Perspective Triangles

Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $a^2 + b^2 + c^2 + d^2 -ab-bc-cd-d +2/5=0$

33 replies

Saucepan_man02

May 26, 2025

lakshya2009

15 minutes ago

classical number theory

vinoth_90_2004 18

N 22 minutes ago by cj13609517288

Source: IMO ShortList 2003, number theory problem 5

An integer $n$ is said to be good if $|n|$ is not the square of an integer. Determine all integers $m$ with the following property: $m$ can be represented, in infinitely many ways, as a sum of three distinct good integers whose product is the square of an odd integer.

Proposed by Hojoo Lee, Korea

18 replies

vinoth_90_2004

May 12, 2004

cj13609517288

22 minutes ago

GCD relation

Kimchiks926 3

N 25 minutes ago by math-olympiad-clown

Source: Baltic Way 2022, Problem 18

Find all pairs $(a, b)$ of positive integers such that $a \le b$ and
$\gcd(x, a) \gcd(x, b) = \gcd(x, 20) \gcd(x, 22)$ holds for every positive integer $x$ .

3 replies

Kimchiks926

Nov 12, 2022

math-olympiad-clown

25 minutes ago

How many friends can sit in that circle at most?

Arytva 5

N 27 minutes ago by Arytva

A group of friends sits in a ring. Each friend picks a different whole number and holds a stone marked with it. Then they pass their stone one seat to the right so everyone ends up with two stones: one they made and one they received. Now they notice something odd: if your original number is $x$ , your right-neighbor’s is $y$ , and the next person over is $z$ , then for every trio in the circle they see

$x + z = (2 - x)\,y.$
They want as many friends as possible before this breaks (since all stones must stay distinct).

How many friends can sit in that circle at most?

5 replies

Arytva

Yesterday at 10:00 AM

Arytva

27 minutes ago

Interesting functional equation with polynomial

Iveela 0

31 minutes ago

Find all functions $f : \mathbb{R}^{+} \to \mathbb{R}^{+}$ and polynomial $P(x) \in \mathbb{R}[x]$ with nonnegative coefficients such that
$f(x + P(x)f(y)) = (y + 1)f(x)$ for all $x, y \in \mathbb{R^{+}}$ .

0 replies

Iveela

31 minutes ago

0 replies

Math high school

Chantria 0

39 minutes ago

for x,y,z positive such that x+y+z=3 prove that (4^x+4^y+4^z)xyz≤12

0 replies

Chantria

39 minutes ago

0 replies

Centrally symmetric polyhedron

genius_007 1

N 44 minutes ago by genius_007

Source: unknown

Does there exist a convex polyhedron with an odd number of sides, where each side is centrally symmetric?

1 reply

genius_007

May 28, 2025

genius_007

44 minutes ago

Reachable Strings

numbertheorist17 22

N an hour ago by cj13609517288

Source: USA TSTST 2014, Problem 1

Let $\leftarrow$ denote the left arrow key on a standard keyboard. If one opens a text editor and types the keys "ab $\leftarrow$ cd $\leftarrow \leftarrow$ e $\leftarrow \leftarrow$ f", the result is "faecdb". We say that a string $B$ is reachable from a string $A$ if it is possible to insert some amount of $\leftarrow$ 's in $A$ , such that typing the resulting characters produces $B$ . So, our example shows that "faecdb" is reachable from "abcdef".

Prove that for any two strings $A$ and $B$ , $A$ is reachable from $B$ if and only if $B$ is reachable from $A$ .

22 replies

numbertheorist17

Jul 16, 2014

cj13609517288

an hour ago

Serbian selection contest for the IMO 2025 - P6

OgnjenTesic 17

N an hour ago by math90

Source: Serbian selection contest for the IMO 2025

For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$ -th row are exactly all the divisors of some natural number $r_i$ ,
- the numbers in the $j$ -th column are exactly all the divisors of some natural number $c_j$ ,
- $r_i \ne r_j$ for every $i \ne j$ .

A prime number $p$ is given. Determine the smallest natural number $n$ , divisible by $p$ , such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović

17 replies

OgnjenTesic

May 22, 2025

math90

an hour ago

Channel name changed

Plane_geometry_youtuber 9

N an hour ago by Phat_23000245

Hi,

Due to the search handle issue in youtube. My channel is renamed to Olympiad Geometry Club. And the new link is as following:

https://www.youtube.com/@OlympiadGeometryClub

Recently I introduced the concept of harmonic bundle. I will move on to the conjugate median soon. In the future, I will discuss more than a thousand theorems on plane geometry and hopefully it can help to the students preparing for the Olympiad competition.

Please share this to the people may need it.

Thank you!

9 replies

Plane_geometry_youtuber

Yesterday at 9:31 PM

Phat_23000245

an hour ago

Perspective Triangles

MNJ2357 2

N Jan 26, 2020 by Pathological

Source: Korea Winter Program Practice Test 2 P4

$\triangle ABC$ and $\triangle A_1B_1C_1$ are perspective triangles. $(ABB_1)$ and $(ACC_1)$ meet at $A_2 (\neq A)$ . Define $B_2,C_2$ analogously. Prove that $AA_2, BB_2,CC_2$ are concurrent.

2 replies

MNJ2357

Jan 23, 2020

Pathological

Jan 26, 2020

Perspective Triangles

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Reveal topic tags

geometry

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Source: Korea Winter Program Practice Test 2 P4

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MNJ2357

644 posts

MNJ2357

#1 h Jan 23, 2020, 3:33 PM • 2 Y

Y by stroller, Adventure10

$\triangle ABC$ and $\triangle A_1B_1C_1$ are perspective triangles. $(ABB_1)$ and $(ACC_1)$ meet at $A_2 (\neq A)$ . Define $B_2,C_2$ analogously. Prove that $AA_2, BB_2,CC_2$ are concurrent.

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stroller

894 posts

stroller

#2 h Jan 23, 2020, 6:03 PM • 3 Y

Y by MNJ2357, Adventure10, Mango247

mOvInG pOiNtS

Lemma: The map $AO \mapsto (BC_1C)$ , $A_1 \mapsto B_2$ is projective.

Proof: After inversion around $B$ , $A_1 \in (ABA_1)$ and $B_2$ is the intersection of line $AA_1$ with line $CC_1$ . Thus the map becomes projective after inversion. Since inversion preserves cross ratios, the map itself is projective, as desired.

Now we induct on $n$ , the number of points among $A_2, B_2, C_2$ that are equal to $O$ . The base case $n = 1$ is true and this follows from the fact that the radical axes of $(AA_1B), (AA_1C),(BOC)$ concur.

Suppose the statement holds for $n-1$ and we prove it for $n (\le 3)$ . WLOG $A_1 \ne O$ . The maps $A_1 \mapsto BB_2, CC_2$ are both projective, hence it suffices to verify the statement for three positions of $A_1$ . We may take $A_1 = O$ (and have the result from induction hypothesis), $\infty$ (in which case the lines concur at $A$ ) and the point on $AO$ such that $AA_1B_1B$ is cyclic.

why am I such a troll

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Pathological

578 posts

Pathological

#4 h Jan 26, 2020, 11:29 PM • 3 Y

Y by stroller, Pluto1708, Adventure10

Easy Trig

Let $O_{ab}$ denote the circumcenter of $\triangle ABB_1$ , and define $O_{ba}, O_{bc}, O_{cb}, O_{ca}, O_{ac}$ analogously.

For a point $P$ and line $\ell$ , let $f(P, \ell)$ denote the line through $P$ perpendicular to $\ell.$

Note that $AA_2$ is the line $f(A, O_{ab}O{ac}).$ Also, $AP$ is the line $f(A, O_{ba}O_{ca}).$ Since it's obvious that $f(A, O_{ba}O_{ca}), f(B, O_{ab}O_{cb}), f(C, O_{ac}O_{bc})$ are concurrent (at $P$ ), we get from sine Ceva in $\triangle ABC$ that $O_{ab} \cdot O_{bc} \cdot O_{ca} = O_{ba} \cdot O_{cb} \cdot O_{ac}$ .

As this condition is clearly symmetric in $a, b, c$ , it follows that $f(A, O_{ab}O_{ac}), f(B, O_{ba}O_{bc}), f(C, O_{ac}O_{bc})$ are concurrent as well.

$\square$

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