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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
The Mother of Cases Problems
KAME06   1
N 2 minutes ago by mathprodigy2011
Source: Ecuador National Olympiad OMEC level U 2024 P5 Day 2
Let $p(n)$ the product of all $n$'s positive divisors, where $n \in \mathbb{Z}$.
Find all $n \in \mathbb{Z}$ such that $\sqrt[2024]{p(n)} \in \mathbb{Z}$ and is not a perfect power of another integer.
1 reply
KAME06
4 hours ago
mathprodigy2011
2 minutes ago
AD=BE implies ABC right
v_Enhance   111
N 9 minutes ago by Marcus_Zhang
Source: European Girl's MO 2013, Problem 1
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$. Prove that, if $AD=BE$, then the triangle $ABC$ is right-angled.
111 replies
v_Enhance
Apr 10, 2013
Marcus_Zhang
9 minutes ago
Equilateral triangles with a parallelogram
kred9   1
N 11 minutes ago by bjump
Source: 2025 Utah Math Olympiad #5
Given parallelogram $ABCD$, we construct equilateral triangle $ABP$ such that $P$ is on the same side of $\overline{AB}$ as $C$ and $D$. It is given that $\overleftrightarrow{CP}$ intersects $\overleftrightarrow{DA}$ at $Q$. Prove that there exists a point $R$ on $\overleftrightarrow{AB}$ such that $\triangle CQR$ is equilateral.
1 reply
+1 w
kred9
2 hours ago
bjump
11 minutes ago
APMO 2016: Sum of power of 2
shinichiman   28
N 16 minutes ago by ray66
Source: APMO 2016, problem 2
A positive integer is called fancy if it can be expressed in the form $$2^{a_1}+2^{a_2}+ \cdots+ 2^{a_{100}},$$where $a_1,a_2, \cdots, a_{100}$ are non-negative integers that are not necessarily distinct. Find the smallest positive integer $n$ such that no multiple of $n$ is a fancy number.

Senior Problems Committee of the Australian Mathematical Olympiad Committee
28 replies
shinichiman
May 16, 2016
ray66
16 minutes ago
No more topics!
AM _|_BF and CFM isosceles wanted, in right isosceles, <AMB=75^o, BF=AB
parmenides51   6
N Sep 1, 2021 by sunken rock
Source: 2017 Kyiv TST2 8.1 for Ukraine MO
In a right isosceles triangle $ABC$ , on the hypotenuse $CB$, the point chosen $M$ is such that $\angle AMB=75^o$. Inside the $\vartriangle ABC$ , a point $F$ lies on the bisector of $\angle CAM$, such that $BF=AB$. Prove that:
a) $AM \perp BF$
b) $\vartriangle CFM$ is isosceles.
6 replies
parmenides51
Jul 14, 2021
sunken rock
Sep 1, 2021
AM _|_BF and CFM isosceles wanted, in right isosceles, <AMB=75^o, BF=AB
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G H BBookmark kLocked kLocked NReply
Source: 2017 Kyiv TST2 8.1 for Ukraine MO
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parmenides51
30629 posts
#1 • 1 Y
Y by donotoven
In a right isosceles triangle $ABC$ , on the hypotenuse $CB$, the point chosen $M$ is such that $\angle AMB=75^o$. Inside the $\vartriangle ABC$ , a point $F$ lies on the bisector of $\angle CAM$, such that $BF=AB$. Prove that:
a) $AM \perp BF$
b) $\vartriangle CFM$ is isosceles.
This post has been edited 1 time. Last edited by parmenides51, Jul 14, 2021, 10:14 AM
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ccmritanobios1
11 posts
#2 • 1 Y
Y by donotoven
bump bump
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tommy2007
265 posts
#3
Y by
/bump $\quad$
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sunken rock
4379 posts
#4
Y by
An easy solution at https://artofproblemsolving.com/community/c6t48f6h2619373_am___bf_and_cfm_isosceles_wanted_in_right_isosceles_ltamb75o_bfab

Best regards,
sunken rock
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tommy2007
265 posts
#5
Y by

umm isn't this just the post itself?
This post has been edited 1 time. Last edited by tommy2007, Aug 30, 2021, 3:52 PM
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AwesomeYRY
579 posts
#6 • 1 Y
Y by tommy2007
We start with some angle chasing. Note that we already have $\angle BAC=90, \angle ABC = 45, \angle ACB=45, \angle AMB = 75$ as givens.

Next, $\angle BAM = 60$ and thus, $\angle MFA=\angle FAC = 15$. Thus, since $BF=AB$, we have $\angle BFA = \angle BAF = 60+15=75$.

Claim: $A,F,M,B$ are cyclic
Proof: $\angle AFB = 75=\angle AMB$.

As a consequence, $\angle FBM = \angle FAM = 15$. Thus, $\angle ABF = 30$, and $\angle AMF = 30$. Now, if we let $AM\cap BF = X$, we have
\[\angle AXB = 180-\angle XAB-\angle XBA = 180-60-30=90\]Thus, part (a) is done, $AM\perp BF$.

If we let $x=\angle FCA$, then trig ceva's on F gives
\[\sin 15 \cdot \sin 30 \cdot \sin (45-x) = \sin 75 \cdot \sin 15 \cdot \sin x \]Solving gives the solution $x=15$ since $\sin 30^2 = \frac14 = \sin 15 \cdot \sin 75$.

Also, $\angle FMC = 180-\angle FMB = \angle BAF = 75$. Thus, $\angle MFC = 180-\angle FMC - \angle FCM = 180-75-(45-x) = 75$. Thus, $\angle FMC = \angle MFC$ and $\triangle CFM$ is isosceles and we are done with (b).
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sunken rock
4379 posts
#7 • 3 Y
Y by Mango247, Mango247, Mango247
tommy2007 wrote:

umm isn't this just the post itself?

Oops! Correct link:[url] https://stanfulger.blogspot.com/2021/08/2017-kyiv-tst-for-ukraine-mo.html[/url]

Best regards,
sunken rock
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