is an obvious solution. Now, let 
.
or 
.
.
Y by
Prove this.
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Existence of perfect squares
egxa 2
N
by pavel kozlov
Source: All Russian 2025 10.3
Find all natural numbers 
for which there exists an even natural number 
such that the number
![\[
(a - 1)(a^2 - 1)\cdots(a^n - 1)
\]](//latex.artofproblemsolving.com/9/c/a/9caf4eeb82ff46b5ba55ab4b6bc28f0cace586ec.png)
is a perfect square.
for which there exists an even natural number 
such that the number![\[
(a - 1)(a^2 - 1)\cdots(a^n - 1)
\]](http://latex.artofproblemsolving.com/9/c/a/9caf4eeb82ff46b5ba55ab4b6bc28f0cace586ec.png)
is a perfect square.
2 replies
1 viewing
IMO 2014 Problem 4
ipaper 169
N
by YaoAOPS
Let 
and 
be on segment 
of an acute triangle 
such that 
and 
. Let 
and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .
Proposed by Giorgi Arabidze, Georgia.
and 
be on segment 
of an acute triangle 
such that 
and 
. Let 
and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .Proposed by Giorgi Arabidze, Georgia.
169 replies
Inequalities
Scientist10 1
N
by Bergo1305
If 
, then prove that the following inequality holds:
![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](//latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
, then prove that the following inequality holds:![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
1 reply
Tangents forms triangle with two times less area
NO_SQUARES 1
N
by Luis González
Source: Kvant 2025 no. 2 M2831
Let 
be triangle, inscribed in parabola. Tangents in points 
forms triangle . Prove that 
. (
is area of triangle 
).
From F.S.Macaulay's book «Geometrical Conics», suggested by M. Panov
be triangle, inscribed in parabola. Tangents in points 
forms triangle . Prove that 
. (
is area of triangle 
).From F.S.Macaulay's book «Geometrical Conics», suggested by M. Panov
1 reply
FE solution too simple?
Yiyj1 9
N
by jasperE3
Source: 101 Algebra Problems from the AMSP
Find all functions 
such that the equality 
holds for all pairs of real numbers 
.
I feel like my solution is too simple. Is there something I did wrong or something I missed?
such that the equality 
holds for all pairs of real numbers 
.I feel like my solution is too simple. Is there something I did wrong or something I missed?
9 replies
interesting function equation (fe) in IR
skellyrah 2
N
by jasperE3
Source: mine
find all function F: IR->IR such that 
2 replies
Complicated FE
XAN4 1
N
by jasperE3
Source: own
Find all solutions for the functional equation 
, in which 
: 
Note: the solution is actually quite obvious -
, but the proof is important.
Note 2: it is likely that the result can be generalized into a more advanced questions, potentially involving more bash.
, in which 
: 
Note: the solution is actually quite obvious -
, but the proof is important.Note 2: it is likely that the result can be generalized into a more advanced questions, potentially involving more bash.
1 reply
Find all sequences satisfying two conditions
orl 34
N
by YaoAOPS
Source: IMO Shortlist 2007, C1, AIMO 2008, TST 1, P1
Let 
be an integer. Find all sequences 
satisfying the following conditions:
![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](//latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](//latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
Author: Dusan Dukic, Serbia
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
Author: Dusan Dukic, Serbia
34 replies
IMO Shortlist 2011, G4
WakeUp 125
N
by Davdav1232
Source: IMO Shortlist 2011, G4
Let be an acute triangle with circumcircle 
. Let 
be the midpoint of 
and let 
be the midpoint of 
. Let 
be the foot of the altitude from 
and let 
be the centroid of the triangle . Let 
be a circle through and that is tangent to the circle at a point 
. Prove that the points 
and 
are collinear.
Proposed by Ismail Isaev and Mikhail Isaev, Russia
be an acute triangle with circumcircle 
. Let 
be the midpoint of 
and let 
be the midpoint of 
. Let 
be the foot of the altitude from 
and let 
be the centroid of the triangle . Let 
be a circle through and that is tangent to the circle at a point 
. Prove that the points 
and 
are collinear.Proposed by Ismail Isaev and Mikhail Isaev, Russia
125 replies
Z[x], P(\sqrt[3]5+\sqrt[3]25)=5+\sqrt[3]5
jasperE3 5
N
by Assassino9931
Source: VJIMC 2013 2.3
Prove that there is no polynomial with integer coefficients such that ![$P\left(\sqrt[3]5+\sqrt[3]{25}\right)=5+\sqrt[3]5$](//latex.artofproblemsolving.com/5/8/e/58ec2a921cd988bd5ae017a8653544ddd758225f.png)
.
with integer coefficients such that ![$P\left(\sqrt[3]5+\sqrt[3]{25}\right)=5+\sqrt[3]5$](http://latex.artofproblemsolving.com/5/8/e/58ec2a921cd988bd5ae017a8653544ddd758225f.png)
.
5 replies
Tangent Spheres and Tangents to Spheres
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G
H
Source: 2002 British Mathematical Olympiad Round 2
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Y by MS_asdfgzxcvb
Let 
be the distance from to the center of 
. Then 
and by the median length formula 
. Hence 
.
be the distance from to the center of 
. Then 
and by the median length formula 
. Hence 
.
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#3
Apr 2, 2025, 9:17 AM
Y by
kiyoras_2001 wrote:
Let be the distance from to the center of . Then and by the median length formula . Hence .
be the distance from to the center of . Then and by the median length formula . Hence .Thank you for this beautiful short solution.
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