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Geometry
srnjbr   1
N 2 hours ago by ricarlos
in triangle abc, we know that bac=60. the circumcircle of the center i is tangent to the sides ab and ac at points e and f respectively. the midpoint of side bc is called m. if lines bi and ci intersect line ef at points p and q respectively, show that pmq is equilateral.
1 reply
srnjbr
Mar 19, 2025
ricarlos
2 hours ago
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A property of divisors
rightways   8
N 2 hours ago by de-Kirschbaum
Source: Kazakhstan NMO 2016, P1
Prove that one can arrange all positive divisors of any given positive integer around a circle so that for any two neighboring numbers one is divisible by another.
8 replies
rightways
Mar 17, 2016
de-Kirschbaum
2 hours ago
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ortho conf DEF, radius MD, intersect ME,MF, collinear H,K,L
star-1ord   0
3 hours ago
Source: Estonia Final Round 2025 12-3
Let $ABC$ be an acute-angled triangle with $|AB|<|AC|$. The altitudes $AD,BE$ and $CF$ intersect at $H$. Let $M$ be the midpoint of $BC$. Point $K$ is chosen on the extension of $EM$ beyond $M$ and point $L$ is chosen on the segment $FM$ such that $|MK|=|ML|=|MD|$. Prove that points $K, L$ and $H$ are collinear.

a little harder version
0 replies
star-1ord
3 hours ago
0 replies
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Funny system of equations in three variables
Tintarn   10
N 4 hours ago by Marcus_Zhang
Source: Baltic Way 2020, Problem 5
Find all real numbers $x,y,z$ so that
\begin{align*} x^2 y + y^2 z + z^2 &= 0 \\ z^3 + z^2 y + z y^3 + x^2 y &= \frac{1}{4}(x^4 + y^4). \end{align*}
10 replies
Tintarn
Nov 14, 2020
Marcus_Zhang
4 hours ago
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a^{2m}+a^{n}+1 is perfect square
kmh1   1
N 4 hours ago by kmh1
Source: own
Find all positive integer triplets $(a,m,n)$ such that $2m>n$ and $a^{2m}+a^{n}+1$ is a perfect square.
1 reply
kmh1
Mar 20, 2025
kmh1
4 hours ago
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Interesting problem
deraxenrovalo   0
4 hours ago
Given $\triangle$$ABC$ with circumcenter $O$$.\;$Let $P$ be an arbitrary point on $(BOC)$ such that $P$ is outside $(ABC)$$.\;$Let $Q$ be an arbitrary point on $(ABC)$$.\;$$AB$ cuts $(ACP)$ again at $E$ and $AC$ cuts $(ABP)$ again at $F$$.\;$The intersection of $BF$ and $CE$ is $R$$.\;$Let $X$ and $Y$ be the intersection of $EF$ with $(PQC)$ and $(PQR)$ respectively such that $X$, $Y$, $P$ are pairwise distinct.
Show that : $(APX)$, $(BPY)$, $(QPE)$ are coaxial circles

hint
0 replies
deraxenrovalo
4 hours ago
0 replies
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Vieta Jumping Unsolved(Reposted)
Eagle116   0
4 hours ago
Source: MONT, Vieta Jumping part
The question is:
Let $x_1$, $x_2$, $\dots$, $x_n$ be $n$ integers. If $k>n$ is an integer, prove that the only solution to
$$x_1^2 + x_2^2 + \dots + x_n^2 = kx_1x_2\dots x_n $$is is $x_1 = x_2 = \dots = x_n = 0$.
0 replies
Eagle116
4 hours ago
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Geometry with parallel lines.
falantrng   32
N 4 hours ago by endless_abyss
Source: RMM 2018,D1 P1
Let $ABCD$ be a cyclic quadrilateral an let $P$ be a point on the side $AB.$ The diagonals $AC$ meets the segments $DP$ at $Q.$ The line through $P$ parallel to $CD$ mmets the extension of the side $CB$ beyond $B$ at $K.$ The line through $Q$ parallel to $BD$ meets the extension of the side $CB$ beyond $B$ at $L.$ Prove that the circumcircles of the triangles $BKP$ and $CLQ$ are tangent .
32 replies
falantrng
Feb 24, 2018
endless_abyss
4 hours ago
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sum divides n-th moment
navi_09220114   1
N 4 hours ago by ja.
Source: Own. Malaysian IMO TST 2025 P9
Given four distinct positive integers $a<b<c<d$ such that $\gcd(a,b,c,d)=1$, find the maximum possible number of integers $1\le n\le 2025$ such that $$a+b+c+d\mid a^n+b^n+c^n+d^n$$
Proposed by Ivan Chan Kai Chin
1 reply
navi_09220114
Yesterday at 1:07 PM
ja.
4 hours ago
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Find all functions
Jackson0423   0
5 hours ago
Find all functions F:R->R such that
1/(F(F(x))-F(x))=F(x)
I know x+1/x works..
0 replies
Jackson0423
5 hours ago
0 replies
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Inequalities
lgx57   1
N Today at 2:52 PM by sqing
Let $a,b,c>0$,$\frac{a^2+b^2+c^2}{ab+bc+ca}=2$, find the minimum of

$$\frac{a^3+b^3+c^3}{abc}$$
1 reply
lgx57
Today at 2:25 PM
sqing
Today at 2:52 PM
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Inequalities
sqing   0
Mar 21, 2025
Let $ a,b,c\geq 0 $ and $ a+b+c\geq 2+abc . $ Prove that
$$a^2+b^2+c^2- \frac{2}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{48}{25}$$$$a^2+b^2+c^2- \frac{3}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{91}{50}$$$$a^2+b^2+c^2- \frac{4}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{42}{25}$$$$a^2+b^2+c^2- \frac{8}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{3(7\sqrt{21}-27)}{25}$$$$a^2+b^2+c^2- \frac{9}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{8}{261+41 \sqrt{41}}$$
0 replies
sqing
Mar 21, 2025
0 replies
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sqing
41195 posts
sqing
#1 h  Mar 21, 2025, 3:58 PM
Y by
Let $ a,b,c\geq 0 $ and $ a+b+c\geq 2+abc . $ Prove that
$$a^2+b^2+c^2- \frac{2}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{48}{25}$$$$a^2+b^2+c^2- \frac{3}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{91}{50}$$$$a^2+b^2+c^2- \frac{4}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{42}{25}$$$$a^2+b^2+c^2- \frac{8}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{3(7\sqrt{21}-27)}{25}$$$$a^2+b^2+c^2- \frac{9}{5}abc-\frac{1}{2}a^2b^2c^2\geq \frac{8}{261+41 \sqrt{41}}$$
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