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n^6 + 5n^3 + 4n + 116 is the product of two or more consecutive numbers
Amir Hossein   2
N an hour ago by KTYC
Source: Bulgaria JBMO TST 2018, Day 1, Problem 3
Find all positive integers $n$ such that the number
$$n^6 + 5n^3 + 4n + 116$$is the product of two or more consecutive numbers.
2 replies
Amir Hossein
Jun 25, 2018
KTYC
an hour ago
IMO Shortlist 2009 - Problem G3
April   49
N 2 hours ago by Ilikeminecraft
Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelogram.
Prove that $GR=GS$.

Proposed by Hossein Karke Abadi, Iran
49 replies
April
Jul 5, 2010
Ilikeminecraft
2 hours ago
Four tangent lines concur on the circumcircle
v_Enhance   36
N 2 hours ago by Ilikeminecraft
Source: USA TSTST 2018 Problem 3
Let $ABC$ be an acute triangle with incenter $I$, circumcenter $O$, and circumcircle $\Gamma$. Let $M$ be the midpoint of $\overline{AB}$. Ray $AI$ meets $\overline{BC}$ at $D$. Denote by $\omega$ and $\gamma$ the circumcircles of $\triangle BIC$ and $\triangle BAD$, respectively. Line $MO$ meets $\omega$ at $X$ and $Y$, while line $CO$ meets $\omega$ at $C$ and $Q$. Assume that $Q$ lies inside $\triangle ABC$ and $\angle AQM = \angle ACB$.

Consider the tangents to $\omega$ at $X$ and $Y$ and the tangents to $\gamma$ at $A$ and $D$. Given that $\angle BAC \neq 60^{\circ}$, prove that these four lines are concurrent on $\Gamma$.

Evan Chen and Yannick Yao
36 replies
v_Enhance
Jun 26, 2018
Ilikeminecraft
2 hours ago
Inspired by m4thbl3nd3r
sqing   1
N 2 hours ago by sqing
Source: Own
Let $  a, b,c>0,b+c>a$. Prove that$$\sqrt{\frac{a}{b+c-a}}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)}\geq 1$$$$\frac{a}{b+c-a}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)} \geq  \frac{4\sqrt 2}{3}-1$$
1 reply
sqing
3 hours ago
sqing
2 hours ago
2-var inequality
sqing   6
N 3 hours ago by sqing
Source: Own
Let $ a,b> 0 $ and $2a+2b+ab=5. $ Prove that
$$ \frac{a^2}{b^2}+\frac{1}{a^2}-a^2\geq  1$$$$ \frac{a^3}{b^3}+\frac{1}{a^3}-a^3\geq  1$$
6 replies
sqing
Yesterday at 1:19 PM
sqing
3 hours ago
Inequality
Amin12   8
N 3 hours ago by A.H.H
Source:  Iran 3rd round-2017-Algebra final exam-P3
Let $a,b$ and $c$ be positive real numbers. Prove that
$$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$
8 replies
1 viewing
Amin12
Sep 2, 2017
A.H.H
3 hours ago
Every subset of size k has sum at most N/2
orl   50
N 4 hours ago by de-Kirschbaum
Source: USAMO 2006, Problem 2, proposed by Dick Gibbs
For a given positive integer $k$ find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k + 1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $\tfrac{N}{2}.$
50 replies
orl
Apr 20, 2006
de-Kirschbaum
4 hours ago
Inspired by a9opsow_
sqing   2
N 4 hours ago by sqing
Source: Own
Let $ a,b > 0  .$ Prove that
$$ \frac{(ka^2 - kab-b)^2 + (kb^2 - kab-a)^2 + (ab-ka-kb )^2}{ (ka+b)^2 + (kb+a)^2+(a - b)^2 }\geq  \frac {1}{(k+1)^2}$$Where $ k\geq 0.37088 .$
$$\frac{(a^2 - ab-b)^2 + (b^2 - ab-a)^2 + ( ab-a-b)^2}{a^2 +b^2+(a - b)^2 } \geq 1$$$$ \frac{(2a^2 - 2ab-b)^2 + (2b^2 - 2ab-a)^2 + (ab-2a-2b )^2}{ (2a+b)^2 + (2b+a)^2+(a - b)^2 }\geq  \frac 19$$
2 replies
sqing
4 hours ago
sqing
4 hours ago
Cute NT Problem
M11100111001Y1R   5
N 5 hours ago by compoly2010
Source: Iran TST 2025 Test 4 Problem 1
A number \( n \) is called lucky if it has at least two distinct prime divisors and can be written in the form:
\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]where \( p_1, \dots, p_k \) are distinct prime numbers that divide \( n \). (Note: it is possible that \( n \) has other prime divisors not among \( p_1, \dots, p_k \).) Prove that for every prime number \( p \), there exists a lucky number \( n \) such that \( p \mid n \).
5 replies
M11100111001Y1R
Tuesday at 7:20 AM
compoly2010
5 hours ago
3 var inequality
SunnyEvan   13
N 5 hours ago by Nguyenhuyen_AG
Let $ a,b,c \in R $ ,such that $ a^2+b^2+c^2=4(ab+bc+ca)$Prove that :$$ \frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48} $$
13 replies
SunnyEvan
May 17, 2025
Nguyenhuyen_AG
5 hours ago
Polar Coordinates
pingpongmerrily   4
N Today at 12:11 AM by K124659
Convert the equation $r=\tan(\theta)$ into rectangular form.
4 replies
pingpongmerrily
Today at 12:02 AM
K124659
Today at 12:11 AM
a