AB^2 +AC^2=\lambda BC^2, medians cut right angle (2011 Flanders Juniors p4)

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0 replies

jlacosta

Apr 2, 2025

0 replies

A problem involving modulus from JEE coaching

AshAuktober 7

N an hour ago by Jhonyboy

Solve over $\mathbb{R}$ :
$|x-1|+|x+2| = 3x.$
(There are two ways to do this, one being bashing out cases. Try to find the other.)

7 replies

AshAuktober

Apr 21, 2025

Jhonyboy

an hour ago

Fun & Simple puzzle

Kscv 6

N 2 hours ago by AbbyWong

$\angle DCA=45^{\circ},$ $\angle BDC=15^{\circ},$ $\overline{AC}=\overline{CB}$

$\angle ADC=?$

6 replies

Kscv

Apr 13, 2025

AbbyWong

2 hours ago

Inequalities from SXTX

sqing 16

N 2 hours ago by DAVROS

T702. Let $a,b,c>0$ and $a+2b+3c=\sqrt{13}.$ Prove that $\sqrt{a^2+1} +2\sqrt{b^2+1} +3\sqrt{c^2+1} \geq 7$ S

$\sqrt{a^2+1} \geq \frac{ \sqrt{13} a +6}{7}$

T703. Let $a,b$ be real numbers such that $a+b\neq 0.$ . Find the minimum of $a^2+b^2+(\frac{1-ab}{a+b} )^2.$
T704. Let $a,b,c>0$ and $a+b+c=3.$ Prove that $\frac{a^2+7}{(c+a)(a+b)} + \frac{b^2+7}{(a+b)(b+c)} +\frac{c^2+7}{(b+c)(c+a)} \geq 6$ S

$\frac{a^2+7}{(a+3)^2} + \frac{b^2+7}{(b+3)^2} +\frac{c^2+7}{(c+3)^2} \geq \frac{3}{2}$

16 replies

sqing

Feb 18, 2025

DAVROS

2 hours ago

FB = BK , circumcircle and altitude related (In the World of Mathematics 516)

parmenides51 5

N 3 hours ago by jasperE3

Let $BT$ be the altitude and $H$ be the intersection point of the altitudes of triangle $ABC$ . Point $N$ is symmetric to $H$ with respect to $BC$ . The circumcircle of triangle $ATN$ intersects $BC$ at points $F$ and $K$ . Prove that $FB = BK$ .

(V. Starodub, Kyiv)

5 replies

parmenides51

Apr 19, 2020

jasperE3

3 hours ago

No more topics!

AB^2 +AC^2=\lambda BC^2, medians cut right angle (2011 Flanders Juniors p4)

parmenides51 2

N Dec 17, 2020 by SlimTune

Prove that there exists a positive number $\lambda$ such that for every triangle $ABC$ , of which the medians from the vertices $B$ and $C$ intersect at right angle, holds $| AB |^2 + | AC |^2 = \lambda | BC |^2$ .

2 replies

parmenides51

Dec 17, 2020

SlimTune

Dec 17, 2020

AB^2 +AC^2=\lambda BC^2, medians cut right angle (2011 Flanders Juniors p4)

G H J

Reveal topic tags

geometry

Medians

right angle

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parmenides51

30631 posts

parmenides51

#1 h Dec 17, 2020, 2:57 AM

Y by

This post has been edited 1 time. Last edited by parmenides51, Dec 17, 2020, 3:02 AM
Reason: added latex to \lambda

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JustinLee2017

1703 posts

JustinLee2017

#2 h Dec 17, 2020, 2:59 AM

Y by

parmenides51 wrote:

FTFY

This post has been edited 1 time. Last edited by JustinLee2017, Dec 17, 2020, 2:59 AM

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SlimTune

353 posts

SlimTune

#3 h Dec 17, 2020, 3:14 AM

Y by

Solution

Let the medians from $B$ and $C$ intersect the opposite sides at $E$ and $F$ respectively.

This means that quadrilateral $EFBC$ is a convex quadrilateral with diagonals intersecting at right angles.

Let $a, b,$ and $c$ represent $|BC|, |AC|,$ and $|AB|$ respectively.

Since $|EF|$ is the distance between the midpoints, we have that $|EF| = \frac{|BC|}{2} = \frac{a}{2}$ .

Also, we have that $|BF| = \frac{c}{2}$ and $|CE| = \frac{a}{2}$ .

By a well-known theorem (which is also easy to prove), we have that the sum of the squares of opposite sides of this particular quadrilateral are equal. From this, we get the following:
$|BF|^2 + |EC|^2 = |EF|^2 + |BC|^2 \Rightarrow \frac{c^2}{4} + \frac{b^2}{4} = \frac{a^2}{4} + a^2$ Multiplying by $4$ , we get that
$c^2 + b^2 = 5a^2 \Rightarrow | AB |^2 + | AC |^2 = 5 | BC |^2$ so we have that $\lambda = 5$ , and we get that a positive number for $\lambda$ does exist. $\blacksquare$

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