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Geometry with fix circle
falantrng   32
N 12 minutes ago by bjump
Source: RMM 2018 Problem 6
Fix a circle $\Gamma$, a line $\ell$ to tangent $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ meet $\ell$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the circumcircle of $XYZ$ is tangent to two fixed circles.
32 replies
falantrng
Feb 25, 2018
bjump
12 minutes ago
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Problem 6
SlovEcience   4
N 16 minutes ago by ZeltaQN2008
Given two points A and B on the unit circle. The tangents to the circle at A and B intersect at point P. Then:
\[ p = \frac{2ab}{a + b} \], \[ p, a, b \in \mathbb{C} \]
4 replies
SlovEcience
May 3, 2025
ZeltaQN2008
16 minutes ago
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Perpendicular passes from the intersection of diagonals, \angle AEB = \angle CED
NO_SQUARES   0
17 minutes ago
Source: 239 MO 2025 10-11 p3
Inside of convex quadrilateral $ABCD$ point $E$ was chosen such that $\angle DAE = \angle CAB$ and $\angle ADE = \angle CDB$. Prove that if perpendicular from $E$ to $AD$ passes from the intersection of diagonals of $ABCD$, then $\angle AEB = \angle CED$.
0 replies
NO_SQUARES
17 minutes ago
0 replies
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A lot of circles
ryan17   7
N 18 minutes ago by SimplisticFormulas
Source: 2019 Polish MO Finals
Denote by $\Omega$ the circumcircle of the acute triangle $ABC$. Point $D$ is the midpoint of the arc $BC$ of $\Omega$ not containing $A$. Circle $\omega$ centered at $D$ is tangent to the segment $BC$ at point $E$. Tangents to the circle $\omega$ passing through point $A$ intersect line $BC$ at points $K$ and $L$ such that points $B, K, L, C$ lie on the line $BC$ in that order. Circle $\gamma_1$ is tangent to the segments $AL$ and $BL$ and to the circle $\Omega$ at point $M$. Circle $\gamma_2$ is tangent to the segments $AK$ and $CK$ and to the circle $\Omega$ at point $N$. Lines $KN$ and $LM$ intersect at point $P$. Prove that $\sphericalangle KAP = \sphericalangle EAL$.
7 replies
ryan17
Jul 9, 2019
SimplisticFormulas
18 minutes ago
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Polyline with increasing links
NO_SQUARES   0
20 minutes ago
Source: 239 MO 2025 10-11 p1
There are $100$ points on the plane, all pairwise distances between which are different. Is there always a polyline with vertices at these points, passing through each point once, in which the link lengths increase monotonously?
0 replies
NO_SQUARES
20 minutes ago
0 replies
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|BX-CX| \geqslant |AD - DX| in a simple construction with incircle
NO_SQUARES   0
22 minutes ago
Source: 239 MO 2025 8-9 p8
The incircle of a right triangle $ABC$ touches its hypotenuse $BC$ at point $D$. The line $AD$ intersects the circumscribed circle at point $X$. Prove that $ |BX-CX| \geqslant |AD - DX|$.
0 replies
NO_SQUARES
22 minutes ago
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Inequality with a,b,c
GeoMorocco   4
N 24 minutes ago by Natrium
Source: Morocco Training
Let $ a,b,c $ be positive real numbers such that : $ ab+bc+ca=3 $ . Prove that : $$\frac{\sqrt{1+a^2}}{1+ab}+\frac{\sqrt{1+b^2}}{1+bc}+\frac{\sqrt{1+c^2}}{1+ca}\ge \sqrt{\frac{3(a+b+c)}{2}}$$
4 replies
1 viewing
GeoMorocco
Apr 11, 2025
Natrium
24 minutes ago
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The chip is aiming for the upper right corner
NO_SQUARES   0
28 minutes ago
Source: 239 MO 2025 8-9 p7
Given a $2025 \times 2025$ board and $k$ chips lying on the table. Initially, the board is empty. It is allowed to place a chip from the table on any free square $A$ if two conditions are met simultaneously:
– the cell next to $A$ from below has a chip (or $A$ is on the bottom edge of the board);
– the cell next to $A$ on the left has a chip (or $A$ is on the left edge of the board).
In addition, it is allowed to remove any chip from the board and put it on the table. At what minimum $k$ can a chip appear in the upper-right corner of the board?
0 replies
NO_SQUARES
28 minutes ago
0 replies
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x^2-x divides by n for some n/\omega(n)+1>x>1
NO_SQUARES   0
32 minutes ago
Source: 239 MO 2025 8-9 p6
Let a positive integer number $n$ has $k$ different prime divisors. Prove that there exists a positive integer number $x \in \left(1, \frac{n}{k}+1 \right)$ such that $x^2-x$ divides by $n$.
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NO_SQUARES
32 minutes ago
0 replies
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9-gon and well-colored in 7 colors plane
NO_SQUARES   0
36 minutes ago
Source: 239 MO 2025 8-9 p5
We will say that a plane is well-colored in several colors if it is divided into convex polygons with an area of at least $1/1000$ and each polygon is colored in one color. Points lying on the border of several polygons can be colored in any of their colors. Are there convex is a $9$-gon $R$ and a good coloring of the plane in $7$ colors such that in any polygon obtained from $R$ by a translate to any vector, all colors occupy the same area ($1/7$ of the area of $R$)?
0 replies
NO_SQUARES
36 minutes ago
0 replies
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\sqrt{2-a}+\sqrt{2-b}+\sqrt{2-c}\geqslant 2+\sqrt{(2-a)(2-b)(2-c)}
NO_SQUARES   0
44 minutes ago
Source: 239 MO 2025 8-9 p4
Positive numbers $a$, $b$ and $c$ are such that $a^2+b^2+c^2+abc=4$. Prove that \[\sqrt{2-a}+\sqrt{2-b}+\sqrt{2-c}\geqslant 2+\sqrt{(2-a)(2-b)(2-c)}.\]
0 replies
NO_SQUARES
44 minutes ago
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Polygons Which Don't Fit
somebodyyouusedtoknow   1
N Apr 27, 2025 by kiyoras_2001
Source: San Diego Honors Math Contest 2025 Part II, Problem 1
Let $P_1,P_2,\ldots,P_n$ be polygons, no two of which are similar. Show that there are polygons $Q_1,Q_2,\ldots,Q_n$ where $Q_i$ is similar to $P_i$ so that for no $i \neq j$ does $Q_i$ contain a polygon that's congruent to $Q_j$.

Note. Here, the word "contain" means for the construction we have, we cannot select a size for $Q_j$ so that $Q_j$ is wholly contained in $Q_i$, and so it does not intersect the edges of $Q_i$ at all.
1 reply
somebodyyouusedtoknow
Apr 26, 2025
kiyoras_2001
Apr 27, 2025
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Polygons Which Don't Fit
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Reveal topic tags
geometry
polygon
similarity
combinatorics
combinatorial geometry
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G H BBookmark kLocked kLocked NReply
Source: San Diego Honors Math Contest 2025 Part II, Problem 1
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somebodyyouusedtoknow
259 posts
somebodyyouusedtoknow
#1 h  Apr 26, 2025, 11:28 PM
Y by
Let $P_1,P_2,\ldots,P_n$ be polygons, no two of which are similar. Show that there are polygons $Q_1,Q_2,\ldots,Q_n$ where $Q_i$ is similar to $P_i$ so that for no $i \neq j$ does $Q_i$ contain a polygon that's congruent to $Q_j$.

Note. Here, the word "contain" means for the construction we have, we cannot select a size for $Q_j$ so that $Q_j$ is wholly contained in $Q_i$, and so it does not intersect the edges of $Q_i$ at all.
This post has been edited 1 time. Last edited by somebodyyouusedtoknow, Apr 26, 2025, 11:41 PM
Reason: Added clarity for the notes
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kiyoras_2001
678 posts
kiyoras_2001
#2 h  Apr 27, 2025, 12:09 PM
Y by
Induct on $n$. The base case $n=2$ is obvious; for fixed $Q_1=P_1$ consider the largest copy of $P_2$ inside it and slightly enlarge it to obtain $Q_2$.

Now the step $n\to n+1$. That is, given $Q_1, \ldots, Q_n$ with $Q_i \not\subset Q_j$ for all $i\ne j$ we seek for appropriate $Q_{n+1}$.

For each $i\in [n]:= \{1, \ldots, n\}$ define $R_i^-$ as the largest copy of $P_{n+1}$ inside $Q_i$. As well as define $R_i^+$ as the smallest copy of $P_{n+1}$ containing $Q_i$.

We want to show that there is a copy of $P_{n+1}$ which is larger than $R_i^-$ and smaller than $R_i^+$ for all $i\in [n]$. That is, we need to prove that $\max_{i\in [n]} R_i^- \subsetneq \min_{i\in [n]}R_i^+$. Suppose not, then there are $i\ne j$ such that $Q_i \subseteq R_i^+ \subseteq R_j^- \subseteq Q_j$. This contradiction proves that we can select $Q_{n+1}$ appropriately.
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