<BMC = 120^o if DE = CE, 2CD =AB (2015 Kurchatov Olympiad 9-10 p4

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0 replies

jlacosta

Apr 2, 2025

0 replies

Inequalities

sqing 5

N an hour ago by pooh123

Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$(1-abc) (1-a)(1-b)(1-c) \ge 208$ $(1+abc) (1-a)(1-b)(1-c) \le -224$ $(1+a^2b^2c^2) (1-a)(1-b)(1-c) \le -5840$

5 replies

sqing

Jul 12, 2024

pooh123

an hour ago

Amc 10 mock

Mathsboy100 4

N an hour ago by pooh123

let $\lfloor x \rfloor$ denote the greatest integer less than or equal to x . What is the sum of the squares of the real numbers x for which $x^2 - 20\lfloor x \rfloor + 19 = 0$

4 replies

Mathsboy100

Oct 9, 2024

pooh123

an hour ago

Inequalities

sqing 0

2 hours ago

Let $a,b,c>0 .$ Prove that
$\left(1 +\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right )\geq 4\left(\frac{a+b}{b+c}+ \frac{b+c}{a+b}\right)$ $\left(1 +\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right )\geq \frac{32}{9}\left(\frac{a+b}{b+c}+ \frac{c+a}{a+b}\right)$ $\left(1 +\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right )\geq \frac{8}{3}\left( \frac{a+b}{b+c}+ \frac{b+c}{c+a}+ \frac{c+a}{a+b}\right)$ $\left(1 +\frac{a^2}{b^2}\right)\left(1+\frac{b^2}{c^2}\right)\left(1+\frac{c^2}{a^2}\right )\geq \frac{8}{3}\left( \frac{a^2+bc}{b^2+ca}+\frac{b^2+ca}{c^2+ab}+\frac{c^2+ab}{a^2+bc}\right)$

0 replies

sqing

2 hours ago

0 replies

Purple Comet High School Math Meet 2024 P1

franklin2013 3

N 2 hours ago by codegirl2013

Joe ate one half of a fifth of a pizza. Gale ate one third of a quarter of that pizza. The difference in the amounts that the two ate was $\frac{1}{n}$ of the pizza, where $n$ is a positive integer. Find $n$ .

3 replies

franklin2013

3 hours ago

codegirl2013

2 hours ago

No more topics!

<BMC = 120^o if DE = CE, 2CD =AB (2015 Kurchatov Olympiad 9-10 p4 - Russia)

parmenides51 3

N Mar 10, 2021 by vanstraelen

On the median $CM$ of triangle $ABC$ , a point $D$ is chosen such that $2CD = AB$ .
Line $BD$ meets side $AC$ at point $E$ . Prove that if $DE = CE$ , then angle $\angle BMC = 120^o.$

3 replies

parmenides51

Mar 8, 2021

vanstraelen

Mar 10, 2021

<BMC = 120^o if DE = CE, 2CD =AB (2015 Kurchatov Olympiad 9-10 p4 - Russia)

G H J

Reveal topic tags

geometry

angles

equal segments

G H BBookmark kLocked kLocked NReply

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parmenides51

30650 posts

parmenides51

#1 h Mar 8, 2021, 7:40 AM

Y by

On the median $CM$ of triangle $ABC$ , a point $D$ is chosen such that $2CD = AB$ .
Line $BD$ meets side $AC$ at point $E$ . Prove that if $DE = CE$ , then angle $\angle BMC = 120^o.$

This post has been edited 2 times. Last edited by parmenides51, Mar 9, 2021, 8:34 AM
Reason: typo corrected according to #2, 2CD=AB instead of 2CD=1/2 AB

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vanstraelen

9004 posts

vanstraelen

#2 h Mar 9, 2021, 7:49 AM • 2 Y

Y by parmenides51, Mango247

parmenides51 wrote:

On the median $CM$ of triangle $ABC$ , a point $D$ is chosen such that $2CD =\frac12 AB$ .
Line $BD$ meets side $AC$ at point $E$ . Prove that if $DE = CE$ , then angle $\angle BMC = 120^o.$

I think $CD =\frac12 AB$ or $2CD =AB$ .

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parmenides51

30650 posts

parmenides51

#3 h Mar 9, 2021, 8:33 AM

Y by

indeed, they had a typo on the official wording

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vanstraelen

9004 posts

vanstraelen

#4 h Mar 10, 2021, 3:10 PM • 1 Y

Y by Mango247

Given $A(-a,0),M(0,0),B(a,0),C(b,c)$ .

Point $D(b-\frac{ab}{w},c-\frac{ac}{w})$ with $w=\sqrt{b^{2}+c^{2}}$ .
Point $E(\frac{a^{2}+2ab-2bw}{a-2w},\frac{2c(a-w}{a-2w})$ .

If $DE = CE$ , then $b=-\frac{c}{\sqrt{3}}$ and $\tan(\angle BMC)=-\sqrt{3}$ .

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