[ Two tangents drawn from one point ]

The sides of a pentagon in order of traversal are equal to 5, 6, 7, 8, and 9. The sides of this pentagon touch one circle. Into what segments does the point of tangency with the side equal to 5 divide this side?
Let \( x \) be one of the desired segments and express sequentially through \( x \) the segments into which the points of tangency divide the sides of the pentagon.

## Solution

Let the tangents from vertex \( A \) be \( x \). "Traverse" our pentagon, expressing sequentially the lengths of the tangents from vertices \( B \) (equal to \( 6 - x \)), \( C \) (equal to \( 7 - (6 - x) = 1 + x \)), \( D \) (equal to \( 8 - (1 + x) = 7 - x \)), and \( E \) (equal to \( 9 - (7 - x) = 2 + x \)). We find that side \( AE \) is divided by the point of tangency into segments \( x \) and \( 2 + x \). From the equation \( x + (2 + x) = 5 \), we find that \( x = \frac{3}{2} \).

## Answer

\(\frac{3}{2}\) and \(\frac{5}{2}\).