PMO Area Stage

Solution:

Let $AB = 12$, $r = 4$ (incircle radius), and $AD : AB = 1 : 3$ so $AD = 4$, $DB = 8$.

Let $AC = b$, $BC = c$.

Let $s$ be the semiperimeter of $\triangle ABC$.

Recall that the incircle touches $AB$ at $D$, and $AD = s - a$, $DB = s - b$, where $a = BC$, $b = AC$, $c = AB$.

But here $AB = 12$, so $c = 12$.

Let $AD = s - b = 4$, $DB = s - a = 8$.

So $s - b = 4$, $s - a = 8$.

But $s = \frac{a + b + c}{2} = \frac{a + b + 12}{2}$.

So:

$s - b = \frac{a + b + 12}{2} - b = \frac{a - b + 12}{2} = 4$

$\implies a - b + 12 = 8$

$\implies a - b = -4$

$\implies a = b - 4$

Similarly,

$s - a = \frac{a + b + 12}{2} - a = \frac{b - a + 12}{2} = 8$

$\implies b - a + 12 = 16$

$\implies b - a = 4$

But from above, $a = b - 4$, so $b - (b - 4) = 4 \implies 4 = 4$ (consistent).

So $a = b - 4$, $c = 12$.

Let $a = x$, $b = x + 4$, $c = 12$.

Then $s = \frac{x + x + 4 + 12}{2} = \frac{2x + 16}{2} = x + 8$.

The area $K = r s = 4(x + 8)$.

But also, by Heron's formula:

$K = \sqrt{s(s - a)(s - b)(s - c)}$

Compute:

$s = x + 8$

$s - a = x + 8 - x = 8$

$s - b = x + 8 - (x + 4) = 4$

$s - c = x + 8 - 12 = x - 4$

So:

$K = \sqrt{(x + 8) \cdot 8 \cdot 4 \cdot (x - 4)}$

But $K = 4(x + 8)$, so:

$4(x + 8) = \sqrt{(x + 8) \cdot 8 \cdot 4 \cdot (x - 4)}$

Square both sides:

$16(x + 8)^2 = (x + 8) \cdot 8 \cdot 4 \cdot (x - 4)$

$16(x + 8)^2 = 32(x + 8)(x - 4)$

Divide both sides by $(x + 8)$ (since $x + 8 > 0$):

$16(x + 8) = 32(x - 4)$

$16(x + 8) = 32x - 128$

$16x + 128 = 32x - 128$

$128 + 128 = 32x - 16x$

$256 = 16x$

$x = 16$

So $a = x = 16$, $b = x + 4 = 20$, $c = 12$.

$s = \frac{16 + 20 + 12}{2} = \frac{48}{2} = 24$

Area $K = r s = 4 \times 24 = 96$

Answer: The area of $\triangle ABC$ is $\boxed{96}$.