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∑(a-b)(a-c)/(2a^2 + (b+c)^2) >= 0
Zhero 24
N
by RevolveWithMe101
Source: ELMO Shortlist 2010, A2
Let 
be positive reals. Prove that
![\[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \]](//latex.artofproblemsolving.com/6/9/3/6937791380a8642edca3e773dc72dbf3f0e0f206.png)
Calvin Deng.
be positive reals. Prove that![\[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \]](http://latex.artofproblemsolving.com/6/9/3/6937791380a8642edca3e773dc72dbf3f0e0f206.png)
Calvin Deng.
24 replies
i am not abel to prove or disprove
frost23 8
N
by frost23
Source: made on my own
sorrrrrry
8 replies
points on sides of a triangle, intersections, extensions, ratio of areas wanted
parmenides51 1
N
by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1997 OMM P5
Let 
be points on the sides 
respectively of a triangle 
. Suppose that 
and 
meet at 
and meet at 
, and 
and meet at 
, such that 
, and 
. Compute the ratio of the area of 
to the area of 
.
be points on the sides 
respectively of a triangle 
. Suppose that 
and 
meet at 
and meet at 
, and 
and meet at 
, such that 
, and 
. Compute the ratio of the area of 
to the area of 
.
1 reply
starting with intersecting circles, line passes through midpoint wanted
parmenides51 2
N
by EmersonSoriano
Source: Peru Ibero TST 2014
Circles 
and 
intersect at different points 
and 
. The straight lines tangents to that pass through and intersect at 
. Let 
be a point on that is out of . The 
line intersects at 
again, the 
line intersects again to in 
and the line 
intersects again to the circumference in 
. Prove that the 
line passes through the midpoint of the 
segment.
and 
intersect at different points 
and 
. The straight lines tangents to that pass through and intersect at 
. Let 
be a point on that is out of . The 
line intersects at 
again, the 
line intersects again to in 
and the line 
intersects again to the circumference in 
. Prove that the 
line passes through the midpoint of the 
segment.
2 replies
An inequality
Rushil 14
N
by frost23
Source: Indian RMO 1994 Problem 8
If are positive real numbers such that 
, prove that ![\[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]](//latex.artofproblemsolving.com/0/b/e/0bedd9ed8362ac5202845f8a4d0d24519423e9b4.png)
are positive real numbers such that 
, prove that ![\[ (1+a)(1+b)(1+c) \geq 8 (1-a)(1-b)(1-c) . \]](http://latex.artofproblemsolving.com/0/b/e/0bedd9ed8362ac5202845f8a4d0d24519423e9b4.png)
14 replies
collinearity as a result of perpendicularity and equality
parmenides51 2
N
by FrancoGiosefAG
Source: Mexican Mathematical Olympiad 1996 OMM P6
In a triangle with 
, points 
are such that 
and 
and 
, and 
and 
, as shown on the picture. Suppose that 
is a right angle. Prove that the points 
are collinear.
with 
, points 
are such that 
and 
and 
, and 
and 
, as shown on the picture. Suppose that 
is a right angle. Prove that the points 
are collinear.
2 replies
3 var inequality
JARP091 6
N
by JARP091
Source: Own
Let 
. Prove that
![\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]](//latex.artofproblemsolving.com/9/0/3/903220c471bdd3028dda2fcb13874a4bcf752872.png)
without using the Rearrangement Inequality or Chebyshev's Inequality.
. Prove that![\[
\sum_{\text{cyc}} \frac{x^3}{y^2 + z^2} \geq \frac{x + y + z}{2}
\]](http://latex.artofproblemsolving.com/9/0/3/903220c471bdd3028dda2fcb13874a4bcf752872.png)
without using the Rearrangement Inequality or Chebyshev's Inequality.6 replies
Helplooo
Bet667 1
N
by Lil_flip38
Let be an acute angled triangle.And altitudes 
and 
intersects at point 
.Let 
be a point on ray such that 
.Circumcircle of 
intersects line 
at and so prove that 
be an acute angled triangle.And altitudes 
and 
intersects at point 
.Let 
be a point on ray such that 
.Circumcircle of 
intersects line 
at and so prove that 
1 reply
Cyclic sum of 1/(a+1/b+1)
v_Enhance 22
N
by Rayvhs
Source: ELMO Shortlist 2013: Problem A2, by David Stoner
Prove that for all positive reals ,
![\[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}. \]](//latex.artofproblemsolving.com/4/3/1/431b0ed375e037ee3d19e48a2bb4099f12f87784.png)
Proposed by David Stoner
,![\[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}. \]](http://latex.artofproblemsolving.com/4/3/1/431b0ed375e037ee3d19e48a2bb4099f12f87784.png)
Proposed by David Stoner
22 replies
geometry problem
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#7
Apr 21, 2025, 12:36 AM
Y by
Pengu14 wrote:
K1mchi_ wrote:
Shan3t wrote:
25 right????
bc
bc
they’re similar triangles
And?
Yeah. We don't have a second side length. Just the ratio.
Z
K
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