# Problem 4. (3 points)

On the side $B C$ of triangle $A B C$, points $A_{1}$ and $A_{2}$ are marked such that $B A_{1}=6, A_{1} A_{2}=8$, $C A_{2}=4$. On the side $A C$, points $B_{1}$ and $B_{2}$ are marked such that $A B_{1}=9, C B_{2}=6$. Segments $A A_{1}$ and $B B_{1}$ intersect at point $K$, and $A A_{2}$ and $B B_{2}$ intersect at point $L$. Points $K, L$ and $C$ lie on the same line. Find $B_{1} B_{2}$.
# Answer: 12

## Solution:

Let $M$ be the point of intersection of line $KL$ and side $AB$. We write two Ceva's theorems, for point $K$ and for point $L$:

$$
\frac{AM}{MB} \cdot \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A}=1 ; \quad \frac{AM}{MB} \cdot \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A}=1
$$

From this, we get $\frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A}=\frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A}$. Four segments in this equality are known to us. $A_1C = A_2C \pm A_1A_2 = A_2C + A_1A_2$, since $A_1A_2 > A_2C$. Similarly, $BA_2 = BA_1 + A_1A_2$. For the other two segments, $AB_2 = AB_1 \pm B_1B_2$ and $CB_1 = CB_2 \pm B_1B_2$ (the signs $\pm$ are the same). Substituting, we get

$$
\begin{gathered}
\frac{BA_1}{A_2C + A_1A_2} \cdot \frac{CB_2 \pm B_1B_2}{B_1A} = \frac{BA_1 + A_1A_2}{A_2C} \cdot \frac{CB_2}{AB_1 \pm B_1B_2} \\
(CB_2 \pm B_1B_2)(AB_1 \pm B_1B_2) = \frac{(BA_1 + A_1A_2)CB_2(A_2C + A_1A_2)B_1A}{A_2C \cdot BA_1} \\
(6 \pm B_1B_2)(9 \pm B_1B_2) = \frac{14 \cdot 6 \cdot 12 \cdot 9}{4 \cdot 6} = 378
\end{gathered}
$$

Since $378 > 6 \cdot 9$, the signs $\pm$ are resolved as +, and we get the quadratic equation $B_1B_2^2 + 15B_1B_2 - 324 = 0$. This equation has roots 12 and -27, of which we are interested in the positive one.