The lengths of the sides of triangle $ABC$ are 4, 6, and 8. The inscribed circle of this triangle touches its sides at points $D, E$, and $F$. Find the area of triangle $DEF$.

#
Solving the system of equations, find the segments into which the points of tangency of the inscribed circle divide the sides of the triangle. Then, using Heron's formula to find the area of the original triangle, use the formula: $S_{\Delta}$

DAF $=\frac{A D}{A B} \cdot \frac{A F}{A C} \cdot S_{\triangle \mathrm{ABC}}$ (if point $D$ lies on side $A B$, and point $F$ lies on side $A C$).

## Solution

Using Heron's formula, we find that $S_{\Delta \mathrm{ABC}}=\sqrt{9 \cdot 5 \cdot 3 \cdot 1}=3 \sqrt{15}$. Let points $D, E$, and $F$ lie on sides $A B=4, B C=6$, and $A C=8$, respectively. Denote $A D=A F=x, B D=B E=y, C E=C F=z$. Then

$$
\left\{\begin{array}{l}
x+y=4 \\
y+z=6 \\
x+z=8
\end{array}\right.
$$

from which we find that $x=3, y=1, z=5$. Then

$$
\begin{aligned}
& S_{\triangle \mathrm{DAF}}=\frac{A D}{A B} \cdot \frac{A F}{A C} \cdot S_{\triangle \mathrm{ABC}}=\frac{3}{4} \cdot \frac{3}{8} \cdot 3 \sqrt{15}=\frac{27 \sqrt{15}}{32} \\
& S_{\triangle \mathrm{DBE}}=\frac{B D}{B A} \cdot \frac{B E}{B C} \cdot S_{\triangle \mathrm{ABC}}=\frac{1}{4} \cdot \frac{1}{6} \cdot 3 \sqrt{15}=\frac{\sqrt{15}}{8} \\
& S_{\triangle \mathrm{ECF}}=\frac{C E}{C B} \cdot \frac{C F}{C A} \cdot S_{\triangle \mathrm{ABC}}=\frac{5}{6} \cdot \frac{5}{8} \cdot 3 \sqrt{15}=\frac{25 \sqrt{15}}{16}
\end{aligned}
$$

Therefore,

$$
S_{\triangle \mathrm{DEF}}=S_{\Delta \mathrm{ABC}}-\left(S_{\triangle \mathrm{DAF}}+S_{\triangle \mathrm{DBE}}+S_{\triangle \mathrm{ECF}}\right)=3 \sqrt{15}-\left(\frac{27 \sqrt{15}}{32}+\frac{\sqrt{15}}{8}+\frac{25 \sqrt{15}}{16}\right)=\frac{15 \sqrt{15}}{32}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-10.jpg?height=277&width=666&top_left_y=-1&top_left_x=702)

## Answer

$\frac{15 \sqrt{15}}{32}$.