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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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0 replies
jlacosta
Apr 2, 2025
0 replies
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Circumcenter of reflection of collinear points over sides
a1267ab   27
N a few seconds ago by Giant_PT
Source: USA TST 2025
Let $ABC$ be a triangle, and let $X$, $Y$, and $Z$ be collinear points such that $AY=AZ$, $BZ=BX$, and $CX=CY$. Points $X'$, $Y'$, and $Z'$ are the reflections of $X$, $Y$, and $Z$ over $BC$, $CA$, and $AB$, respectively. Prove that if $X'Y'Z'$ is a nondegenerate triangle, then its circumcenter lies on the circumcircle of $ABC$.

Michael Ren
27 replies
a1267ab
Jan 11, 2025
Giant_PT
a few seconds ago
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a+b+c=abc
KhuongTrang   1
N 9 minutes ago by KhuongTrang
Source: own
Problem. Let $a,b,c$ be three positive real numbers satisfying $a+b+c=abc.$ Prove that$$\sqrt{a^2+b^2+3}+\sqrt{b^2+c^2+3}+\sqrt{c^2+a^2+3}\ge4\cdot \frac{a^2b^2c^2-3}{ab+bc+ca-3}-7.$$There is a very elegant proof :-D Could anyone think of it?
1 reply
1 viewing
KhuongTrang
Wednesday at 11:51 AM
KhuongTrang
9 minutes ago
J
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multiple of 15-15 positive factors
britishprobe17   0
14 minutes ago
Source: KTOM Maret 2025
Find the sum of all natural numbers $n$ such that $n$ is a multiple of $15$ and has exactly $15$ positive factors.
0 replies
britishprobe17
14 minutes ago
0 replies
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general form
pennypc123456789   0
19 minutes ago
If $a,b,c$ are positive real numbers, $k \ge 3$ then
$$ \frac{a + b}{a + kb + c} + \dfrac{b + c}{b + kc + a}+\dfrac{c + a}{c + ka + b} \geq \dfrac{6}{k+2}$$
0 replies
pennypc123456789
19 minutes ago
0 replies
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No more topics!
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ratio of sidelengths in a right triangle, given ratio of line through incenter
parmenides51   4
N Aug 5, 2018 by Sen_25
Source: Quantum - English M73
In a right triangle one half of the hypotenuse (from a vertex to the midpoint of the hypotenuse) subtends a right angle at
the triangle's incenter. Find the ratio of side lengths of the triangle.

(B. Pitskel)
4 replies
parmenides51
Aug 5, 2018
Sen_25
Aug 5, 2018
J
ratio of sidelengths in a right triangle, given ratio of line through incenter
G H J
Reveal topic tags
ratio
right triangle
geometry
incenter
Quantum English
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Source: Quantum - English M73
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parmenides51
30629 posts
parmenides51
#1 h  Aug 5, 2018, 11:08 AM • 2 Y
Y by Adventure10, Mango247
In a right triangle one half of the hypotenuse (from a vertex to the midpoint of the hypotenuse) subtends a right angle at
the triangle's incenter. Find the ratio of side lengths of the triangle.

(B. Pitskel)
Z K Y
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NikoIsLife
9657 posts
NikoIsLife
#2 h  Aug 5, 2018, 11:51 AM • 1 Y
Y by Adventure10
Does this problem have a diagram? I didn't quite get what it meant by "one half of the hypotenuse subtends a right angle at
the triangle's incenter"
Z K Y
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TuZo
19351 posts
TuZo
#3 h  Aug 5, 2018, 12:05 PM • 2 Y
Y by Adventure10, Mango247
If $M$ is the midpoint of the $BC$, and I the incenter, we have angle(IBM)=angle(ICM)$, so angleB=angleC=45°, it is right?
This post has been edited 1 time. Last edited by TuZo, Aug 5, 2018, 12:05 PM
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parmenides51
30629 posts
parmenides51
#4 h  Aug 5, 2018, 12:29 PM • 1 Y
Y by Adventure10
no figure given, it means that $<BIM=90^o$ where $I, M$ incenter and midpoint
Z K Y
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Sen_25
140 posts
Sen_25
#5 h  Aug 5, 2018, 1:22 PM • 2 Y
Y by Adventure10, Mango247
TuZo wrote:
If $M$ is the midpoint of the $BC$, and I the incenter, we have angle(IBM)=angle(ICM), so angleB=angleC=45°, it is right?

How can you say that $ < IBM = < ICM? $
If $ < IBM = < ICM \implies \triangle BIM \cong \triangle CIM \implies <BIM=<CIM=90^{\circ}????$
This post has been edited 1 time. Last edited by Sen_25, Aug 5, 2018, 1:22 PM
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