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Tangents forms triangle with two times less area
NO_SQUARES   1
N 12 minutes ago by Luis González
Source: Kvant 2025 no. 2 M2831
Let $DEF$ be triangle, inscribed in parabola. Tangents in points $D,E,F$ forms triangle $ABC$. Prove that $S_{DEF}=2S_{ABC}$. ($S_T$ is area of triangle $T$).
From F.S.Macaulay's book «Geometrical Conics», suggested by M. Panov
1 reply
NO_SQUARES
Today at 9:08 AM
Luis González
12 minutes ago
Complicated FE
XAN4   1
N 17 minutes ago by jasperE3
Source: own
Find all solutions for the functional equation $f(xyz)+\sum_{cyc}f(\frac{yz}x)=f(x)\cdot f(y)\cdot f(z)$, in which $f$: $\mathbb R^+\rightarrow\mathbb R^+$
Note: the solution is actually quite obvious - $f(x)=x^n+\frac1{x^n}$, but the proof is important.
Note 2: it is likely that the result can be generalized into a more advanced questions, potentially involving more bash.
1 reply
+2 w
XAN4
Today at 11:53 AM
jasperE3
17 minutes ago
Find all sequences satisfying two conditions
orl   34
N 35 minutes ago by YaoAOPS
Source: IMO Shortlist 2007, C1, AIMO 2008, TST 1, P1
Let $ n > 1$ be an integer. Find all sequences $ a_1, a_2, \ldots a_{n^2 + n}$ satisfying the following conditions:
\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]

\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]
Author: Dusan Dukic, Serbia
34 replies
orl
Jul 13, 2008
YaoAOPS
35 minutes ago
IMO Shortlist 2011, G4
WakeUp   125
N 43 minutes ago by Davdav1232
Source: IMO Shortlist 2011, G4
Let $ABC$ be an acute triangle with circumcircle $\Omega$. Let $B_0$ be the midpoint of $AC$ and let $C_0$ be the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$ and let $G$ be the centroid of the triangle $ABC$. Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circle $\Omega$ at a point $X\not= A$. Prove that the points $D,G$ and $X$ are collinear.

Proposed by Ismail Isaev and Mikhail Isaev, Russia
125 replies
WakeUp
Jul 13, 2012
Davdav1232
43 minutes ago
Z[x], P(\sqrt[3]5+\sqrt[3]25)=5+\sqrt[3]5
jasperE3   5
N an hour ago by Assassino9931
Source: VJIMC 2013 2.3
Prove that there is no polynomial $P$ with integer coefficients such that $P\left(\sqrt[3]5+\sqrt[3]{25}\right)=5+\sqrt[3]5$.
5 replies
jasperE3
May 31, 2021
Assassino9931
an hour ago
IMO problem 1
iandrei   77
N an hour ago by YaoAOPS
Source: IMO ShortList 2003, combinatorics problem 1
Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,1000000\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets \[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100  \] are pairwise disjoint.
77 replies
iandrei
Jul 14, 2003
YaoAOPS
an hour ago
Divisibility on 101 integers
BR1F1SZ   4
N an hour ago by BR1F1SZ
Source: Argentina Cono Sur TST 2024 P2
There are $101$ positive integers $a_1, a_2, \ldots, a_{101}$ such that for every index $i$, with $1 \leqslant i \leqslant 101$, $a_i+1$ is a multiple of $a_{i+1}$. Determine the greatest possible value of the largest of the $101$ numbers.
4 replies
BR1F1SZ
Aug 9, 2024
BR1F1SZ
an hour ago
2^x+3^x = yx^2
truongphatt2668   2
N 2 hours ago by CM1910
Prove that the following equation has infinite integer solutions:
$$2^x+3^x = yx^2$$
2 replies
truongphatt2668
Yesterday at 3:38 PM
CM1910
2 hours ago
Prove perpendicular
shobber   29
N 2 hours ago by zuat.e
Source: APMO 2000
Let $ABC$ be a triangle. Let $M$ and $N$ be the points in which the median and the angle bisector, respectively, at $A$ meet the side $BC$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $NA$ meets $MA$ and $BA$, respectively. And $O$ the point in which the perpendicular at $P$ to $BA$ meets $AN$ produced.

Prove that $QO$ is perpendicular to $BC$.
29 replies
shobber
Apr 1, 2006
zuat.e
2 hours ago
The smallest of sum of elements
hlminh   1
N 2 hours ago by nguyenhuybao_06
Let $S=\{1,2,...,2014\}$ and $X\subset S$ such that for all $a,b\in X,$ if $a+b\leq 2014$ then $a+b\in X.$ Find the smallest of sum of all elements of $X.$
1 reply
hlminh
2 hours ago
nguyenhuybao_06
2 hours ago
NT from ukr contest
mshtand1   3
N 2 hours ago by ravengsd
Source: Ukrainian TST for RMM 2021(2) and EGMO 2022 P2
Find the greatest positive integer $n$ such that there exist positive integers $a_1, a_2, ..., a_n$ for which the following holds $a_{k+2} = \dfrac{(a_{k+1}+a_k)(a_{k+1}+1)}{a_k}$ for all $1 \le k \le n-2$.
Proposed by Mykhailo Shtandenko and Oleksii Masalitin
3 replies
mshtand1
Oct 2, 2021
ravengsd
2 hours ago
Cute inequality in equilateral triangle
Miquel-point   1
N Apr 7, 2025 by Quantum-Phantom
Source: Romanian IMO TST 1981, Day 3 P5
Let $ABC$ be an equilateral triangle, $M$ be a point inside it, and $A',B',C'$ be the intersections of $AM,\; BM,\; CM$ with the sides of $ABC$. If $A'',\; B'',\; C''$ are the midpoints of $BC$, $CA$, $AB$, show that there is a triangle with sides $A'A''$, $B'B''$ and $C'C''$.

Laurențiu Panaitopol
1 reply
Miquel-point
Apr 6, 2025
Quantum-Phantom
Apr 7, 2025
Cute inequality in equilateral triangle
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Source: Romanian IMO TST 1981, Day 3 P5
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Miquel-point
472 posts
#1 • 2 Y
Y by PikaPika999, kiyoras_2001
Let $ABC$ be an equilateral triangle, $M$ be a point inside it, and $A',B',C'$ be the intersections of $AM,\; BM,\; CM$ with the sides of $ABC$. If $A'',\; B'',\; C''$ are the midpoints of $BC$, $CA$, $AB$, show that there is a triangle with sides $A'A''$, $B'B''$ and $C'C''$.

Laurențiu Panaitopol
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Quantum-Phantom
267 posts
#2
Y by
I think we should not allow $M$ to be on the three medians of the triangle.

It suffices to prove that $A'A''+B'B''>C'C''$. By Ceva's theorem, if $\tfrac{AB'}{B'C}=a$ and $\tfrac{BA'}{A'C}=b$, then $\tfrac{AC'}{C'B}=\tfrac ab$ ($a$, $b\ne1$, $a\ne b$). Let the side length be $1$, then
\[A'A''=\left|BA'-BA''\right|=\left|\frac b{b+1}-\frac12\right|=\frac{|b-1|}{2(b+1)}.\]Similarly we obtain $B'B''=\tfrac{|a-1|}{2(a+1)}$ and $C'C''=\tfrac{|a-b|}{2(a+b)}$. Multiplying by $2$, we need to show that
\[\frac{|a-1|}{a+1}+\frac{|b-1|}{b+1}=\frac{(a+1)|b-1|+(b+1)|a-1|}{(a+1)(b+1)}>\frac{|a-b|}{a+b}.\tag{*}\]On the one hand,
\begin{align*}
\frac{(a+1)|b-1|+(b+1)|a-1|}{(a+1)(b+1)}&\ge\frac{|(a+1)(b-1)+(b+1)(a-1)|}{(a+1)(b+1)}\\&=\frac{2|ab-1|}{(a+1)(b+1)}\overset{(1)}>\frac{|a-b|}{a+b},
\end{align*}where
\begin{align*}
(1)&\Leftrightarrow4(ab-1)^2(a+b)^2>(a-b)^2(a+1)^2(b+1)^2\\
&\Leftrightarrow(a-1)(b-1)\left(b^2+3ab+a+3b\right)\left(a^2+3ab+b+3a\right)>0.
\end{align*}On the other hand,
\begin{align*}
\frac{(a+1)|b-1|+(b+1)|a-1|}{(a+1)(b+1)}&\ge\frac{|(a+1)(b-1)-(b+1)(a-1)|}{(a+1)(b+1)}\\&=\frac{2|a-b|}{(a+1)(b+1)}\overset{(2)}>\frac{|a-b|}{a+b},
\end{align*}where
\[(2)\Leftrightarrow2(a+b)>(a+1)(b+1)\Leftrightarrow(a-1)(b-1)<0.\]At least one of $(1)$ and $(2)$ holds, so we are done.

Are there any other nice proofs of $(^*)$?
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