5. In a triangle $ABC$, the angle bisector from $A$, the median from $B$, and the altitude from $C$ are concurrent, and moreover, the angle bisector from $A$ and the median from $B$ are perpendicular. If the side $AB$ measures one unit, find the lengths of the other two sides.
Solution. We provide two different solutions.

Solution 1. Let $P, M$, and $Q$ be the feet of the angle bisector from $A$, the median from $B$, and the altitude from $C$, respectively, which intersect at point $X$. In triangle $ABM$, the angle bisector from $A, AX$, is perpendicular to $BM$ (since by hypothesis the median and the angle bisector of $ABC$ are perpendicular), so $\angle ABX = \angle AMX$, which means $ABM$ is isosceles and $AM = AB = 1$, hence $AC = 2$.

Let $BP = x$ and $BQ = y$. By the Angle Bisector Theorem, $BP / AB = PC / AC$, which gives $PC = 2x$. Now, by Ceva's Theorem,

$$
1 = \frac{MC}{CP} \cdot \frac{PB}{BQ} \cdot \frac{QA}{AM} = \frac{1}{2x} \cdot \frac{x}{y} \cdot \frac{1-y}{1}
$$

from which $y = 1 / 3$. We draw the perpendiculars to $AC$ from $B$ and $X$, respectively, with feet $R$ and $S$. Triangles $XQB$ and $XSM$ are congruent (they are right triangles, $XB = XM$ because $AX$ is the altitude of the isosceles triangle $ABM$, and $XQ = XS$ because they are perpendiculars from a point on the angle bisector to the sides $AB$ and $AC$), so $SM = BQ = y = 1 / 3$. On the other hand, by Thales' Theorem, $BX = XM$ implies $RS = SM$, so $RS = 1 / 3$ and $AR = AM - RS - SM = 1 / 3$. Finally, by the Pythagorean Theorem,

$$
BC^2 = BR^2 + RC^2 = AB^2 - AR^2 + RC^2 = 1 - \frac{1}{9} + \left(2 - \frac{1}{3}\right)^2 = \frac{11}{3}
$$

Thus, the other two sides measure 2 and $\frac{\sqrt{33}}{3}$.

![](https://cdn.mathpix.com/cropped/2024_04_26_8d8943c65e3c25ed875fg-4.jpg?height=531&width=714&top_left_y=1979&top_left_x=685)

Solution 2. As before, $AC = 2$. We fix a coordinate system with origin at $A$ such that $\overrightarrow{AC} = (2, 0)$. Let $\overrightarrow{AB} = (x, y)$. Then

$$
\overrightarrow{CX} = \overrightarrow{CA} + \overrightarrow{AX} = (-2, 0) + \left(\frac{x+1}{2}, \frac{y}{2}\right) = \left(\frac{x-3}{2}, \frac{y}{2}\right)
$$

(note that $X$ is the midpoint of $BM$). But the vectors $\overrightarrow{AB}$ and $\overrightarrow{CX}$ are orthogonal, so $x(x-3) + y^2 = 0$. Since $x^2 + y^2 = 1$, we have $x = 1 / 3$ and $y = \sqrt{8} / 3$. By the Pythagorean Theorem, $BC^2 = y^2 + (2 - x)^2 = 11 / 3$, which means $BC = \sqrt{33} / 3$.