In triangle $A B C$, such that $A B=B C=4$ and $A C=2$, the bisector $A A_{1}$, median $B B_{1}$, and altitude $C C_{1}$ are drawn.

Find the area of the triangle formed by the intersection of the lines: a) $A C, A A_{1}$ and $C C_{1}$; b) $A A_{1}, B B_{1}$ and $C C_{1}$.
Let the lines $A A_{1}$ and $B B_{1}$ intersect at point $K$, the lines $B B_{1}$ and $C C_{1}$ at point $L$, and the lines $A A_{1}$ and $C C_{1}$ at point $M$.

a) By the Pythagorean theorem, the height $B B_{1}=\sqrt{15}$. Therefore, the height $C C_{1}$ to the lateral side is half of that, i.e., $C C_{1}=\frac{\sqrt{15}}{2}$. Thus, $A C_{1}=\sqrt{4-\frac{15}{4}}=\frac{1}{2}$, and $S_{A C C_{1}}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\sqrt{15}}{2}=\frac{\sqrt{15}}{8}$. The angle bisector $A M$ divides this area in the ratio $1: 4$, so $S_{A M C}=\frac{4}{5} \cdot \frac{\sqrt{15}}{8}=\frac{\sqrt{15}}{10}$.

b) The right triangles $A K B_{1}$ and $A M C_{1}$ are similar with a ratio of $A B_{1}: A C_{1}=2$, so $A K=2 A M$. $K B_{1}=1 / 5 B B_{1}$. The right triangles $B C B_{1}$ and $C L B_{1}$ are similar with a ratio of $B B_{1}: C B_{1}=\sqrt{15}$, so $L B_{1}=\frac{1}{\sqrt{15}}=\frac{1}{15} B B_{1}=\frac{1}{3} K B_{1}$. Therefore,

$$
S_{K M L}=\frac{1}{2} \cdot \frac{2}{3} S_{A K B_{1}}=\frac{1}{3} \cdot \frac{1}{2} \cdot 1 \cdot \frac{\sqrt{15}}{5}=\frac{\sqrt{15}}{30}
$$

## Answer

$$
\text { a) } \frac{\sqrt{15}}{10} \text {; b) } \frac{\sqrt{15}}{30} \text {. }
$$