, such that 
Prove that :
Y by
Given 
such that 
. Prove that
such that 
. Prove that
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4 lines concurrent
Zavyk09 4
N
by pingupignu
Source: Homework
Let 
be triangle with circumcenter 
and orthocenter 
. 
intersect again at 
respectively. Lines through parallel to 
intersects 
at 
respectively. Point 
such that 
is a parallelogram. Prove that lines 
and 
are concurrent at a point on 
.
be triangle with circumcenter 
and orthocenter 
. 
intersect again at 
respectively. Lines through parallel to 
intersects 
at 
respectively. Point 
such that 
is a parallelogram. Prove that lines 
and 
are concurrent at a point on 
.
4 replies
Two circles and Three line concurrency
mofidy 0
Two circles 
and 
with equal radii intersect at P and Q. Points B and C are located on the circles and so that they are inside the circles and , respectively. Also, points X and Y distinct from P are located on and , respectively, so that:
The intersection point of the circumcircles of triangles XPC and YPB is called S. Prove that BC, XY and QS are concurrent.
Thanks.
and 
with equal radii intersect at P and Q. Points B and C are located on the circles and so that they are inside the circles and , respectively. Also, points X and Y distinct from P are located on and , respectively, so that:
The intersection point of the circumcircles of triangles XPC and YPB is called S. Prove that BC, XY and QS are concurrent.Thanks.
0 replies
Tilted Students Thoroughly Splash Tiger part 2
DottedCaculator 17
N
by HoRI_DA_GRe8
Source: ELMO 2024/5
In triangle with 
and 
, let 
be the midpoint of 
. Choose point 
on the extension of 
past 
and point 
on segment 
such that lies on 
. Let 
be on the opposite side of 
from 
such that 
and 
. Let 
intersect the circumcircle of 
again at 
, and let 
intersect the circumcircle of 
again at 
. Prove that , 
, and 
are collinear.
Tiger Zhang
with 
and 
, let 
be the midpoint of 
. Choose point 
on the extension of 
past 
and point 
on segment 
such that lies on 
. Let 
be on the opposite side of 
from 
such that 
and 
. Let 
intersect the circumcircle of 
again at 
, and let 
intersect the circumcircle of 
again at 
. Prove that , 
, and 
are collinear.Tiger Zhang
17 replies
radii relationship
steveshaff 0
Two externally tangent circles with radii a and b are each internally tangent to a semicircle and its diameter. The two points of tangency on the semicircle and the two points of tangency on its diameter lie on a circle of radius r. Prove that r^2 = 3ab.
0 replies
NT Function with divisibility
oVlad 3
N
by sangsidhya
Source: Romanian District Olympiad 2023 9.4
Determine all strictly increasing functions 
which satisfy ![\[f(x)\cdot f(y)\mid (1+2x)\cdot f(y)+(1+2y)\cdot f(x)\]](//latex.artofproblemsolving.com/6/4/f/64f2720634f57b4d88db823b9931379386ac1c27.png)
for all non-negative integers 
and 
.
which satisfy ![\[f(x)\cdot f(y)\mid (1+2x)\cdot f(y)+(1+2y)\cdot f(x)\]](http://latex.artofproblemsolving.com/6/4/f/64f2720634f57b4d88db823b9931379386ac1c27.png)
for all non-negative integers 
and 
.
3 replies
Minimum with natural numbers
giangtruong13 1
N
by Ianis
Let 
be natural numbers such that: 
and 
. Find the min: 
be natural numbers such that: 
and 
. Find the min: 
1 reply
angle wanted, right ABC, AM=CB , CN=MB
parmenides51 3
N
by Mathzeus1024
Source: 2022 European Math Tournament - Senior First + Grand League - Math Battle 1.3
In a right-angled triangle , points and 
are taken on the legs 
and 
, respectively, so that 
and 
. Find the acute angle between line segments 
and 
.
, points and 
are taken on the legs 
and 
, respectively, so that 
and 
. Find the acute angle between line segments 
and 
.
3 replies
polonomials
Ducksohappi 0
Let 
be the real polonomial such that all roots are real and distinct. Prove that there is a rational number 
that all roots of 
are real numbers
be the real polonomial such that all roots are real and distinct. Prove that there is a rational number 
that all roots of 
are real numbers
0 replies
IMO ShortList 2002, algebra problem 4
orl 62
N
by Ihatecombin
Source: IMO ShortList 2002, algebra problem 4
Find all functions 
from the reals to the reals such that ![\[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz) \]](//latex.artofproblemsolving.com/6/1/0/610e13a6f27341b5eed393923bb98e182537e254.png)
for all real .
from the reals to the reals such that ![\[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz) \]](http://latex.artofproblemsolving.com/6/1/0/610e13a6f27341b5eed393923bb98e182537e254.png)
for all real .
62 replies
An inequality problem
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Y by Arithmetic_fighter
Clearly 
yields a stronger inequality, so let 
, 
, 
. We need to show that
![\[\sum_{\rm cyc}\frac{ab}{c^2+a^2+1}=\sum_{\rm cyc}\frac{3ab}{4c^2+4a^2+b^2}\le1,\]](//latex.artofproblemsolving.com/9/c/2/9c26ff3a0d26422b2e6b8737508ce7c6ceb898ff.png)
or
![\[\sum_{\rm cyc}\left(\frac34-\frac{3ab}{4c^2+4a^2+b^2}\right)=\sum_{\rm cyc}\frac{12c^2+3(2a-b)^2}{4\left(4c^2+4a^2+b^2\right)}\ge3\times\frac34-1=\frac54.\]](//latex.artofproblemsolving.com/9/d/f/9df0e5e43eb4aa5bc7ed6840c91f1b5407f30cdc.png)
By C-S, we have
![\[\sum_{\rm cyc}\frac{12c^2+3(2a-b)^2}{4c^2+4a^2+b^2}\ge\frac{\left\{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\right\}^2}{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\left(4c^2+4a^2+b^2\right)}.\]](//latex.artofproblemsolving.com/9/7/d/97d8d081d238446616e436860ae189d00e82adfd.png)
Therefore, it suffices to show that
![\[\left\{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\right\}^2\ge5\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\left(4c^2+4a^2+b^2\right).\]](//latex.artofproblemsolving.com/7/1/8/71827e7b9f7dac09b44e21ddbd0ce942c317f3ff.png)
This is a fourth degree polynomial inequality, so we have
yields a stronger inequality, so let 
, 
, 
. We need to show that![\[\sum_{\rm cyc}\frac{ab}{c^2+a^2+1}=\sum_{\rm cyc}\frac{3ab}{4c^2+4a^2+b^2}\le1,\]](http://latex.artofproblemsolving.com/9/c/2/9c26ff3a0d26422b2e6b8737508ce7c6ceb898ff.png)
or![\[\sum_{\rm cyc}\left(\frac34-\frac{3ab}{4c^2+4a^2+b^2}\right)=\sum_{\rm cyc}\frac{12c^2+3(2a-b)^2}{4\left(4c^2+4a^2+b^2\right)}\ge3\times\frac34-1=\frac54.\]](http://latex.artofproblemsolving.com/9/d/f/9df0e5e43eb4aa5bc7ed6840c91f1b5407f30cdc.png)
By C-S, we have![\[\sum_{\rm cyc}\frac{12c^2+3(2a-b)^2}{4c^2+4a^2+b^2}\ge\frac{\left\{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\right\}^2}{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\left(4c^2+4a^2+b^2\right)}.\]](http://latex.artofproblemsolving.com/9/7/d/97d8d081d238446616e436860ae189d00e82adfd.png)
Therefore, it suffices to show that![\[\left\{\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\right\}^2\ge5\sum\limits_{\rm cyc}\left[12c^2+3(2a-b)^2\right]\left(4c^2+4a^2+b^2\right).\]](http://latex.artofproblemsolving.com/7/1/8/71827e7b9f7dac09b44e21ddbd0ce942c317f3ff.png)
This is a fourth degree polynomial inequality, so we have
![\begin{align*}
&m \sum_{\rm cyc} a^{4}+n \sum_{\rm cyc} b^{2} c^{2}+p \sum_{\rm cyc} a^{3} b+q \sum_{\rm cyc} a b^{3}-(m+n+p+q) \sum_{\rm cyc} a^2bc \\
={}&\frac{1}{18 m} \sum_{\rm cyc}\left[3 m\left(a^{2}-b^{2}\right)+(p+2 q) c a+(p-q) a b-(2 p+q) b c\right]^{2}\\
&+\frac{3 m(m+n)-\left(p^{2}+p q+q^{2}\right)}{18 m\left(p^{2}+p q+q^{2}\right)} \sum_{\rm cyc}[(p+2 q) c a+(p-q) a b-(2 p+q) b c]^{2}.
\end{align*}](http://latex.artofproblemsolving.com/7/e/f/7ef2fe85183f92f06e1decea6e259a27ec2da3bd.png)
This post has been edited 1 time. Last edited by Quantum-Phantom, Apr 3, 2025, 6:52 AM
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#3
Apr 3, 2025, 5:45 AM
Y by
Quantum-Phantom wrote:
This is a fourth-degree polynomial inequality, so we have
Could you please clarify how you obtained this equation? It seems complicated.
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N Quick Reply
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