![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
is an obvious solution. Now, let 
.
or 
.
.
Y by Adventure10, Mango247
inscribe 
. 
is the midpoint of 
. 
is the intersection of the tangent of at 
and 
. The tangent of at 
intersects 
at 
. Simillarity, we have 
.Prove that
have second common point (not 
).
G
Topic
First Poster
Last Poster
Estonian Math Competitions 2005/2006
STARS 2
N
by jasperE3
Source: Juniors Problem 4
A 
square is divided into unit squares. Is it possible to fill each unit square with a number 
in such a way that, whenever one places the tile so that it fully covers nine unit squares, the tile will cover nine different numbers?
square is divided into unit squares. Is it possible to fill each unit square with a number 
in such a way that, whenever one places the tile so that it fully covers nine unit squares, the tile will cover nine different numbers?
2 replies
Sum of whose elements is divisible by p
nntrkien 43
N
by lpieleanu
Source: IMO 1995, Problem 6, Day 2, IMO Shortlist 1995, N6
Let 
be an odd prime number. How many -element subsets 
of 
are there, the sum of whose elements is divisible by ?
be an odd prime number. How many -element subsets 
of 
are there, the sum of whose elements is divisible by ?
43 replies
+1 wArrangement of integers in a row with gcd
egxa 2
N
by Qing-Cloud
Source: All Russian 2025 10.5 and 11.5
Let 
be a natural number. The numbers 
are written in a row in some order. For each pair of adjacent numbers, their greatest common divisor (GCD) is calculated and written on a sheet. What is the maximum possible number of distinct values among the 
GCDs obtained?
be a natural number. The numbers 
are written in a row in some order. For each pair of adjacent numbers, their greatest common divisor (GCD) is calculated and written on a sheet. What is the maximum possible number of distinct values among the 
GCDs obtained?2 replies
Integer representation
RL_parkgong_0106 1
N
by Jackson0423
Source: Own
Show that for any positive integer 
, there exists some positive integer 
that makes the following equation have no integer root 
.
, there exists some positive integer 
that makes the following equation have no integer root 
.
1 reply
Factor of P(x)
Brut3Forc3 19
N
by xytunghoanh
Source: 1976 USAMO Problem 5
If 
, and 
are all polynomials such that ![\[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\]](//latex.artofproblemsolving.com/1/6/a/16a71abc110ece558d427a07d7be2d1f72148024.png)
prove that 
is a factor of 
.
, and 
are all polynomials such that ![\[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\]](http://latex.artofproblemsolving.com/1/6/a/16a71abc110ece558d427a07d7be2d1f72148024.png)
prove that 
is a factor of 
.
19 replies
2^x+3^x = yx^2
truongphatt2668 1
N
by Jackson0423
Prove that the following equation has infinite integer solutions:
1 reply
FE solution too simple?
Yiyj1 7
N
by ariopro1387
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?
7 replies
A cyclic inequality
KhuongTrang 2
N
by NguyenVanDucThang
Source: own-CRUX
IMAGE
https://cms.math.ca/.../uploads/2025/04/Wholeissue_51_4.pdf
https://cms.math.ca/.../uploads/2025/04/Wholeissue_51_4.pdf
2 replies
Iran second round 2025-q1
mohsen 3
N
by Parsia--
Find all positive integers n>2 such that sum of n and any of its prime divisors is a perfect square.
3 replies
Geometry
G
H
G
H
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Y by Adventure10
noway wrote:
LKira wrote:
My geogebra say no, it not true, may you check again your problem ?
I have checked carefully, it's true ^^
Then see VMO 2018 P7
This post has been edited 1 time. Last edited by LKira, Dec 5, 2019, 2:26 PM
Z
K
Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Y by Adventure10, Mango247
Your second common point of the three circles has the first baricentric coordinate
a^2 (2 a^16-6 a^14 (b^2+c^2)+a^12 (-4 b^4+38 b^2 c^2-4 c^4)+2 a^10 (9 b^6-20 b^4 c^2-20 b^2 c^4+9 c^6)+a^8 (-67 b^6 c^2+174 b^4 c^4-67 b^2 c^6)+4 a^4 (b^2-c^2)^2 (b^8+9 b^6 c^2-32 b^4 c^4+9 b^2 c^6+c^8)-(b^4-c^4)^2 (2 b^8-9 b^6 c^2+10 b^4 c^4-9 b^2 c^6+2 c^8)+2 a^2 (b^2-c^2)^2 (3 b^10-18 b^8 c^2+11 b^6 c^4+11 b^4 c^6-18 b^2 c^8+3 c^10)-2 a^6 (9 b^10-51 b^8 c^2+46 b^6 c^4+46 b^4 c^6-51 b^2 c^8+9 c^10))
This triangle center is not currently in https://faculty.evansville.edu/ck6/encyclopedia/ETC.html
a^2 (2 a^16-6 a^14 (b^2+c^2)+a^12 (-4 b^4+38 b^2 c^2-4 c^4)+2 a^10 (9 b^6-20 b^4 c^2-20 b^2 c^4+9 c^6)+a^8 (-67 b^6 c^2+174 b^4 c^4-67 b^2 c^6)+4 a^4 (b^2-c^2)^2 (b^8+9 b^6 c^2-32 b^4 c^4+9 b^2 c^6+c^8)-(b^4-c^4)^2 (2 b^8-9 b^6 c^2+10 b^4 c^4-9 b^2 c^6+2 c^8)+2 a^2 (b^2-c^2)^2 (3 b^10-18 b^8 c^2+11 b^6 c^4+11 b^4 c^6-18 b^2 c^8+3 c^10)-2 a^6 (9 b^10-51 b^8 c^2+46 b^6 c^4+46 b^4 c^6-51 b^2 c^8+9 c^10))
This triangle center is not currently in https://faculty.evansville.edu/ck6/encyclopedia/ETC.html
Z
K
Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Y by Adventure10, Mango247
Is this correct?
We begin with the main claim
: Given a 
and point 
on its euler line. Let line 
,
,
intersect the 
again at 
,
, and 
respectively. Define 
to be the midpoint of 
,
and 
respectively and define also 
,
,
to be the midpont of 
,
and 
respectively. Then we have that 
,
and 
are concurrent.
:
Animate point on the euler line, therefore we will get that the degree of point 
are all 2. Hence the degree of 
are also 2. Since point are all fixed we must have that the degree of line 
are all 2. So if we want to prove that are concurrent we only need to check that for 
possibilities of point , are always concurrent. It's easy to check that for is the circumcenter, centroid the two intersections of euler line with the circumcircle of 
, intersection of euler line with ,
,, the three lines always concurrent. Therefore, 
We go back to the original problem
Define
be the point on 
different from 
such that 
, 
and 
are all tangent to . Now define 
to be the midpoint of 
, 
and 
respectively.
Invert the figure WRT to, then the original problem is equivalent to proving that 
,
and 
are concur at one point.
However, it's a classic result from IMO 2011 G4 that
and 
are concurrent at point 
, the isogonal conjugate of isotomic conjugate of orthocenter 
of . Moreover it's well known that this point is on the euler line. Therefore by the first claim, the problem is done.
We begin with the main claim
: Given a and point 
on its euler line. Let line 
,
,
intersect the 
again at 
,
, and 
respectively. Define 
to be the midpoint of 
,
and 
respectively and define also 
,
,
to be the midpont of 
,
and 
respectively. Then we have that 
,
and 
are concurrent.
:Animate point
on the euler line, therefore we will get that the degree of point 
are all 2. Hence the degree of 
are also 2. Since point are all fixed we must have that the degree of line 
are all 2. So if we want to prove that are concurrent we only need to check that for 
possibilities of point , are always concurrent. It's easy to check that for is the circumcenter, centroid the two intersections of euler line with the circumcircle of 
, intersection of euler line with ,
,, the three lines always concurrent. Therefore, 
We go back to the original problem
Define
be the point on different from 
such that 
, 
and 
are all tangent to . Now define 
to be the midpoint of 
, 
and 
respectively.Invert the figure WRT to
, then the original problem is equivalent to proving that 
,
and 
are concur at one point.However, it's a classic result from IMO 2011 G4 that
and 
are concurrent at point 
, the isogonal conjugate of isotomic conjugate of orthocenter 
of . Moreover it's well known that this point is on the euler line. Therefore by the first claim, the problem is done.This post has been edited 5 times. Last edited by Sugiyem, Dec 23, 2019, 10:11 PM
Z
K
Y
N Quick Reply
G
H
