Cevius Maximus

Since stretching a triangle preserves ratios of areas, stretch the triangle so that the vertices are at $A(0, 0)$ , $B(12, 0)$ , $C(0, 12)$ , and $P(p, q)$ .

$AA'$ has the equation $y = \cfrac{q}{p}x$ , and intersects $BC$ or $y = 12 - x$ at $A'\bigg(\cfrac{12p}{p + q}, \cfrac{12q}{p + q}\bigg)$ .

$BB'$ has the equation $y = \cfrac{q}{p - 12}(x - 12)$ , and has a $y$ -intercept of $B'\bigg(0, \cfrac{12q}{12 - p}\bigg)$ .

$CC'$ has the equation $y = \cfrac{q - 12}{p}x + 12$ , and has an $x$ -intercept of $C'\bigg(\cfrac{12p}{12 - q}, 0\bigg)$ .

$\overrightarrow{A'B'} = \bigg(-\cfrac{12p}{p + q}, \cfrac{12q}{12 - p} - \cfrac{12q}{p + q}\bigg)$ and $\overrightarrow{A'C'} = \bigg(\cfrac{12q}{12 - q} - \cfrac{12p}{p + q}, -\cfrac{12q}{p + q}\bigg)$ .

The area of $\triangle A'B'C'$ is $A_{\triangle A'B'C'} = \cfrac{1}{2}|\overrightarrow{A'B'}_x \cdot \overrightarrow{A'C'}_y - \overrightarrow{A'B'}_y \cdot \overrightarrow{A'C'}_x| = \cfrac{144pq(12 - p - q)}{(12 - p)(12 - q)(p + q)}$ .

The maximum area of $\triangle A'B'C'$ occurs when $\cfrac{dA_{\triangle A'B'C'}}{dp} = \cfrac{1728q^2(12 - 2p - q)}{(12 - p)^2(12 - q)(p + q)^2} = 0$ and $\cfrac{dA_{\triangle A'B'C'}}{dq} = \cfrac{1728p^2(12 - p - 2q)}{(12 - p)(12 - q)^2(p + q)^2} = 0$ .

For $p \neq 0$ and $q \neq 0$ , this means that $12 - 2p - q = 0$ and $12 - p - 2q = 0$ , which solves to $p = q = 4$ (the centroid) and makes the maximum area $A_{\triangle A'B'C' \text{ max}} = \cfrac{144 \cdot 4 \cdot 4 \cdot (12 - 4 - 4)}{(12 - 4)(12 - 4)(4 + 4)} = 18$ .

Therefore, since the area of $\triangle ABC$ is $A_{\triangle ABC} = \cfrac{1}{2} \cdot 12 \cdot 12 = 72$ , the maximum ratio is $\cfrac{A_{\triangle A'B'C' \text{ max}}}{A_{\triangle ABC}} = \cfrac{18}{72} = \cfrac{1}{4}$ , so $p = 1$ , $q = 4$ , and $p + q = \boxed{5}$ .