A triangle has sides of length 3, 4, and 5. Determine the radius of the inscribed circle (circle inside the triangle and tangent to all three sides).
If we call $I$ the center of the inscribed circle, and $r$ the radius of the inscribed circle, since $r$ is the height of each of the triangles $I A B, I B C$ and $I C A$, the area of these triangles is: $\frac{1}{2} r \times A B$, $\frac{1}{2} r \times B C, \frac{1}{2} r \times C A$, so their sum is $r p$ by calling $p$ the semi-perimeter $\frac{1}{2}(A B+B C+C A)$.

## II. FIRST DAY

1. WARM-UP EXERCISES

The sum of these three areas is precisely the area $S$ of the triangle $A B C$, regardless of the point I inside the triangle, and by the Pythagorean theorem, since $3^{2}+4^{2}=5^{2}$, this triangle is right-angled. Its area is $S=\frac{1}{2}(3 \times 4)=6$, its semi-perimeter, $p=\frac{1}{2}(3+4+5)=6$, so the radius of its inscribed circle is: $r=\frac{S}{p}=1$.