5. The bisectors $\mathrm{AD}$ and $\mathrm{BE}$ of triangle $\mathrm{ABC}$ intersect at point I. It turns out that the area of triangle ABI is equal to the area of quadrilateral CDIE. Find $AB$, if $CA=9, CB=4$.
Answer: 6.

Solution. Let $\mathrm{S}(\mathrm{CDIE})=\mathrm{S}_{1}, \mathrm{~S}(\mathrm{ABI})=\mathrm{S}_{2}$, $S(B D I)=S_{3}, S(A I E)=S_{4}$ (see figure). Since the ratio of the areas of triangles with a common height is equal to the ratio of the bases, and the angle bisector divides the opposite side in the ratio of the adjacent sides, we have $\left(\mathrm{S}_{1}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{2}+\mathrm{S}_{3}\right)=\mathrm{CD} / \mathrm{BD}$ $=\mathrm{AC} / \mathrm{AB} . \quad$ Similarly, $\left(\mathrm{S}_{2}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{1}+\mathrm{S}_{3}\right)$ $=A E / E C=A B / B C$. Since $S_{1}=S_{2}$, then $\left(\mathrm{S}_{1}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{2}+\mathrm{S}_{3}\right)=\quad\left(\mathrm{S}_{2}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{1}+\mathrm{S}_{3}\right) \quad$ from which $A B / B C=A C / A B . \quad A B / 4=9 / A B . \quad A B^{2}=36$,

![](https://cdn.mathpix.com/cropped/2024_05_06_209f947ed7e329235966g-2.jpg?height=508&width=458&top_left_y=436&top_left_x=942)
$\mathrm{AB}=6$ (since the length of a segment is a positive number). It is not difficult to verify that such a triangle exists $(4+6>9)$.

Criteria. The length of the side is found correctly, but there is no check for the existence of the triangle: 6 points.