sum of areas (PAF) + (PBD) + (PCE) is constant.

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0 replies

jlacosta

Apr 2, 2025

0 replies

BMO Shortlist 2021 A5

Lukaluce 17

N 7 minutes ago by jasperE3

Source: BMO Shortlist 2021

Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$f(xf(x + y)) = yf(x) + 1$ holds for all $x, y \in \mathbb{R}^{+}$ .

Proposed by Nikola Velov, North Macedonia

17 replies

Lukaluce

May 8, 2022

jasperE3

7 minutes ago

Function equation

luci1337 3

N 25 minutes ago by jasperE3

find all function $f:R \rightarrow R$ such that:
$2f(x)f(x+y)-f(x^2)=\frac{x}{2}(f(2x)+f(f(y)))$ with all $x,y$ is real number

3 replies

luci1337

Yesterday at 3:01 PM

jasperE3

25 minutes ago

Circumcenter of reflection of collinear points over sides

a1267ab 27

N 25 minutes ago by Giant_PT

Source: USA TST 2025

Let $ABC$ be a triangle, and let $X$ , $Y$ , and $Z$ be collinear points such that $AY=AZ$ , $BZ=BX$ , and $CX=CY$ . Points $X'$ , $Y'$ , and $Z'$ are the reflections of $X$ , $Y$ , and $Z$ over $BC$ , $CA$ , and $AB$ , respectively. Prove that if $X'Y'Z'$ is a nondegenerate triangle, then its circumcenter lies on the circumcircle of $ABC$ .

Michael Ren

27 replies

a1267ab

Jan 11, 2025

Giant_PT

25 minutes ago

a+b+c=abc

KhuongTrang 1

N 34 minutes ago by KhuongTrang

Source: own

Problem. Let $a,b,c$ be three positive real numbers satisfying $a+b+c=abc.$ Prove that $\sqrt{a^2+b^2+3}+\sqrt{b^2+c^2+3}+\sqrt{c^2+a^2+3}\ge4\cdot \frac{a^2b^2c^2-3}{ab+bc+ca-3}-7.$ There is a very elegant proof :-D Could anyone think of it?

1 reply

KhuongTrang

Wednesday at 11:51 AM

KhuongTrang

34 minutes ago

No more topics!

sum of areas (PAF) + (PBD) + (PCE) is constant.

parmenides51 1

N Oct 26, 2021 by vanstraelen

Source: 1997 Argentina IberoAmerican Training List p8 https://artofproblemsolving.com/community/c2434876_argentina_training_lists__geom

Let $ABC$ be a triangle, $G$ be the centroid, and $P$ be a variable point in the interior of $ABC$ . Let $D, E, F$ be points on the sides $BC, CA, AB$ respectively, such that $PD\parallel AG$ , $PE\parallel BG$ , $PF\parallel CG$ . Prove that the sum of areas $(PAF) + (PBD) + (PCE)$ is constant.

1 reply

parmenides51

Oct 14, 2021

vanstraelen

Oct 26, 2021

sum of areas (PAF) + (PBD) + (PCE) is constant.

G H J

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Source: 1997 Argentina IberoAmerican Training List p8 https://artofproblemsolving.com/community/c2434876_argentina_training_lists__geom

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parmenides51

30629 posts

parmenides51

#1 h Oct 14, 2021, 3:11 PM

Y by

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vanstraelen

8967 posts

vanstraelen

#2 h Oct 26, 2021, 4:51 PM • 1 Y

Y by parmenides51

Given $\triangle ABC\ :\ A(0,3a),B(-3b,0),C(3c,0)$ ; point $G(c-b,a)$ .
Choose $P(k,l)$ .

The line $PD\ :\ y-l=\frac{2a}{b-c}(x-k)$ cuts $BC$ in the point $D(k+\frac{l(c-b)}{2a},0)$
and the area $\triangle PBD$ equals $\frac{1}{2}[3bl+\frac{l^{2}(c-b)}{2a}+kl]$ .

Same method, twice.
Sum of the three areas $=\frac{9a(b+c)}{4}=\frac{1}{2} \cdot$ area of $\triangle ABC$ .

Z K Y

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