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Grouping angles in a pentagon with bisectors
Assassino9931   1
N 23 minutes ago by nabodorbuco2
Source: Al-Khwarizmi International Junior Olympiad 2025 P2
Let $ABCD$ be a convex quadrilateral with \[\angle ADC = 90^\circ, \ \ \angle BCD = \angle ABC > 90^\circ, \mbox{ and } AB = 2CD.\]The line through \(C\), parallel to \(AD\), intersects the external angle bisector of \(\angle ABC\) at point \(T\). Prove that the angles $\angle ATB$, $\angle TBC$, $\angle BCD$, $\angle CDA$, $\angle DAT$ can be divided into two groups, so that the angles in each group have a sum of $270^{\circ}$.

Miroslav Marinov, Bulgaria
1 reply
Assassino9931
May 9, 2025
nabodorbuco2
23 minutes ago
UC Berkeley Integration Bee 2025 Bracket Rounds
Silver08   56
N 2 hours ago by Aiden-1089
Regular Round

Quarterfinals

Semifinals

3rd Place Match

Finals
56 replies
Silver08
May 9, 2025
Aiden-1089
2 hours ago
Polynomial with integer coefficients
smartvong   1
N 3 hours ago by alexheinis
Source: UM Mathematical Olympiad 2024
Prove that there is no polynomial $f(x)$ with integer coefficients, such that $f(p) = \dfrac{p + q}{2}$ and $f(q) = \dfrac{p - q}{2}$ for some distinct primes $p$ and $q$.
1 reply
smartvong
3 hours ago
alexheinis
3 hours ago
Existence of scalars
smartvong   0
3 hours ago
Source: UM Mathematical Olympiad 2024
Let $U$ be a finite subset of $\mathbb{R}$ such that $U = -U$. Let $f,g : \mathbb{R} \to \mathbb{R}$ be functions satisfying
$$g(x) - g(y ) = (x - y)f(x + y)$$for all $x,y \in \mathbb{R} \backslash U$.
Show that there exist scalars $\alpha, \beta, \gamma \in \mathbb{R}$ such that
$$f(x) = \alpha x + \beta$$for all $x \in \mathbb{R}$,
$$g(x) = \alpha x^2 + \beta x + \gamma$$for all $x \in \mathbb{R} \backslash U$.
0 replies
smartvong
3 hours ago
0 replies
Invertible matrices in F_2
smartvong   1
N 4 hours ago by alexheinis
Source: UM Mathematical Olympiad 2024
Let $n \ge 2$ be an integer and let $\mathcal{S}_n$ be the set of all $n \times n$ invertible matrices in which their entries are $0$ or $1$. Let $m_A$ be the number of $1$'s in the matrix $A$. Determine the minimum and maximum values of $m_A$ in terms of $n$, as $A$ varies over $S_n$.
1 reply
smartvong
Today at 12:41 AM
alexheinis
4 hours ago
ISI UGB 2025 P3
SomeonecoolLovesMaths   13
N 4 hours ago by iced_tea
Source: ISI UGB 2025 P3
Suppose $f : [0,1] \longrightarrow \mathbb{R}$ is differentiable with $f(0) = 0$. If $|f'(x) | \leq f(x)$ for all $x \in [0,1]$, then show that $f(x) = 0$ for all $x$.
13 replies
SomeonecoolLovesMaths
May 11, 2025
iced_tea
4 hours ago
Group Theory
Stephen123980   3
N Yesterday at 9:01 PM by BadAtMath23
Let G be a group of order $45.$ If G has a normal subgroup of order $9,$ then prove that $G$ is abelian without using Sylow Theorems.
3 replies
Stephen123980
May 9, 2025
BadAtMath23
Yesterday at 9:01 PM
calculus
youochange   2
N Yesterday at 7:46 PM by tom-nowy
$\int_{\alpha}^{\theta} \frac{d\theta}{\sqrt{cos\theta-cos\alpha}}$
2 replies
youochange
Yesterday at 2:26 PM
tom-nowy
Yesterday at 7:46 PM
ISI UGB 2025 P1
SomeonecoolLovesMaths   6
N Yesterday at 5:10 PM by SomeonecoolLovesMaths
Source: ISI UGB 2025 P1
Suppose $f \colon \mathbb{R} \longrightarrow \mathbb{R}$ is differentiable and $| f'(x)| < \frac{1}{2}$ for all $x \in \mathbb{R}$. Show that for some $x_0 \in \mathbb{R}$, $f \left( x_0 \right) = x_0$.
6 replies
SomeonecoolLovesMaths
May 11, 2025
SomeonecoolLovesMaths
Yesterday at 5:10 PM
Cute matrix equation
RobertRogo   3
N Yesterday at 2:23 PM by loup blanc
Source: "Traian Lalescu" student contest 2025, Section A, Problem 2
Find all matrices $A \in \mathcal{M}_n(\mathbb{Z})$ such that $$2025A^{2025}=A^{2024}+A^{2023}+\ldots+A$$Edit: Proposed by Marian Vasile (congrats!).
3 replies
RobertRogo
May 9, 2025
loup blanc
Yesterday at 2:23 PM
Integration Bee Kaizo
Calcul8er   63
N Yesterday at 1:50 PM by MS_asdfgzxcvb
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
63 replies
Calcul8er
Mar 2, 2025
MS_asdfgzxcvb
Yesterday at 1:50 PM
Weird Geo
Anto0110   2
N Apr 21, 2025 by Anto0110
In a trapezium $ABCD$, the sides $AB$ and $CD$ are parallel and the angles $\angle ABC$ and $\angle BAD$ are acute. Show that it is possible to divide the triangle $ABC$ into 4 disjoint triangle $X_1. . . , X_4$ and the triangle $ABD$ into 4 disjoint triangles $Y_1,. . . , Y_4$ such that the triangles $X_i$ and $Y_i$ are congruent for all $i$.
2 replies
Anto0110
Apr 20, 2025
Anto0110
Apr 21, 2025
Weird Geo
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Anto0110
103 posts
#1
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In a trapezium $ABCD$, the sides $AB$ and $CD$ are parallel and the angles $\angle ABC$ and $\angle BAD$ are acute. Show that it is possible to divide the triangle $ABC$ into 4 disjoint triangle $X_1. . . , X_4$ and the triangle $ABD$ into 4 disjoint triangles $Y_1,. . . , Y_4$ such that the triangles $X_i$ and $Y_i$ are congruent for all $i$.
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cooljoseph
1451 posts
#2
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Triangles $ABC$ and $ABD$ have the same height and base. Break them up in the following way:
[asy]
import graph;

size(300);

// Define points
pair A = (211.01433, -221.79339);
pair B = (200.96219, -141.37629); // A - (10.05214, 80.4171)
pair C = (120.54509, -221.79339); // A - (80.4171, 0)

pair D = (195.96219, -221.37629);
pair E = (205.96219, -181.37629); // D + (10, -40)
pair F = (160.96219, -181.37629); // D + (-45, -40)

pair G = (40.67437, -141.04735);
pair H = (20.592134, -221.37629);
pair I = (110.96219, -221.37629);

pair J = (35.962184, -221.37629);
pair K = (30.962194, -181.37629); // J + (-5, -40)
pair L = (75.962184, -181.37629); // J + (40, -40)

draw(A--B--C--cycle);
draw(D--E--F--cycle);
draw(G--H--I--cycle);
draw(J--K--L--cycle);
[/asy]

The middle triangle on the left is constructed to be similar to the the entire triangle on the right, except half as large and rotated 180 degrees. Similar for the middle triangle on the right.

Observe the following:
1. The left triangle on the left is congruent to the right triangle on the right.
2. The right triangle on the left is congruent to the left triangle on the right.
3. The top triangle on the left is congruent to the middle triangle on the right.
4. The middle triangle on the left is congruent to the top triangle on the right.

So, this decomposition can transform any acute triangle with some base and height into any other acute triangle with the same base and height. QED
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Anto0110
103 posts
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Great solution nice
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