equal segments wanted, circles passing through incenter related

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0 replies

jlacosta

Apr 2, 2025

0 replies

function

CarlFriedrichGauss-1777 4

N 3 minutes ago by jasperE3

Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that:
$f(2021+xf(y))=yf(x+y+2021)$

4 replies

CarlFriedrichGauss-1777

Jun 4, 2021

jasperE3

3 minutes ago

Find all functions f with f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y)

Martin N. 10

N 20 minutes ago by jasperE3

Source: (4th Middle European Mathematical Olympiad, Individual Competition, Problem 1)

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x, y\in\mathbb{R}$ , we have
$f(x+y)+f(x)f(y)=f(xy)+(y+1)f(x)+(x+1)f(y).$

10 replies

Martin N.

Sep 11, 2010

jasperE3

20 minutes ago

interesting fe

skellyrah 1

N 30 minutes ago by jasperE3

find all functions $f :\mathbb{R} \to \mathbb{R}$ such that $xf(x+yf(xy)) + f(f(x)) = f(xf(y))^2 + (x+1)f(x)$

1 reply

+1 w

skellyrah

39 minutes ago

jasperE3

30 minutes ago

International FE olympiad P3

Functional_equation 22

N 31 minutes ago by jasperE3

Source: IFEO Day 1 P3

Find all functions $f:\mathbb R^+\rightarrow \mathbb R^+$ such that $f(f(x)f(f(x))+y)=xf(x)+f(y)$ for all $x,y\in \mathbb R^+$

$\textit{Proposed by Functional\_equation, Mr.C and TLP.39}$

22 replies

Functional_equation

Feb 6, 2021

jasperE3

31 minutes ago

No more topics!

equal segments wanted, circles passing through incenter related

parmenides51 1

N Mar 5, 2020 by Pluto04

Source: 2019 Austrian Federal Competition For Advanced Students, Part 1 p2

Let $ABC$ be a triangle and $I$ its incenter. The circle passing through $A, C$ and $I$ intersect the line $BC$ for second time at point $X$ . The circle passing through $B, C$ and $I$ intersects the line $AC$ for second time at point $Y$ . Show that the segments $AY$ and $BX$ have equal length.

1 reply

parmenides51

Mar 4, 2020

Pluto04

Mar 5, 2020

equal segments wanted, circles passing through incenter related

G H J

Reveal topic tags

G H BBookmark kLocked kLocked NReply

Source: 2019 Austrian Federal Competition For Advanced Students, Part 1 p2

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parmenides51

30650 posts

parmenides51

#1 h Mar 4, 2020, 9:11 PM • 1 Y

Y by Mango247

Z K Y

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Pluto04

797 posts

Pluto04

#4 h Mar 5, 2020, 8:52 PM • 1 Y

Y by Mango247

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Solution:Since B,I,Y,C are concyclic.
$\implies \angle AYI=\angle IBC=\frac{\angle B}{2}$
Also, $\angle IAY=\frac{\angle A}{2}$
$\implies \angle AIY=\angle AIB \implies \triangle DXB \equiv \triangle CYB$
Hence, AB = AY.
Similarly, AB = BX.
Hence, AY = BX.

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