AoPS - Problem

Source: IGO 2014 Junior 5

Tags: geometry, circumcircle, arc midpoint, geometric inequality

parmenides51

22.07.2018 15:44

Two points $X, Y$ lie on the arc $BC$ of the circumcircle of $\triangle ABC$ (this arc does not contain $A$ ) such that $\angle BAX = \angle CAY$ . Let $M$ denotes the midpoint of the chord $AX$ . Show that $BM +CM > AY$ . by Mahan Tajrobekar

XbenX

22.07.2018 17:40

Double post

parmenides51

22.07.2018 18:23

because it is 3 years before, we can say that this one is that problem's source

Gold_man

30.11.2020 09:38

Let $O$ be the circumcenter of the circumcircle of triangle $ABC$ . Then $OM \perp AX$ . We draw a normal line from the point $B$ at $OM$ and let it intersect the circumcircle in the point $Z$ . Since $BZ \perp OM$ we have that $OM$ is a line of symmetry of $BZ$ . According to this, $MZ = MB$ . Now, from the triangle inequality we have that $BM+MC=ZM+MC>CZ$ . But, $BZ||AX$ , so $\widehat{AZ}=\widehat{BX}=\widehat{CY}$ where from we get $\widehat{ZAC}=\widehat{YCA}$ i.e. $CZ=AY$ . That is why $BM+CM>AY$ .

Assassino9931

25.09.2023 16:29

Hmm, why do I think that equality actually can hold (particularly checking Gold_man's approach, in which the triangle inequality application the equality case has not been ruled out)?

The condition

$\angle BAX = \angle CAY$ with

$|a| = |b| = |c| = |x| = |y|$ is equivalent to

$y = \frac{cx}{b}$ , hence we wish to show

$\left|b - \frac{a+x}{2}\right| + \left|c - \frac{a+x}{2}\right| > \left|a - \frac{cx}{b}\right| = |ab-cx| = \left|\frac{ab}{x} - c\right| = \left|c- \frac{ab}{x}\right|$ . Checking (by expansion) that

$\left|\frac{a+x}{2} - \frac{ab}{x}\right| = \left|b-\frac{a+x}{2}\right|$ and the triangle inequality imply that the left hand side of the desired inequality is at least

$\left|c- \frac{ab}{x}\right|$ .

Can anybody give an even more natural complex numbers approach?

R8kt

25.09.2023 20:42

The equality holds when X and Y are the same point.