$, the number $ab+1$ is also in $S$. Show that the set of primes that do not divide any element of $S$ is finite.

Proposed by Carl Schildkraut
28 replies
+1 w
pieater314159
Jun 25, 2019
john0512
22 minutes ago
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Zack likes Moving Points
pinetree1   72
N an hour ago by endless_abyss
Source: USA TSTST 2019 Problem 5
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. A line through $H$ intersects segments $AB$ and $AC$ at $E$ and $F$, respectively. Let $K$ be the circumcenter of $\triangle AEF$, and suppose line $AK$ intersects $\Gamma$ again at a point $D$. Prove that line $HK$ and the line through $D$ perpendicular to $\overline{BC}$ meet on $\Gamma$.

Gunmay Handa
72 replies
pinetree1
Jun 25, 2019
endless_abyss
an hour ago
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Find all primes of the form n^n + 1 less than 10^{19}
Math5000   2
N 2 hours ago by SomeonecoolLovesMaths
Find all primes of the form $n^n + 1$ less than $10^{19}$

The first two primes are obvious: $n = 1, 2$ yields the primes $2, 5$. After that, it is clear that $n$ has to be even to yield an odd number.

So, $n = 2k \implies p = (2k)^{2k} + 1 \implies p-1 = (2k)^{k^2} = 2^{k^2}k^{k^2}$. All of these transformations don't seem to help. Is there any theorem I can use? Or is there something I'm missing?

2 replies
Math5000
Oct 15, 2019
SomeonecoolLovesMaths
2 hours ago
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No more topics!
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trapezoid AD//BC, AB = BC = BD, altitude BK cuts AC at M, <CDM?
parmenides51   2
N May 11, 2023 by Bexultan
Source: 2012 Oral Moscow Geometry Olympiad grades 8-9 p1
In trapezoid $ABCD$, the sides $AD$ and $BC$ are parallel, and $AB = BC = BD$. The height $BK$ intersects the diagonal $AC$ at $M$. Find $\angle CDM$.
2 replies
parmenides51
Sep 8, 2019
Bexultan
May 11, 2023
J
trapezoid AD//BC, AB = BC = BD, altitude BK cuts AC at M, <CDM?
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Source: 2012 Oral Moscow Geometry Olympiad grades 8-9 p1
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parmenides51
30630 posts
parmenides51
#1 h  Sep 8, 2019, 9:39 AM • 1 Y
Y by Adventure10
In trapezoid $ABCD$, the sides $AD$ and $BC$ are parallel, and $AB = BC = BD$. The height $BK$ intersects the diagonal $AC$ at $M$. Find $\angle CDM$.
This post has been edited 1 time. Last edited by parmenides51, Feb 25, 2024, 7:23 AM
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tait1k27
490 posts
tait1k27
#2 h  Sep 9, 2019, 12:57 PM • 2 Y
Y by Adventure10, Mango247
$\angle{CMD}=2\angle{CAD}=\angle{CBD}$
$\implies BMDC$ is a cyclic quadrilateral
$\implies \angle{CDM}=90^{\circ}$
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Bexultan
177 posts
Bexultan
#3 h  May 11, 2023, 5:06 PM
Y by
The answer is $\angle CDM=90^{\circ}$. To prove this, let's prove that the quadrilateral $BMDC$ is cyclic. Let $\angle CAD=\alpha$. Then $\angle DBC=2\alpha$, since $B$ is the circumcenter of triangle $ADC$ (from $AB=BC=BD$). The triangle $AMD$ is isosceles, so $\angle DMC=2\alpha$. So we have $\angle MDC=2\alpha=\angle DBC$, meaning the quadrilateral $BMDC$ is cyclic.
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