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Isosceles Triangle Geo
oVlad   4
N 2 hours ago by Double07
Source: Romania Junior TST 2025 Day 1 P2
Consider the isosceles triangle $ABC$ with $\angle A>90^\circ$ and the circle $\omega$ of radius $AC$ centered at $A.$ Let $M$ be the midpoint of $AC.$ The line $BM$ intersects $\omega$ a second time at $D.$ Let $E$ be a point on $\omega$ such that $BE\perp AC.$ Let $N$ be the intersection of $DE$ and $AC.$ Prove that $AN=2\cdot AB.$
4 replies
oVlad
Apr 12, 2025
Double07
2 hours ago
Geometry
Lukariman   1
N 2 hours ago by Primeniyazidayi
Given acute triangle ABC ,AB=b,AC=c . M is a variable point on side AB. The circle circumscribing triangle BCM intersects AC at N.

a)Let I be the center of the circle circumscribing triangle AMN. Prove that I always lies on a fixed line.

b)Let J be the center of the circle circumscribing triangle MBC. Prove that line segment IJ has a constant length.
1 reply
Lukariman
6 hours ago
Primeniyazidayi
2 hours ago
Kingdom of Anisotropy
v_Enhance   24
N 2 hours ago by deduck
Source: IMO Shortlist 2021 C4
The kingdom of Anisotropy consists of $n$ cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from $X$ to $Y$ is a sequence of roads such that one can move from $X$ to $Y$ along this sequence without returning to an already visited city. A collection of paths is called diverse if no road belongs to two or more paths in the collection.

Let $A$ and $B$ be two distinct cities in Anisotropy. Let $N_{AB}$ denote the maximal number of paths in a diverse collection of paths from $A$ to $B$. Similarly, let $N_{BA}$ denote the maximal number of paths in a diverse collection of paths from $B$ to $A$. Prove that the equality $N_{AB} = N_{BA}$ holds if and only if the number of roads going out from $A$ is the same as the number of roads going out from $B$.

Proposed by Warut Suksompong, Thailand
24 replies
v_Enhance
Jul 12, 2022
deduck
2 hours ago
Incentre-excentre geometry
oVlad   2
N 3 hours ago by Double07
Source: Romania Junior TST 2025 Day 2 P2
Consider a scalene triangle $ABC$ with incentre $I$ and excentres $I_a,I_b,$ and $I_c$, opposite the vertices $A,B,$ and $C$ respectively. The incircle touches $BC,CA,$ and $AB$ at $E,F,$ and $G$ respectively. Prove that the circles $IEI_a,IFI_b,$ and $IGI_c$ have a common point other than $I$.
2 replies
oVlad
Yesterday at 12:54 PM
Double07
3 hours ago
Great similarity
steven_zhang123   4
N 3 hours ago by khina
Source: a friend
As shown in the figure, there are two points $D$ and $E$ outside triangle $ABC$ such that $\angle DAB = \angle CAE$ and $\angle ABD + \angle ACE = 180^{\circ}$. Connect $BE$ and $DC$, which intersect at point $O$. Let $AO$ intersect $BC$ at point $F$. Prove that $\angle ACE = \angle AFC$.
4 replies
steven_zhang123
Today at 2:13 PM
khina
3 hours ago
Unexpected FE
Taco12   18
N 3 hours ago by lpieleanu
Source: 2023 Fall TJ Proof TST, Problem 3
Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that for all integers $x$ and $y$, \[ f(2x+f(y))+f(f(2x))=y. \]
Calvin Wang and Zani Xu
18 replies
Taco12
Oct 6, 2023
lpieleanu
3 hours ago
Powers of a Prime
numbertheorist17   33
N 4 hours ago by OronSH
Source: USA TSTST 2014, Problem 6
Suppose we have distinct positive integers $a, b, c, d$, and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence \begin{align*}
		ca &- db \\
		ca^2 &- db^2 \\
		ca^3 &- db^3 \\
		ca^4 &- db^4 \\
&\vdots
	\end{align*} and looks at the highest power of $p$ that divides each of them, these powers are not all zero, and are all at most $M$. Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$) such that whenever $p$ divides an element of this sequence, the maximum power of $p$ that divides that element is exactly $p^T$.
33 replies
numbertheorist17
Jul 16, 2014
OronSH
4 hours ago
Expected Intersections from Random Pairing on a Circle
tom-nowy   2
N 4 hours ago by lele0305
Let $n$ be a positive integer. Consider $2n$ points on the circumference of a circle.
These points are randomly divided into $n$ pairs, and $n$ line segments are drawn connecting the points in each pair.
Find the expected number of intersection points formed by these segments, assuming no three segments intersect at a single point.
2 replies
tom-nowy
5 hours ago
lele0305
4 hours ago
question4
sahadian   5
N 5 hours ago by Mamadi
Source: iran tst 2014 first exam
Find the maximum number of Permutation of set {$1,2,3,...,2014$} such that for every 2 different number $a$ and $b$ in this set at last in one of the permutation
$b$ comes exactly after $a$
5 replies
sahadian
Apr 14, 2014
Mamadi
5 hours ago
Find all functions $f$: \(\mathbb{R^+}\) \(\rightarrow\) \(\mathbb{R^+}\) such
guramuta   5
N 5 hours ago by jasperE3
Source: Balkan MO SL 2021
A5: Find all functions $f$: \(\mathbb{R^+}\) \(\rightarrow\) \(\mathbb{R^+}\) such that:
$$f(xf(x+y)) = xf(y) + 1 $$
5 replies
guramuta
Today at 3:28 PM
jasperE3
5 hours ago
Minimum with natural numbers
giangtruong13   1
N Apr 10, 2025 by Ianis
Let $x,y,z,t$ be natural numbers such that: $x^2-y^2+t^2=21$ and $x^2+3y^2+4z^2=101$. Find the min: $$M=x^2+y^2+2z^2+t^2$$
1 reply
giangtruong13
Apr 10, 2025
Ianis
Apr 10, 2025
Minimum with natural numbers
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giangtruong13
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Let $x,y,z,t$ be natural numbers such that: $x^2-y^2+t^2=21$ and $x^2+3y^2+4z^2=101$. Find the min: $$M=x^2+y^2+2z^2+t^2$$
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Ianis
411 posts
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We have$$x^2-y^2\equiv x^2+3y^2+4z^2=101\equiv 1\pmod 4,$$so $x$ is odd and $y$ is even. Say $y=2b$, with $b\in \mathbb{N}$, then$$x^2+12b^2+4z^2=101,$$so $b=1$ or $b=2$.
If $b=1$ then$$x^2+4z^2=89,$$so $z\leq 4$, and only $z=4$ gives a solution in natural numbers. In this case we get $(x,y,z)=(5,2,4)$, which gives $t=0$, so this doesn't work.
Hence $b=2$, and so $y=4$. Then$$x^2+4z^2=53,$$so $z\leq 3$, and only $z=1$ gives a solution in natural numbers. In this case we get $(x,y,z)=(7,4,1)$, which gives $t^2=-12$, so this doesn't work.

Hence I assume you mean whole numbers rather than natural numbers.
In this case $b=1$ gives the solution $(x,y,z,t)=(5,2,4,0)$ and $M=61$.
We also have to check the case $b=0$. Here we have$$x^2+4z^2=101,$$so $z\leq 5$, and only $z=5$ gives a solution in whole numbers. In this case we get $(x,y,z)=(1,0,5)$, which gives $t^2=20$, so this doesn't work.

Hence there are no solutions in natural numbers and only one solution in whole numbers.
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