Belarus Team Selection Test 2010

[rsoldo]

Points $H$ and $T$ are marked respectively on the sides $BC$ abd $AC$ of triangle $ABC$ so that $AH$ is the altitude and $BT$ is the bisectrix $ABC$. It is known that the gravity center of $ABC$ lies on the line $HT$.
a) Find $AC$ if $BC$=a and $AB$=c.
b) Determine all possible values of $\frac{c}{a}$ for all triangles $ABC$ satisfying the given condition.

[drzewo]

the key point of the problem is that the gravity center of a triangle is situated in the intersection of the medians, I guess

-- 07.08.2024, 16:14 --

introduce a rectangular coordinate frame $Hxy$ such that $BC\subset Hx$
and let $A=(0,\alpha),\quad B=(-\beta,0),\quad C=(\sigma,0)$

[provincialka]

Может, применить несколько раз теорему Менелая? К разным треугольникам.

Впрочем, это будет громоздко, наверное, не лучше, чем координатный метод...

[rsoldo]

a) ( the drawing is missing )
From $m_a=\sqrt{\frac{2b^2+2c^2-a^2}{4}}$, and the cosine rule from triangle $ABM$ we have:

$$c^2=\frac{2b^2+2c^2-a^2}{4}+\frac{a^2}{4}-2\cdot\frac{a}{2}\cdot \sqrt{\frac{2b^2+2c^2-a^2}{4}}\cdot \cos \angle BMD$$

$$\cos\angle BMD=\frac{b^2-c^2}{a\cdot \sqrt{2b^2+2c^2-a^2}},$$

and because $\cos\angle BMD=\frac{HM}{m_a}$ we have

$$HM=\frac{b^2-c^2}{2a} \implies HC = \frac{{b^2} - {c^2}}{2a} + \frac{a}{2} = \frac{{b^2} + {a^2} - {c^2}}{2a}.$$

From Menelaus theorem in $AMC$ with transversal $MGA$ we get: $\frac{{HM}}{{HC}} \cdot \frac{{CT}}{{TA}} \cdot \frac{{AG}}{{GM}} = 1 \Leftrightarrow $

$\frac{{{b^2} - {c^2}}}{{{b^2} + {a^2} - {c^2}}} \cdot \frac{a}{c} \cdot 2 = 1 \Leftrightarrow $ $\boxed{b = \sqrt {\frac{{c(2ac + {a^2} - {c^2})}}{{2a - c}}} }.$