XIII OM - III - Task 3

What condition should the angles of triangle $ ABC $ satisfy so that the angle bisector of $ A $, the median drawn from vertex $ B $, and the altitude drawn from vertex $ C $ intersect at one point?
The common point of the bisector $AD$, the median $BS$, and the altitude $CH$ of triangle $ABC$ can only be an internal point of the triangle, since $AD$ and $BS$ intersect inside the triangle. Such a point can therefore exist only if the line $CH$ contains a segment lying inside the triangle, i.e., if angles $A$ and $B$ are acute.
Suppose this condition is satisfied and that the altitude $CH$ intersects the bisector $AD$ at point $K$ (Fig. 35), and the median $BS$ at point $L$ (Fig. 36).
In triangles $AHC$ and $AHK$






therefore






In triangles $BHC$ and $BHL$






where $\lambda$ denotes the acute angle $HBL$, hence






The lines $AD$, $BS$, and $CH$ intersect at one point if and only if $HK = HL$, i.e., if






The desired condition can be obtained by transforming equation (1) in such a way as to obtain a relationship between the angles of the triangle.
Applying the Law of Sines to triangle $ASB$ we get






hence






therefore






From (1) and (2) it follows that






Substituting in equation (3) $a = 2R\sin A$, $b = 2R \sin B$, $c = 2R \sin C$, we get






We can simplify relation (4). Substituting $\tan \frac{A}{2} = \frac{\sin A}{1 + \cos A}$, dividing both sides by $\sin A$, and multiplying by the denominator, we get






Therefore






and finally






We have thus obtained the necessary conditions for the existence of a common point of the lines $AD$, $BS$, and $CH$: the inequalities $\angle A < 90^\circ$, $\angle B < 90^\circ$, and equation (5).
However, note that these inequalities already follow from condition (5). Since $\sin C > 0$, according to (5) $\tan A \cdot \cos B > 0$, and since the numbers $\tan A$ and $\cos B$ cannot both be negative, $\tan A > 0$ and $\cos B > 0$, and angles $A$ and $B$ are acute.
Reversing the reasoning, we conclude that if $A$, $B$, $C$ satisfy condition (5), then equation (1) holds, and $AD$, $BS$, and $CH$ pass through one point.
The answer to the posed question is therefore: In triangle $ABC$, the bisector, median, and altitude drawn from vertices $A$, $B$, $C$ respectively intersect at one point if and only if the angles $A$, $B$, $C$ satisfy equation (5).