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
Problem
Source: IGO 2014 Junior 5
Tags: geometry, circumcircle, arc midpoint, geometric inequality
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.