Hard Combi Geo

by AbbyWong, Mar 30, 2025, 11:03 PM

A (possibly non-convex) planar polygon P is good if no two sides of P are parallel.
For any good polygon P, we may take any three sides of P and extend them into lines. These lines
intersect to form a triangle. Such a triangle is called a peritriangle of P. Let f(P) denote the minimal
number of peritriangles of P whose union completely cover P.
For each positive integer n, find all possible values of f(P) as P ranges over all good n-gons.

combinatorial geometry

combinatorics

ded
comment

Best bid given the other's bid

by fancyfairy, Mar 30, 2025, 10:04 PM

Consider sets $N = \{1,2\}$ (player) and $M = \{a,b,c,d\}$ (items), with valuations given by matrix
$V = \left[\begin{matrix} v_{1a} & v_{1b} & v_{1c} & v_{1d} \\ v_{2a} & v_{2b} & v_{2c} & v_{2d} \end{matrix}\right]$
where $v_{ij} \in [0, \infty]$ .

A player’s total valuation is the sum of item values won, with half-value for half-won items.

Given player $N \backslash \{i\}$ 's bid, player $i$ 's bid $B_i$ is strictly better than another possible bid $B_i'$ of his if the total valuation derived from $B_i$ is strictly more than that derived from ${B_i}'$ .

Each player simultaneously bids on items with a bid vector $B_i = [b^i_1, b^i_2, b^i_3, b^i_4]$ such that $b^i_j \in [0,1]$ and $\sum_{j \in M} b^i_j \leq 1$ . The highest bidder wins an item; ties split ownership equally.

Given valuation matrix $V$ , can we always find bid vectors $B_1$ and $B_2$ such that given $B_1$ , the second player can not submit a bid strictly better than $B_2$ and vice-versa?

combinatorics

combinatorial optimization

Easy problem

by Hip1zzzil, Mar 30, 2025, 1:18 PM

$(C,M,S)$ is a pair of real numbers such that

$2C+M+S-2C^{2}-2CM-2MS-2SC=0$
$C+2M+S-3M^{2}-3CM-3MS-3SC=0$
$C+M+2S-4S^{2}-4CM-4MS-4SC=0$

Find $2C+3M+4S$ .

This post has been edited 1 time. Last edited by Hip1zzzil, Yesterday at 1:20 PM
Reason: Rr

algebra

My problem

by hacbachvothuong, Mar 29, 2025, 10:10 AM

Let $a, b, c$ be positive real numbers such that $ab+bc+ca=3$ . Prove that:
$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\ge1$

inequalities

fifth power

by mathbetter, Mar 25, 2025, 1:11 PM

$\text{Find all prime numbers } (p, q) \text{ such that } p^2 + 3pq + q^2 \text{ is a fifth power of an integer.}$

number theory

prime numbers

Incenter and midpoint geom

by sarjinius, Jul 17, 2024, 12:41 PM

Let $ABC$ be a triangle with $AB < AC < BC$ . Let the incenter and incircle of triangle $ABC$ be $I$ and $\omega$ , respectively. Let $X$ be the point on line $BC$ different from $C$ such that the line through $X$ parallel to $AC$ is tangent to $\omega$ . Similarly, let $Y$ be the point on line $BC$ different from $B$ such that the line through $Y$ parallel to $AB$ is tangent to $\omega$ . Let $AI$ intersect the circumcircle of triangle $ABC$ at $P \ne A$ . Let $K$ and $L$ be the midpoints of $AC$ and $AB$ , respectively.
Prove that $\angle KIL + \angle YPX = 180^{\circ}$ .

Proposed by Dominik Burek, Poland

This post has been edited 4 times. Last edited by sarjinius, Jul 17, 2024, 4:20 PM

parallel tangent line

Bishops and permutations

by Assassino9931, Feb 29, 2024, 7:57 PM

Let $n$ be a positive integer. Initially, a bishop is placed in each square of the top row of a $2^n \times 2^n$
chessboard; those bishops are numbered from $1$ to $2^n$ from left to right. A jump is a simultaneous move made by all bishops such that each bishop moves diagonally, in a straight line, some number of squares, and at the end of the jump, the bishops all stand in different squares of the same row.

Find the total number of permutations $\sigma$ of the numbers $1, 2, \ldots, 2^n$ with the following property: There exists a sequence of jumps such that all bishops end up on the bottom row arranged in the order $\sigma(1), \sigma(2), \ldots, \sigma(2^n)$ , from left to right.

Israel

This post has been edited 1 time. Last edited by Assassino9931, Mar 4, 2024, 10:59 AM

Orthocenter madness once again!

by MathLuis, Oct 22, 2023, 10:58 PM

Let $ABC$ be an acute triangle with orthocenter $H$ . Points $A_1$ , $B_1$ , $C_1$ are chosen in the interiors of sides $BC$ , $CA$ , $AB$ , respectively, such that $\triangle A_1B_1C_1$ has orthocenter $H$ . Define $A_2 = \overline{AH} \cap \overline{B_1C_1}$ , $B_2 = \overline{BH} \cap \overline{C_1A_1}$ , and $C_2 = \overline{CH} \cap \overline{A_1B_1}$ .

Prove that triangle $A_2B_2C_2$ has orthocenter $H$ .

Ankan Bhattacharya

This post has been edited 2 times. Last edited by v_Enhance, Oct 22, 2023, 11:44 PM

IMO 2016 Problem 1

by quangminhltv99, Jul 11, 2016, 6:23 AM

Triangle $BCF$ has a right angle at $B$ . Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$ . Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$ . Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$ . Let $M$ be the midpoint of $CF$ . Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.

This post has been edited 2 times. Last edited by sseraj, Jul 11, 2016, 4:41 PM

Prove that $\angle FAC = \angle EDB$

by micliva, Apr 18, 2013, 7:06 PM

Points $E$ and $F$ are given on side $BC$ of convex quadrilateral $ABCD$ (with $E$ closer than $F$ to $B$ ). It is known that $\angle BAE = \angle CDF$ and $\angle EAF = \angle FDE$ . Prove that $\angle FAC = \angle EDB$ .

M. Smurov

geometry proposed

geometry

Submit

done!
$~~~~$
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pupitre

by AbbyWong, Mar 30, 2025, 11:03 PM

by fancyfairy, Mar 30, 2025, 10:04 PM

by Hip1zzzil, Mar 30, 2025, 1:18 PM

by hacbachvothuong, Mar 29, 2025, 10:10 AM

by mathbetter, Mar 25, 2025, 1:11 PM

by sarjinius, Jul 17, 2024, 12:41 PM

by Assassino9931, Feb 29, 2024, 7:57 PM

by MathLuis, Oct 22, 2023, 10:58 PM

by quangminhltv99, Jul 11, 2016, 6:23 AM

by micliva, Apr 18, 2013, 7:06 PM