Summer is a great time to explore cool problems to keep your skills sharp!  Schedule a class today!

G
Topic
First Poster
Last Poster
Complex number
ronitdeb   1
N 2 hours ago by alexheinis
Let $z_1, ... ,z_5$ be vertices of regular pentagon inscribed in a circle whose radius is $2$ and center is at $6+i8$. Find all possible values of $z_1^2+z_2^2+...+z_5^2$
1 reply
ronitdeb
Yesterday at 6:13 PM
alexheinis
2 hours ago
Set of Integers
billzhao   41
N 2 hours ago by endless_abyss
Source: USAMO 2004, problem 2
Suppose $a_1, \dots, a_n$ are integers whose greatest common divisor is 1. Let $S$ be a set of integers with the following properties:

(a) For $i=1, \dots, n$, $a_i \in S$.
(b) For $i,j = 1, \dots, n$ (not necessarily distinct), $a_i - a_j \in S$.
(c) For any integers $x,y \in S$, if $x+y \in S$, then $x-y \in S$.

Prove that $S$ must be equal to the set of all integers.
41 replies
billzhao
Apr 29, 2004
endless_abyss
2 hours ago
ai+aj is the multiple of n
Jackson0423   1
N 2 hours ago by alexheinis

Consider an increasing sequence of integers \( a_n \).
For every positive integer \( n \), there exist indices \( 1 \leq i < j \leq n \) such that \( a_i + a_j \) is divisible by \( n \).
Given that \( a_1 \geq 1 \), find the minimum possible value of \( a_{100} \).
1 reply
Jackson0423
Today at 12:41 AM
alexheinis
2 hours ago
Circumscribed Quadrilateral
billzhao   17
N 3 hours ago by endless_abyss
Source: USAMO 2004, problem 1
Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that
\[ 
\frac{1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|. 
\]
When does equality hold?
17 replies
billzhao
Apr 29, 2004
endless_abyss
3 hours ago
Own made functional equation
Primeniyazidayi   2
N 3 hours ago by JARP091
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
2 replies
Primeniyazidayi
May 26, 2025
JARP091
3 hours ago
IMO Shortlist 2008, Geometry problem 2
April   43
N 3 hours ago by ezpotd
Source: IMO Shortlist 2008, Geometry problem 2, German TST 2, P1, 2009
Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE = \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$.

Proposed by Charles Leytem, Luxembourg
43 replies
April
Jul 9, 2009
ezpotd
3 hours ago
In Cyclic Quadrilateral ABCD, find AB^2+BC^2-CD^2-AD^2
Darealzolt   0
4 hours ago
Source: KTOM April 2025 P8
Given Cyclic Quadrilateral \(ABCD\) with an area of \(2025\), with \(\angle ABC = 45^{\circ}\). If \( 2AC^2 = AB^2+BC^2+CD^2+DA^2\), Hence find the value of \(AB^2+BC^2-CD^2-DA^2\).
0 replies
Darealzolt
4 hours ago
0 replies
Plz give me the solution
Madunglecha   1
N 4 hours ago by top1vien
For given M
h(n) is defined as the number of which is relatively prime with M, and 1 or more and n or less.
As B is h(M)/M, prove that there are at least M/3 or more N such that satisfying the below inequality
|h(N)-BN| is under 1+sqrt(B×2^((the number of prime factor of M)-3))
1 reply
Madunglecha
Today at 1:32 AM
top1vien
4 hours ago
King's Constrained Walk
Hellowings   1
N 5 hours ago by Hellowings
Source: Own
Given an n x n chessboard, with a king starting at any square, the king's task is to visit each square in the board exactly once (essentially an open path); this king moves how a king in chess would.
However, we are allowed to place k numbers on the board of any value such that for each number A we placed on the board, the king must be in the position of that number A on its Ath square in its journey, with the starting square as its 1st square.
Suppose after we placed k numbers, there is one and only one way to complete the king's task (this includes placing the king in a starting square), find the minimum value of k set by n.

Didn't know I could post it here xd; I'm unsure how hard this question could be.
1 reply
Hellowings
Today at 1:35 AM
Hellowings
5 hours ago
Inspired by qrxz17
sqing   9
N 6 hours ago by sqing
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
9 replies
sqing
Yesterday at 8:50 AM
sqing
6 hours ago
Iran Team Selection Test 2016
MRF2017   9
N Apr 23, 2025 by SimplisticFormulas
Source: TST3,day1,P2
Let $ABC$ be an arbitrary triangle and $O$ is the circumcenter of $\triangle {ABC}$.Points $X,Y$ lie on $AB,AC$,respectively such that the reflection of $BC$ WRT $XY$ is tangent to circumcircle of $\triangle {AXY}$.Prove that the circumcircle of triangle $AXY$ is tangent to circumcircle of triangle $BOC$.
9 replies
MRF2017
Jul 15, 2016
SimplisticFormulas
Apr 23, 2025
Iran Team Selection Test 2016
G H J
Source: TST3,day1,P2
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MRF2017
237 posts
#1 • 6 Y
Y by baopbc, doxuanlong15052000, Tawan, Adventure10, Mango247, Rounak_iitr
Let $ABC$ be an arbitrary triangle and $O$ is the circumcenter of $\triangle {ABC}$.Points $X,Y$ lie on $AB,AC$,respectively such that the reflection of $BC$ WRT $XY$ is tangent to circumcircle of $\triangle {AXY}$.Prove that the circumcircle of triangle $AXY$ is tangent to circumcircle of triangle $BOC$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
baopbc
225 posts
#2 • 6 Y
Y by buratinogigle, Trafalgar0246, HECAM-CA-CEBEPA, Tawan, Adventure10, Mango247
$\textbf{Proof.}$ Let $\Omega$ be the symmetric of $\odot (AXY)$ through $XY$. Hence, $\Omega$ tangents to $BC$ at $R$.
Let $T$ be the intersection of $\odot (BRY)$ and $\odot (CRX)$. We have : $\angle BTC=\angle BYR+\angle RXC$
But from $\Omega$ is the symmetric of $(AXY)$ through $XY$ so $\angle XRY=\angle XAY.$
Hence $\angle BYR+\angle RXC=2\angle BAC=\angle BOC$ which gives $B,T,O,C$ are concyclic.
Now, by simple angle chasing and we are done. $\blacksquare$
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
andria
824 posts
#3 • 4 Y
Y by rightways, soroush.MG, Tawan, Adventure10
The iranians can see my solution in the following pdf :) :
Attachments:
proof.pdf (27kb)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ismael10
25 posts
#4 • 2 Y
Y by Adventure10, Mango247
Can you clarify andria in your proof why the angle condition imply the tangent result . Thanks in advance.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
andria
824 posts
#6 • 2 Y
Y by Adventure10, Mango247
Ismael10 wrote:
Can you clarify andria in your proof why the angle condition imply the tangent result . Thanks in advance.

Because of the following lemma:

Lemma
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ismael10
25 posts
#7 • 2 Y
Y by Adventure10, Mango247
To prove the lemma we can consider the tangent to one of the circle at X . For the implication it is clear .
For the reciproqual, we find that the condition and the tangency for one of the circle imply the tangency to the other circle.
This post has been edited 1 time. Last edited by Ismael10, Apr 11, 2017, 3:27 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
nixon0630
31 posts
#8 • 1 Y
Y by javohirsultanov20
A really hard problem, I wouldn't have solved it without geogebra. Here's my solution:
Let the reflection of $\odot{(AXY)}$ over $\overline{XY}$ touches $\overline{BC}$ at $T$ and let $Q$ be the Miquel point of $\{X, T, Y\}$ w.r.t. $\triangle{ABC}$.
Claim. $Q$, $B$, $C$, $O$ are concyclic.
Proof.
Finally: $$\measuredangle{BCQ} + \measuredangle{QYX} = \measuredangle{TCQ} + \measuredangle{QYX} = \measuredangle{TYQ} + \measuredangle{QYX} = \measuredangle{TYQ} = \measuredangle{TYX} = \measuredangle{BTX} = \measuredangle{BQX}$$hence, $\odot{AXY}$ and $\odot{BOC}$ are tangent to each other at $Q$. $\blacksquare$
Attachments:
This post has been edited 3 times. Last edited by nixon0630, Oct 22, 2021, 1:28 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
JAnatolGT_00
559 posts
#9
Y by
Oh no, seems like I spent too much time on this, thinking that it would be harder...

Let reflection of $(AXY)$ wrt $XY$ tangents to $BC$ at $S.$ By the Miquel $(AXY),$ $(BSX),$ $(CSY)$ concur at some point $Z.$
$$\measuredangle BZC=\measuredangle BZS+\measuredangle SZC=\measuredangle XAY+\measuredangle YSX=2\measuredangle BAC\implies Z
\in (BOC),$$$$\measuredangle XZB=\measuredangle XSB=\measuredangle XYS=\measuredangle XYZ+\measuredangle ZYS=\measuredangle XZY+\measuredangle ZYS.$$Hence $(AXY),$ $(BOC)$ tangent at $Z.$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
guptaamitu1
658 posts
#10
Y by
One motivation of considering the Miquel configuration is the following:

Suppose $\overline{BC}$ is tangent to reflection of $\odot(AXY)$ in $\overline{XY}$ at $T$ ; and $Q$ is the required tangency point of $\odot(BOC)$ and $\odot(AXY)$ (notations are same as in the diagram above, post #8).If points $B,C,Y,X$ are concyclic, we are just directly done by inversion at $A$ swapping $\{B,X\}$ and $\{C,Y\}$, as it also swaps $$\odot(BOC) \longleftrightarrow  \odot(XYT) \qquad , \qquad \overline{BC} \longleftrightarrow \odot(AXY)$$In that case, $Q$ is just the image of $T$ under inversion, and the only possible way to define $Q$ without inversion or point $A$, is that
$$ Q = \odot(BXT) \cap \odot(CYT) \ne T $$So this motivates us to define $Q$ in general like that too. Luckily, we are just done after that.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SimplisticFormulas
124 posts
#11
Y by
@above $thank$ $you$
Z K Y
N Quick Reply
G
H
=
a