midpoint is foot of altitude (Kharkiv City X 2012

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0 replies

jlacosta

Apr 2, 2025

0 replies

Inequalities

sqing 2

N 3 minutes ago by sqing

Let $a,b,c> 0$ and $\frac{a}{a^2+ab+c}+\frac{b}{b^2+bc+a}+\frac{c}{c^2+ca+b} \geq 1$ . Prove that
$a+b+c\leq 3$

2 replies

1 viewing

sqing

Yesterday at 3:52 AM

sqing

3 minutes ago

Inequalities

sqing 0

28 minutes ago

Let $a,b$ be real numbers such that $a^2+b^2+a^3 +b^3=4 .$ Prove that
$a+b \leq 2$ Let $a,b$ be real numbers such that $a+b + a^2+b^2+a^3 +b^3=6 .$ Prove that
$a+b \leq 2$

0 replies

sqing

28 minutes ago

0 replies

that statement is true

pennypc123456789 3

N an hour ago by sqing

we have $a^3+b^3 = 2$ and $3(a^4+b^4)+2a^4b^4 \le 8$ , then we can deduce $a^2+b^2$ \le 2 $ ?

3 replies

pennypc123456789

Mar 23, 2025

sqing

an hour ago

Distance vs time swimming problem

smalkaram_3549 1

N 2 hours ago by Lankou

How should I approach a problem where we deal with velocities becoming negative and stuff. I know that they both travel 3 Lengths of the pool before meeting a second time.

1 reply

smalkaram_3549

Today at 2:57 AM

Lankou

2 hours ago

No more topics!

midpoint is foot of altitude (Kharkiv City X 2012 - Ukr)

parmenides51 3

N Mar 14, 2020 by DarthVader2618

In the acute-angled triangle $ABC$ on the sides $AC$ and $BC$ , points $D$ and $E$ are chosen such that points $A, B, E$ , and $D$ lie on one circle. The circumcircle of triangle $DEC$ intersects side $AB$ at points $X$ and $Y$ . Prove that the midpoint of segment $XY$ is the foot of the altitude of the triangle, drawn from point $C$ .

3 replies

parmenides51

Mar 10, 2020

DarthVader2618

Mar 14, 2020

midpoint is foot of altitude (Kharkiv City X 2012 - Ukr)

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parmenides51

30629 posts

parmenides51

#1 h Mar 10, 2020, 4:48 PM

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wassupbros

83 posts

wassupbros

#2 h Mar 11, 2020, 3:57 PM • 1 Y

Y by myh2910

Not every time does the circumcircle of triangle DEC intersects side AB. You need some conditions

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Mario_Leopoldo

26 posts

Mario_Leopoldo

#3 h Mar 11, 2020, 5:04 PM • 1 Y

Y by Mango247

It just means that $CO \perp AB$ (not difficult!).
$O$ : circumcenter of triangle $DEC$ .

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DarthVader2618

20 posts

DarthVader2618

#4 h Mar 14, 2020, 11:15 AM

Y by

solution

Proof:
Let $M$ be the midpoint of $XY$ and the $(DEC)$ be $W$
Power of A w.r.t $W$ ={ $AM$ + $MX$ }×{ $AM$ - $MX$ }= $AC$ ( $AC$ - $DC$ ).
Therefore $AM^2$ - $MX^2$ = $AC^2$ - $AC$ × $DC$ eqn 1
Similarly
Power of B w.r.t $W$ = $BM^2$ - $MY^2$ = $BC^2$ - $BC$ × $EC$ eqn 2
Let $(ADEB)$ be $R$
Power of C w.r.t $R$ = $CD$ × $CA$ = $CE$ × $CB$ . eqn 3
Eqn1-Eqn2= $AM^2$ - $BM^2$ = $AC^2$ - $BC^2$ . Eqn4
Let $CM'$ be perpendicular to $AB$
Applying Pythagoras in $AM'C$ & $BM'C$
$AC^2$ - $BC^2$ = $AM'^2$ - $BM'^2$
Eqn5
Comparing equations 4 and 5, we obtain that $M$ & $M'$ must coincide hence $CM$ is perpendicular to $AB$

This post has been edited 4 times. Last edited by DarthVader2618, Apr 22, 2020, 8:03 PM

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