A box in the shape of a triangle with side lengths $13$, $14$, and $15$ is being used to
pack a pizza pi. The pizza is in the shape of a circle and just touches each of the
three sides of the triangular box. What is the area of the circle?
Proposed as the problem of the week #23 of year 2025/26, by the Centre for Education in Mathematics and Computing of the University of Waterloo.
In the geometry lingo, this $\pi$e is what is called the circle inscribed to the triangle, and
the way to measure its radius – which is the essential parameter from which we will derive
its area – is by leveraging precisely this condition.
For the inexperienced reader, pause and try to convince yourself that such a circle intersects with its bounding shape forming $90^\circ$-angles on such points; using this result we will now divide the original box into three smaller ones, which share the property that their height is the radius we are looking for: With this key insight in mind, we can now relate the area of the big triangle, to that of the smaller ones, and thus to $r$; that is, $$ [\triangle ABC] = \frac{1}{2}13 r + \frac{1}{2}14 r + \frac{1}{2}15 r = 21r\quad \implies\quad r = \frac{[\triangle ABC]}{21}. $$
However, we still have not figured out $[\triangle ABC]$, and it is Heron's formula the one that tells us exactly this. Letting $s$ be half of the perimeter, then $$ [\triangle ABC] = \sqrt{s(s-13)(s-14)(s-15)} = \sqrt{21\cdot 8 \cdot 7 \cdot 6} = 84, $$ so finally, $$ r = \frac{84}{21} = 4 \quad\implies\quad [\pi] = \pi \cdot 4^2 = 16\pi. $$
For the inexperienced reader, pause and try to convince yourself that such a circle intersects with its bounding shape forming $90^\circ$-angles on such points; using this result we will now divide the original box into three smaller ones, which share the property that their height is the radius we are looking for: With this key insight in mind, we can now relate the area of the big triangle, to that of the smaller ones, and thus to $r$; that is, $$ [\triangle ABC] = \frac{1}{2}13 r + \frac{1}{2}14 r + \frac{1}{2}15 r = 21r\quad \implies\quad r = \frac{[\triangle ABC]}{21}. $$
However, we still have not figured out $[\triangle ABC]$, and it is Heron's formula the one that tells us exactly this. Letting $s$ be half of the perimeter, then $$ [\triangle ABC] = \sqrt{s(s-13)(s-14)(s-15)} = \sqrt{21\cdot 8 \cdot 7 \cdot 6} = 84, $$ so finally, $$ r = \frac{84}{21} = 4 \quad\implies\quad [\pi] = \pi \cdot 4^2 = 16\pi. $$