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Hagge circle, Thomson cubic, coaxal
kosmonauten3114   0
7 minutes ago
Source: My own (maybe well-known)
Let $\triangle{ABC}$ be a triangle, $\triangle{M_AM_BM_C}$ its medial triangle, and $P$ a point on the Thomson cubic (= $\text{K002}$) of $\triangle{ABC}$. (Suppose that $P \notin \odot(ABC)$ ).
Let $\triangle{A'B'C'}$ be the circumcevian triangle of $P$ wrt $\triangle{ABC}$.
Let $\triangle{P_AP_BP_C}$ be the pedal triangle of $P$ wrt $\triangle{ABC}$.
Let $A_1$ be the reflection in $BC$ of $A'$. Define $B_1$, $C_1$ cyclically.
Let $A_2$ be the reflection in $M_A$ of $A'$. Define $B_2$, $C_2$ cyclically.
Let $A_3$ be the reflection in $P_A$ of $A'$. Define $B_3$, $C_3$ cyclically.

Prove that $\odot(A_1B_1C_1)$, $\odot(A_2B_2C_2)$, $\odot(A_3B_3C_3)$ and the orthocentroidal circle of $\triangle{ABC}$ are coaxal.
0 replies
+1 w
kosmonauten3114
7 minutes ago
0 replies
Geometry One Problem Bounty Hunt Contest: Trapezium Geometry
anantmudgal09   14
N 13 minutes ago by ihategeo_1969
Source: zephyrcrush78
Let $ABC$ be an acute-angled scalene triangle with incircle $\omega$ and circumcircle $\Gamma$. Suppose $\omega$ touches line $BC$ at $D$ and the tangent to $\Gamma$ at $A$ meets line $BC$ at $T$. Two circles passing through $A$ and $T$ tangent to $\omega$ meet line $AD$ again at $X$ and $Y$.

Prove that $BXCY$ is a trapezium.

14 replies
anantmudgal09
Jun 3, 2024
ihategeo_1969
13 minutes ago
Max and Min
Butterfly   0
39 minutes ago

Let $a_1,a_2,\cdots,a_n$ be an arrangement of $\{1,2,3,\cdots,n\}$. Find the maximum and minimum values of $$\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}.$$
0 replies
+1 w
Butterfly
39 minutes ago
0 replies
Cup of Combinatorics
M11100111001Y1R   4
N 43 minutes ago by sami1618
Source: Iran TST 2025 Test 4 Problem 2
There are \( n \) cups labeled \( 1, 2, \dots, n \), where the \( i \)-th cup has capacity \( i \) liters. In total, there are \( n \) liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.

$a)$ Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most \( \frac{4n}{3} \) steps.

$b)$ Prove that in at most \( \frac{5n}{3} \) steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
4 replies
M11100111001Y1R
May 27, 2025
sami1618
43 minutes ago
(c^n+1)/(2^na+b) is an integer for all n
parmenides51   2
N an hour ago by Assassino9931
Source: Ukraine TST 2010 p6
Find all pairs of odd integers $a$ and $b$ for which there exists a natural number$ c$ such that the number $\frac{c^n+1}{2^na+b}$ is integer for all natural $n$.
2 replies
parmenides51
May 4, 2020
Assassino9931
an hour ago
Nine point circle + Perpendicularities
YaoAOPS   18
N an hour ago by AndreiVila
Source: 2025 CTST P2
Suppose $\triangle ABC$ has $D$ as the midpoint of $BC$ and orthocenter $H$. Let $P$ be an arbitrary point on the nine point circle of $ABC$. The line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$. The line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$. If $M$ is the midpoint of $AQ$, show that $HX \perp DM$.
18 replies
YaoAOPS
Mar 5, 2025
AndreiVila
an hour ago
Inequality conjecture
RainbowNeos   0
an hour ago
Show (or deny) that there exists an absolute constant $C>0$ that, for all $n$ and $n$ positive real numbers $x_i ,1\leq i \leq n$, there is
\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]
0 replies
RainbowNeos
an hour ago
0 replies
inequality 2905
pennypc123456789   0
an hour ago
Consider positive real numbers \( x, y, z \) that satisfy the condition
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3.
\]Find the maximum value of the expression
\[
P = \dfrac{yz}{\sqrt[3]{3y^2z^2+ 3x^2y^2z^2+ x^2z^2 + x^2y^2}}
+ \frac{xz}{\sqrt[3]{3x^2z^2 + 3x^2y^2z^2 + x^2y^2 + y^2z^2}}
+ \frac{xy}{\sqrt[3]{3x^2y^2 + 3x^2y^2z^2 +y^2z^2 + x^2z^2}}.
\]
0 replies
pennypc123456789
an hour ago
0 replies
Inspired by m4thbl3nd3r
sqing   3
N 2 hours ago by sqing
Source: Own
Let $  a, b,c>0,b+c>a$. Prove that$$\sqrt{\frac{a}{b+c-a}}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)}\geq 1$$$$\frac{a}{b+c-a}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)} \geq  \frac{4\sqrt 2}{3}-1$$
3 replies
sqing
Today at 3:43 AM
sqing
2 hours ago
Inspired by qrxz17
sqing   7
N 2 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 $$a+b+c\geq 3\sqrt {3}$$
7 replies
sqing
6 hours ago
sqing
2 hours ago
Geometry problem
Whatisthepurposeoflife   2
N 2 hours ago by Whatisthepurposeoflife
Source: Derived from MEMO 2024 I3
Triangle ∆ABC is scalene the circle w that goes through the points A and B intersects AC at E BC at D let the Lines BE and AD intersect at point F. And let the tangents A and B of circle w Intersect at point G.
Prove that C F and G are collinear
2 replies
Whatisthepurposeoflife
Yesterday at 1:45 PM
Whatisthepurposeoflife
2 hours ago
A Sequence of +1's and -1's
ike.chen   36
N 2 hours ago by maromex
Source: ISL 2022/C1
A $\pm 1$-sequence is a sequence of $2022$ numbers $a_1, \ldots, a_{2022},$ each equal to either $+1$ or $-1$. Determine the largest $C$ so that, for any $\pm 1$-sequence, there exists an integer $k$ and indices $1 \le t_1 < \ldots < t_k \le 2022$ so that $t_{i+1} - t_i \le 2$ for all $i$, and $$\left| \sum_{i = 1}^{k} a_{t_i} \right| \ge C.$$
36 replies
ike.chen
Jul 9, 2023
maromex
2 hours ago
Path within S which does not meet itself
orl   5
N Apr 6, 2025 by atdaotlohbh
Source: IMO 1982, Day 2, Problem 6
Let $S$ be a square with sides length $100$. Let $L$ be a path within $S$ which does not meet itself and which is composed of line segments $A_0A_1,A_1A_2,A_2A_3,\ldots,A_{n-1}A_n$ with $A_0=A_n$. Suppose that for every point $P$ on the boundary of $S$ there is a point of $L$ at a distance from $P$ no greater than $\frac {1} {2}$. Prove that there are two points $X$ and $Y$ of $L$ such that the distance between $X$ and $Y$ is not greater than $1$ and the length of the part of $L$ which lies between $X$ and $Y$ is not smaller than $198$.
5 replies
orl
Nov 11, 2005
atdaotlohbh
Apr 6, 2025
Path within S which does not meet itself
G H J
Source: IMO 1982, Day 2, Problem 6
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orl
3647 posts
#1 • 1 Y
Y by Adventure10
Let $S$ be a square with sides length $100$. Let $L$ be a path within $S$ which does not meet itself and which is composed of line segments $A_0A_1,A_1A_2,A_2A_3,\ldots,A_{n-1}A_n$ with $A_0=A_n$. Suppose that for every point $P$ on the boundary of $S$ there is a point of $L$ at a distance from $P$ no greater than $\frac {1} {2}$. Prove that there are two points $X$ and $Y$ of $L$ such that the distance between $X$ and $Y$ is not greater than $1$ and the length of the part of $L$ which lies between $X$ and $Y$ is not smaller than $198$.
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SCP
1502 posts
#3 • 2 Y
Y by Adventure10, Mango247
There is sth wrong ith my idea:

The length of $X,Y>4*99-1$ is quite trivial in one direction, but if they ean the shortest distance, we just take points very close to the boundary and with angles close to $180$ degrees except in the corners, what is the mistake in those ideas?
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mavropnevma
15142 posts
#4 • 2 Y
Y by Adventure10, Mango247
See John Scholes' solution at http://mks.mff.cuni.cz/kalva/.
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Phorphyrion
398 posts
#6
Y by
The condition should be $A_0\neq A_n$. That's the statement appearing on the official IMO website.
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Lukariman
417 posts
#7
Y by
The solution is difficult to understand because it involves compact sets. Can anyone explain the properties of compact sets
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atdaotlohbh
193 posts
#8
Y by
The first one to post a solution!
Start walking from $A_0$ to $A_n$ along $L$. Paint in red all points on the boundary of $S$ that are within distance $\frac{1}{2}$ from the current point. Obviously at any point the red points are a finite union of red segments (with possibly length zero). Suppose such points $X$ and $Y$ do not exist.

The key observation: If $AB$ and $CD$ are opposite sides of squares, then we must completely paint $AB$ before we start painting $CD$ (or vice versa).
Proof: Suppose it is false and $AB$ was partly painted when we firstly painted a point on $CD$. Let $X$ be a red point on $AB$ such that on some side of it on $AB$ its close is not painted (More simply, an endpoint of a segment). Say $X$ was painted when passing $M$ on $L$. We must return to paint the neighbourhood of $X$, and as we always paint segments, we will point $X$ once again as well, say when passing through $N$. Notice that from triangle inequality $MN \leq MX+NX \leq \frac{1}{2}+\frac{1}{2}=1$. Also, to get from $M$ to a point which painted $CD$ a bit we must have travelled at least $99$, and to get back to $N$ we need $99$ again. Thus, the distance travelled between $M$ and $N$ is at least 198, and so $M$ and $N$ work as $X$ and $Y$ in the problem statement.

Now we finish by proving that the key observation is impossible. To do that, let the square be $ABCD$ and $AB$ painted before $CD$, $AD$ painted before $BC$. WLOG $B$ was painted before $D$, but $B$ is a part of $BC$, thus before $B$ was painted $AD$ was completely done already, and hence $D$ also, which is the desired contradiction.

Notice we never used the fact that $L$ doesn't intersect itself.
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