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Not so classic orthocenter problem
m4thbl3nd3r   6
N 40 minutes ago by maths_enthusiast_0001
Source: own?
Let $O$ be circumcenter of a non-isosceles triangle $ABC$ and $H$ be a point in the interior of $\triangle ABC$. Let $E,F$ be foots of perpendicular lines from $H$ to $AC,AB$. Suppose that $BCEF$ is cyclic and $M$ is the circumcenter of $BCEF$, $HM\cap AB=K,AO\cap BE=T$. Prove that $KT$ bisects $EF$
6 replies
1 viewing
m4thbl3nd3r
Yesterday at 4:59 PM
maths_enthusiast_0001
40 minutes ago
J
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A weird inequality
Eeightqx   0
2 hours ago
For all $a,\,b,\,c>0$, find the maximum $\lambda$ which satisfies
$$\sum_{cyc}a^2(a-2b)(a-\lambda b)\ge 0.$$hint
0 replies
Eeightqx
2 hours ago
0 replies
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Finding maximum sum of consecutive ten numbers in circle.
Goutham   3
N 2 hours ago by FarrukhKhayitboyev
Each of $999$ numbers placed in a circular way is either $1$ or $-1$. (Both values appear). Consider the total sum of the products of every $10$ consecutive numbers.
$(a)$ Find the minimal possible value of this sum.
$(b)$ Find the maximal possible value of this sum.
3 replies
Goutham
Feb 8, 2011
FarrukhKhayitboyev
2 hours ago
J
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Something nice
KhuongTrang   25
N 4 hours ago by Zuyong
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
25 replies
KhuongTrang
Nov 1, 2023
Zuyong
4 hours ago
J
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Equality occurs in strange points
arqady   10
N 4 hours ago by sqing
Let $a$, $b$ and $c$ are non-negatives such that $a+b+c\neq0$. Prove that:
1) \[\sum_{cyc}\sqrt{a^2+b^2}\leq\frac{11(a^2+b^2+c^2)+7(ab+ac+bc)}{3\sqrt2(a+b+c)}\]
2)\[\sum_{cyc}\sqrt{a^2+7ab+b^2}\leq\frac{2(a^2+b^2+c^2)+7(ab+ac+bc)}{a+b+c}\]
3)\[\sum_{cyc}\sqrt{4a^2+ab+4b^2}\leq\frac{5(a^2+b^2+c^2)+4(ab+ac+bc)}{a+b+c}\]
10 replies
arqady
Aug 28, 2011
sqing
4 hours ago
J
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An interesting inequality
JK1603JK   4
N 4 hours ago by Nguyenhuyen_AG
Source: unknown
Let a,b,c>=0 and a^2+b^2+c^2+abc=4 then prove \frac{1}{a+b+2}+\frac{1}{b+c+2}+\frac{1}{c+a+2} \le \frac{6-(a+b+c)}{4}
When does equality occur?
4 replies
JK1603JK
Mar 21, 2025
Nguyenhuyen_AG
4 hours ago
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My problem
hacbachvothuong   2
N 6 hours ago by arqady
Let $a, b, c$ be positive real numbers such that $ab+bc+ca=3$. Prove that:
$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\ge1$
2 replies
hacbachvothuong
Today at 10:10 AM
arqady
6 hours ago
J
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Let $a,b,c\ge 0$
khanhsy   1
N Today at 11:44 AM by arqady
Let $a,b,c\ge 0$ and $ab+bc+ca>0$. Prove
$$\sum_{cyc}\dfrac{1}{a^3+bc(b+c)} \ge \dfrac{1}{a^3+b^3+c^3}+\dfrac{4}{ab(a+b)+bc(b+c)+ca(c+a)}.$$
1 reply
khanhsy
Today at 11:16 AM
arqady
Today at 11:44 AM
J
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In triangle ABC prove that:
khanhsy   0
Today at 11:12 AM
$$\dfrac{1}{cC+aB+bA}+\dfrac{1}{bB+aC+cA}+\dfrac{1}{aA+bC+cB}\ge \dfrac{1}{aA+bB+cC}+\dfrac{4}{a(B+C)+b(C+A)+c(A+B)}.$$
0 replies
khanhsy
Today at 11:12 AM
0 replies
J
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Let $a,b,c$ be nonnegative real numbers. Prove that
khanhsy   1
N Today at 10:56 AM by KhuongTrang
Let $a,b,c,$ be nonnegative real numbers such that $a+b+c=3$. Prove that:
$$\dfrac{1}{\sqrt{a^3+3bc}}+\dfrac{1}{\sqrt{b^3+3ca}}+\dfrac{1}{\sqrt{c^3+3ab}}\ge \dfrac{1}{\sqrt{a^3+b^3+c^3+abc}}+\dfrac{2\sqrt{2}}{\sqrt{(a+b)(b+c)(c+a)}}.$$
1 reply
khanhsy
Today at 9:21 AM
KhuongTrang
Today at 10:56 AM
J
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An inequality on triangles sides
nAalniaOMliO   6
N Today at 7:09 AM by arqady
Source: Belarusian National Olympiad 2025
Numbers $a,b,c$ are lengths of sides of some triangle. Prove the inequality$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \geq \frac{a+b}{2c}+\frac{b+c}{2a}+\frac{c+a}{2b}$$
6 replies
nAalniaOMliO
Yesterday at 8:26 PM
arqady
Today at 7:09 AM
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J
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Tangent circles
oVlad   6
N Sep 15, 2024 by Helixglich
Source: Russian TST 2018, Day 9 P2 (Group NG), P4 (Groups A & B)
The point $K{}$ is the middle of the arc $BAC$ of the circumcircle of the triangle $ABC$. The point $I{}$ is the center of its inscribed circle $\omega$. The line $KI$ intersects the circumcircle of the triangle $ABC$ at $T{}$ for the second time. Prove that the circle passing through the midpoints of the segments $BC, BT$ and $CT$ is tangent to the circle which is symmetric to $\omega$ with respect to $BC$.
6 replies
oVlad
Mar 30, 2023
Helixglich
Sep 15, 2024
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Tangent circles
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Reveal topic tags
geometry
tangency
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Source: Russian TST 2018, Day 9 P2 (Group NG), P4 (Groups A & B)
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oVlad
1721 posts
oVlad
#1 h  Mar 30, 2023, 4:41 PM • 2 Y
Y by NO_SQUARES, Rounak_iitr
The point $K{}$ is the middle of the arc $BAC$ of the circumcircle of the triangle $ABC$. The point $I{}$ is the center of its inscribed circle $\omega$. The line $KI$ intersects the circumcircle of the triangle $ABC$ at $T{}$ for the second time. Prove that the circle passing through the midpoints of the segments $BC, BT$ and $CT$ is tangent to the circle which is symmetric to $\omega$ with respect to $BC$.
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hef4875
131 posts
hef4875
#2 h  Apr 3, 2023, 6:04 AM
Y by
Bump for this
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nguyenducmanh2705
26 posts
nguyenducmanh2705
#4 h  Feb 11, 2024, 2:07 PM
Y by
Bump....
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JafoNoksy
3 posts
JafoNoksy
#5 h  May 6, 2024, 9:48 PM • 1 Y
Y by NO_SQUARES
In file you can see my solution.
Attachments:
something.pdf (383kb)
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NO_SQUARES
1073 posts
NO_SQUARES
#6 h  May 8, 2024, 7:01 PM • 1 Y
Y by JafoNoksy
JafoNoksy wrote:
In file you can see my solution.

Very nice solution and problem!
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math_comb01
662 posts
math_comb01
#7 h  Aug 18, 2024, 2:34 PM
Y by
We claim this circle is ninepoint circle reflected over $MI$ which finishes by feurbach tangency, and so it suffices to show that $MI$ passes through the midpoint of $AT'$ call it $W$.It suffices to show $WMI$ are collinear and $W$ is on NPC, the latter follows by homoethety of $2$ at $A$ and the former, let nagel cevian meet farther incircle at $L$ then well known $AL \parallel MI$, so suffices to show reflection of $T'$ over $I$ lies on nagellian but follows as $MT'$ is refl of nagellian in $I$.
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Helixglich
113 posts
Helixglich
#8 h  Sep 15, 2024, 12:57 PM
Y by
Perhaps a more reasonable solution. Let $T$' be the reflection of $T$ over $BC$. Now take homothety at $T$' with factor $2$. This leads to proving that this homothetized circle is tangent to the circumcircle. Work with classical incentre setup.
$t = \frac{k-p}{k\overline{p}-1} = -\frac{2abc+a^2b+a^2c}{2a+b+c}$. Now we compute $t' = b^2+c^2+b^2c^2\frac{2a+b+c}{2abc+a^2b+a^2c}$ and hence $-i' = 2(ab+bc+ca)+b^2+c^2+b^2c^2\frac{2a+b+c}{2abc+a^2b+a^2c} = \frac{(2a+b+c)(ab+bc+ca)^2}{a(2bc+ab+ac)}$.
Hence we conclude that the power of $I$' with respect to the circumcircle is $i'\overline{i'} = \frac{(a+b+c)^2(ab+bc+ca)^2}{a^2b^2c^2} = (\overline{i}i)^2$ and thus we are done as this clearly implies that $Pow(I',(ABC)) = (2Rr)^2$.
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