No triangle with heights as sides

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies

jlacosta

May 1, 2025

0 replies

Sequence of rational numbers

mojyla222 2

N 10 minutes ago by Assassino9931

Source: Iran 2024 3rd round number theory exam P1

Given a sequence $x_1,x_2,x_3,\cdots$ of positive integers, Ali proceed the following algorythm: In the i-th step he markes all rational numbers in the interval $[0,1]$ which have denominator equal to $x_i$ . Then he write down the number $a_i$ equal to the length of the smallest interval in $[0,1]$ which both two ends of that is a marked number. Find all sequences $x_1,x_2,x_3,\cdots$ with $x_5=5$ and such that for all $n\in \mathbb N$ we have
$a_1+a_2+\cdots+a_n= 2-\dfrac{1}{x_n}.$
Proposed by Mojtaba Zare

2 replies

mojyla222

Aug 27, 2024

Assassino9931

10 minutes ago

Generic Real-valued FE

lucas3617 6

N 22 minutes ago by Audreyma0321

$f: \mathbb{R} -> \mathbb{R}$ , find all functions where $f(2x+f(2y-x))+f(-x)+f(y)=2f(x)+f(y-2x)+f(2y)$ for all $x$ , $y \in \mathbb{R}$

6 replies

lucas3617

Apr 25, 2025

Audreyma0321

22 minutes ago

Another FE

M11100111001Y1R 2

N 34 minutes ago by AblonJ

Source: Iran TST 2025 Test 2 Problem 3

Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for all $x,y>0$ we have:
$f(f(f(xy))+x^2)=f(y)(f(x)-f(x+y))$

2 replies

M11100111001Y1R

Today at 8:03 AM

AblonJ

34 minutes ago

Midpoint in a weird configuration

Gimbrint 1

N 34 minutes ago by Beelzebub

Source: Own

Let $ABC$ be an acute triangle ( $AB<BC$ ) with circumcircle $\omega$ . Point $L$ is chosen on arc $AC$ , not containing $B$ , so that, letting $BL$ intersect $AC$ at $S$ , one has $AS<CS$ . Points $D$ and $E$ lie on lines $AB$ and $BC$ respectively, such that $BELD$ is a parallelogram. Point $P$ is chosen on arc $BC$ , not containing $A$ , such that $\angle CBP=\angle BDE$ . Line $AP$ intersects $EL$ at $X$ , and line $CP$ intersects $DL$ at $Y$ . Line $XY$ intersects $AB$ , $BC$ and $BP$ at points $M$ , $N$ and $T$ respectively.

Prove that $TN=TM$ .

1 reply

Gimbrint

May 23, 2025

Beelzebub

34 minutes ago

No more topics!

No triangle with heights as sides

bigant146 3

N Jan 20, 2022 by Asilbek_Abduqayumov18-06

Source: VI Caucasus Mathematical Olympiad

A triangle $\Delta$ with sidelengths $a\leq b\leq c$ is given. It appears that it is impossible to construct a triangle from three segments whose lengths are equal to the altitudes of $\Delta$ . Prove that $b^2>ac$ .

3 replies

bigant146

Mar 14, 2021

Asilbek_Abduqayumov18-06

Jan 20, 2022

No triangle with heights as sides

G H J

Reveal topic tags

geometry

Geometry inequality

G H BBookmark kLocked kLocked NReply

Source: VI Caucasus Mathematical Olympiad

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bigant146

111 posts

bigant146

#1 h Mar 14, 2021, 11:45 AM

Y by

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RagvaloD

4922 posts

RagvaloD

#2 h Mar 14, 2021, 11:56 AM • 1 Y

Y by Illuzion

So $\frac{1}{a}>\frac{1}{b}+\frac{1}{c}$

$bc>ab+ac$
$ac<bc-ab=b(c-a)<b^2$

Z K Y

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rafaello

1079 posts

rafaello

#5 h Sep 5, 2021, 1:03 PM

Y by

Blah blah storage,
We know that $h_a>h_b+h_c$ , where $h_a,h_b,h_c$ are the altitudes of $\Delta$ , this is equivalent to
$\frac{2S}{a}>\frac{2S}{b}+\frac{2S}{c}\Longleftrightarrow bc>ab+ac.$ Suppose $ac\geq b^2$ , then we would have $c>a+\frac{ac}{b}\geq a+b$ , which contradicts the triangle inequality, we are done.

Z K Y

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Asilbek_Abduqayumov18-06

6 posts

Asilbek_Abduqayumov18-06

#6 h Jan 20, 2022, 10:13 AM

Y by

$ac\geq b^2$

Z K Y

N Quick Reply