CM = AB wanted, equilateral square (Oral Moscow 2021 VII p8)

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jlacosta

Apr 2, 2025

0 replies

Eazy equation clap

giangtruong13 1

N an hour ago by iniffur

Find all $x,y,z$ satisfy that: $\frac{x}{y+z}=2x-1; \frac{y}{x+z}=3y-1;\frac{z}{x+y}=5x-1$

1 reply

giangtruong13

Yesterday at 4:03 PM

iniffur

an hour ago

Inequalities

nhathhuyyp5c 0

3 hours ago

Prove that for all positive real numbers $a, b, c$ , the following inequality holds:

$\sqrt{a + b} + \sqrt{b + c} + \sqrt{c + a} \geq \frac{4(ab + bc + ca)}{\sqrt{(a + b)(b + c)(c + a)}}$

0 replies

nhathhuyyp5c

3 hours ago

0 replies

Inequalities

hn111009 3

N 5 hours ago by James4623009

Let $a,b,c>0$ satisfied $a^2+b^2+c^2=9.$ Find the minimum of $P=\dfrac{a}{bc}+\dfrac{2b}{ca}+\dfrac{5c}{ab}.$

3 replies

hn111009

6 hours ago

James4623009

5 hours ago

Olympiad

sasu1ke 3

N Today at 1:00 AM by sasu1ke

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3 replies

sasu1ke

Yesterday at 11:52 PM

sasu1ke

Today at 1:00 AM

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CM = AB wanted, equilateral square (Oral Moscow 2021 VII p8)

parmenides51 2

N May 26, 2021 by DAVROS

Points $E$ and $F$ are marked on sides $BC$ and $CD$ of square $ABCD$ , respectively, so that $AEF$ is an equilateral triangle. Point $M$ is the midpoint of segment $AE$ . Prove that $CM = AB$ .

2 replies

parmenides51

May 26, 2021

DAVROS

May 26, 2021

CM = AB wanted, equilateral square (Oral Moscow 2021 VII p8)

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parmenides51

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parmenides51

#1 h May 26, 2021, 2:13 PM • 4 Y

Y by centslordm, Mango247, Mango247, Mango247

Points $E$ and $F$ are marked on sides $BC$ and $CD$ of square $ABCD$ , respectively, so that $AEF$ is an equilateral triangle. Point $M$ is the midpoint of segment $AE$ . Prove that $CM = AB$ .

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Qu4ckqu4cko_1_1

287 posts

Qu4ckqu4cko_1_1

#2 h May 26, 2021, 2:47 PM • 1 Y

Y by centslordm

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Let's say $C=\left(0,0\right)$ , $B=\left(1,0\right)$ , $A=\left(1,1\right)$ , and $D=\left(0,1\right)$ . If we let the length of $CE=x$ , we can say that $x\sqrt{2}=\sqrt{\left(1-x\right)^{2}+1}$ . If we simplify, we get $2x^{2}=x^{2}-2x+2$ , $x^{2}+2x-2=0$ , and $x=\sqrt{3}-1$ . That will be the $x$ -coordinate of $E$ , with the $y$ -coordinate as $0$ . Therefore, the coordinates of $M$ are $\left(\left(\frac{\sqrt{3}}{2}\right),0.5\right)$ . We are trying to prove the distance from this to $\left(0,0\right)$ is equal to $AB$ , which is exactly $1$ . We use the Pythagorean theorem to solve this; with $\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}+0.5^{2}}$ being the length of $CM$ , $\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}+0.5^{2}}=\sqrt{\frac{3}{4}+0.25}=\sqrt{1}=1$ .

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DAVROS

1642 posts

DAVROS

#3 h May 26, 2021, 9:03 PM

Y by

Let $AC$ meet $FE$ at $N$ then $N$ is mid-point $FE$ and $AN\perp FE$
Circle with $AE$ as diameter contains $B$ and $N$ hence $MB=ME=MN=NE$ since $\angle MEN =60$
Chasing angles gives $\angle EMB=30, \angle MBE=75$
Circle on diamerter $FE$ contains $M$ and $C \implies \angle CME = \angle CFE =45$
Hence $\angle CMB=30+45=75$ so $\triangle CMB$ is isosceles giving $CM=CB=AB$

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