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Trapezium with two right-angles: prove < AKB = 90° and more
Leonardo   6
N 23 minutes ago by FrancoGiosefAG
Source: Mexico 2002
Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$.

[Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .]
6 replies
Leonardo
May 8, 2004
FrancoGiosefAG
23 minutes ago
Nice concurrency
navi_09220114   4
N 30 minutes ago by quacksaysduck
Source: TASIMO 2025 Day 1 Problem 2
Four points $A$, $B$, $C$, $D$ lie on a semicircle $\omega$ in this order with diameter $AD$, and $AD$ is not parallel to $BC$. Points $X$ and $Y$ lie on segments $AC$ and $BD$ respectively such that $BX\parallel AD$ and $CY\perp AD$. A circle $\Gamma$ passes through $D$ and $Y$ is tangent to $AD$, and intersects $\omega$ again at $Z\neq D$. Prove that the lines $AZ$, $BC$ and $XY$ are concurrent.
4 replies
navi_09220114
May 19, 2025
quacksaysduck
30 minutes ago
Numbers on a circle
navi_09220114   3
N 41 minutes ago by quacksaysduck
Source: TASIMO 2025 Day 1 Problem 1
For a given positive integer $n$, determine the smallest integer $k$, such that it is possible to place numbers $1,2,3,\dots, 2n$ around a circle so that the sum of every $n$ consecutive numbers takes one of at most $k$ values.
3 replies
navi_09220114
May 19, 2025
quacksaysduck
41 minutes ago
D1033 : A problem of probability for dominoes 3*1
Dattier   1
N 42 minutes ago by Dattier
Source: les dattes à Dattier
Let $G$ a grid of 9*9, we choose a little square in $G$ of this grid three times, we can choose three times the same.

What the probability of cover with 3*1 dominoes this grid removed by theses little squares (one, two or three) ?
1 reply
Dattier
May 15, 2025
Dattier
42 minutes ago
f(n)=f(n-1)+1
Seventh   9
N 2 hours ago by cursed_tangent1434
Source: Problem 3, Brazilian MO 2015
Given a natural $n>1$ and its prime fatorization $n=p_1^{\alpha 1}p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, its false derived is defined by $$f(n)=\alpha_1p_1^{\alpha_1-1}\alpha_2p_2^{\alpha_2-1}...\alpha_kp_k^{\alpha_k-1}.$$Prove that there exist infinitely many naturals $n$ such that $f(n)=f(n-1)+1$.
9 replies
Seventh
Oct 20, 2015
cursed_tangent1434
2 hours ago
Functional equation meets inequality condition
Lukaluce   1
N 3 hours ago by sarjinius
Source: 2025 Macedonian Balkan Math Olympiad TST Problem 3
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfy
\[f(xf(y) + f(x)) = f(x)f(y) + 2f(x) + f(y) - 1,\]for every $x, y \in \mathbb{R}$, and $f(kx) > kf(x)$ for every $x \in \mathbb{R}$ and $k \in \mathbb{R}$, such that $k > 1$.
1 reply
Lukaluce
Apr 14, 2025
sarjinius
3 hours ago
Moving points in a plane
IAmTheHazard   2
N 3 hours ago by shanelin-sigma
Source: ELMO Shortlist 2024/C5
Let $\mathcal{S}$ be a set of $10$ points in a plane that lie within a disk of radius $1$ billion. Define a $move$ as picking a point $P \in \mathcal{S}$ and reflecting it across $\mathcal{S}$'s centroid. Does there always exist a sequence of at most $1500$ moves after which all points of $\mathcal{S}$ are contained in a disk of radius $10$?

Advaith Avadhanam
2 replies
IAmTheHazard
Jun 22, 2024
shanelin-sigma
3 hours ago
thank you !
Nakumi   0
3 hours ago
Given two non-constant polynomials $P(x),Q(x)$ such that for every real number $c$, $P(c)$ is a perfect square if and only if $Q(c)$ is a perfect square. Prove that $P(x)Q(x)$ is the square of a polynomial with real coefficients.
0 replies
Nakumi
3 hours ago
0 replies
Same divisor
sam-n   16
N 3 hours ago by AbdulWaheed
Source: IMO Shortlist 1997, Q14, China TST 2005
Let $ b, m, n$ be positive integers such that $ b > 1$ and $ m \neq n.$ Prove that if $ b^m - 1$ and $ b^n - 1$ have the same prime divisors, then $ b + 1$ is a power of 2.
16 replies
sam-n
Mar 6, 2004
AbdulWaheed
3 hours ago
sum of gcd over sets is more then sum of gcd over union
Miquel-point   3
N 4 hours ago by Jupiterballs
Source: KoMaL A. 882
Let $H_1, H_2,\ldots, H_m$ be non-empty subsets of the positive integers, and let $S$ denote their union. Prove that
\[\sum_{i=1}^m \sum_{(a,b)\in H_i^2}\gcd(a,b)\ge\frac1m \sum_{(a,b)\in S^2}\gcd(a,b).\]Proposed by Dávid Matolcsi, Berkeley
3 replies
Miquel-point
Jun 11, 2024
Jupiterballs
4 hours ago
Erasing the difference of two numbers
BR1F1SZ   5
N 4 hours ago by Jupiterballs
Source: Austria National MO Part 1 Problem 3
Consider the following game for a positive integer $n$. Initially, the numbers $1, 2, \ldots, n$ are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers $3,6,11$ and $17$ are on the board, then $3$ can be erased as $6 - 3=3$, or $6$ as $17 - 11=6$, or $11$ as $17 - 6=11$.)

For which values of $n$ is it possible to end with only one number remaining on the board?

(Michael Reitmeir)
5 replies
BR1F1SZ
May 5, 2025
Jupiterballs
4 hours ago
complex bashing in angles??
megahertz13   2
N Apr 18, 2025 by ali123456
Source: 2013 PUMAC FA2
Let $\gamma$ and $I$ be the incircle and incenter of triangle $ABC$. Let $D$, $E$, $F$ be the tangency points of $\gamma$ to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ and let $D'$ be the reflection of $D$ about $I$. Assume $EF$ intersects the tangents to $\gamma$ at $D$ and $D'$ at points $P$ and $Q$. Show that $\angle DAD' + \angle PIQ = 180^\circ$.
2 replies
megahertz13
Nov 5, 2024
ali123456
Apr 18, 2025
complex bashing in angles??
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G H BBookmark kLocked kLocked NReply
Source: 2013 PUMAC FA2
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megahertz13
3183 posts
#1 • 1 Y
Y by Rayanelba
Let $\gamma$ and $I$ be the incircle and incenter of triangle $ABC$. Let $D$, $E$, $F$ be the tangency points of $\gamma$ to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ and let $D'$ be the reflection of $D$ about $I$. Assume $EF$ intersects the tangents to $\gamma$ at $D$ and $D'$ at points $P$ and $Q$. Show that $\angle DAD' + \angle PIQ = 180^\circ$.
Z K Y
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TestX01
341 posts
#2
Y by
Polars or Lahire gives that $IQ\perp AD'$, $AD\perp IP$. Let $PI\cap DQ=X$. We claim $\triangle AD'D\sim\triangle IQX$. This follows from $\angle IDA=\angle DPI=\angle QXI$, and $180^\circ-\angle AD'I=180^\circ-\angle IQX$ from the perpendiculars.

The similarity easily finishes.
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ali123456
52 posts
#3 • 1 Y
Y by Rayanelba
My solution
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