angle chasing in an isosceles triangle, equal segments and bisector related

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0 replies

jlacosta

Apr 2, 2025

0 replies

hard problem....

Cobedangiu 1

N 3 minutes ago by arqady

let $a,b,c$ be the lengths of the sides of the triangle. Prove that:
$(a+b+c)(\dfrac{3a-b}{a^2+ab}+\dfrac{3b-c}{b^2+bc}+\dfrac{3c-a}{c^2+ac})\le 9$

1 reply

Cobedangiu

an hour ago

arqady

3 minutes ago

Prove the inequality with the condition (a+1)(b+1)(c+1)=8

hlminh 1

N 4 minutes ago by quacksaysduck

Let $a,b,c>0$ such that $(a+1)(b+1)(c+1)=8.$ Prove that $abc(a+b+c)\leq 3.$

1 reply

+2 w

hlminh

2 hours ago

quacksaysduck

4 minutes ago

integer functional equation

ABCDE 147

N 9 minutes ago by Adywastaken

Source: 2015 IMO Shortlist A2

Determine all functions $f:\mathbb{Z}\rightarrow\mathbb{Z}$ with the property that $f(x-f(y))=f(f(x))-f(y)-1$ holds for all $x,y\in\mathbb{Z}$ .

147 replies

ABCDE

Jul 7, 2016

Adywastaken

9 minutes ago

number theory FE

pomodor_ap 0

26 minutes ago

Source: Own, PDC002-P7

Let $f : \mathbb{Z}^+ \to \mathbb{Z}^+$ be a function such that
$f(m) + mn + n^2 \mid f(m)^2 + m^2 f(n) + f(n)^2$ for all $m, n \in \mathbb{Z}^+$ . Find all such functions $f$ .

0 replies

pomodor_ap

26 minutes ago

0 replies

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angle chasing in an isosceles triangle, equal segments and bisector related

parmenides51 1

N Sep 28, 2018 by leon.tyumen

Source: Almaty Olympiad 2012 p3 , Kazakhstan

In the isosceles triangle $ABC$ $(BC = AC)$ on the bisector of $BN$ there was a point $K$ such that $BK = KC$ and $KN = NA$ . Find the angles of the triangle $ABC$ .

1 reply

parmenides51

Sep 28, 2018

leon.tyumen

Sep 28, 2018

angle chasing in an isosceles triangle, equal segments and bisector related

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Source: Almaty Olympiad 2012 p3 , Kazakhstan

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parmenides51

30630 posts

parmenides51

#1 h Sep 28, 2018, 6:07 AM • 2 Y

Y by Adventure10, Mango247

In the isosceles triangle $ABC$ $(BC = AC)$ on the bisector of $BN$ there was a point $K$ such that $BK = KC$ and $KN = NA$ . Find the angles of the triangle $ABC$ .

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leon.tyumen

184 posts

leon.tyumen

#2 h Sep 28, 2018, 9:51 AM • 1 Y

Y by Adventure10

Let $CK$ intersect $AB$ in point $S$ .If $\angle ABK=\angle KBC=x$ , we have that $\angle ANK+\angle ASK=180-\angle CAB-\angle ABK+\angle SCB+\angle SBC=180-2x+x+x=180$ and $ANKS$ is concyclic. Then, $SN$ bisect angle $ACK$ , but $BN$ bisect angle $SBC$ and we have that $N$ is $B$ -excircle in triangle $SBC$ . $=180-5x=180-\angle KNC-2\angle KBC=180-\angle KNC-\angle NKC=\angle NCK=90-\frac{1}{2}\angle KCB=90-\frac{1}{2}x$ , and $\angle ABC=\angle BAC=2x=40$

This post has been edited 1 time. Last edited by leon.tyumen, Sep 28, 2018, 3:02 PM

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