Estonian Math Competitions

Answer: $\frac{7}{2}$.

Solution: Introduce a Cartesian coordinate system with origin $O$ at one vertex of the rectangle. Let the other vertices of the rectangle be $P=(10,0)$, $Q=(10,7)$ and $R=(0,7)$. Given the distances from a point to three sides of the rectangle, two of these sides must be opposite sides and the corresponding distances sum up to the length of the perpendicular side of the rectangle. Thus $A$ must lie at distances $3$ and $4$ from the horizontal sides and at distances $2$ and $8$ from the vertical sides. W.l.o.g., let $A=(2,3)$. Similarly, $B$ must lie at distances $3$ and $4$ from the horizontal sides and at distances $5$ and $5$ from the vertical sides. Thus $B=B_1=(5,3)$ or $B=B_2=(5,4)$ (Fig. 16). Finally, $C$ must lie at distances $4$ and $6$ from the vertical sides and at distances $2$ and $5$ from the horizontal sides. Thus $C=C_1=(4,2)$ or $C=C_2=(4,5)$ or $C=C_3=(6,2)$ or $C=C_4=(6,5)$ (Fig. 17).

Altogether, the triangle $ABC$ can be located in $8$ ways. We study which one yields the largest area. If $B=B_1=(5,3)$, take the side $AB$ with length $3$ as base. The corresponding altitude has length $1$ if $C=C_1=(4,2)$ or $C=C_3=(6,2)$ and $2$ if $C=C_2=(4,5)$ or $C=C_4=(6,5)$ (Fig. 18). Thus the largest area in the observed cases is $3$. Consider now the case $B=B_2=(5,4)$. Reflecting the points $C_1$ and $C_3$ from the midpoint of the side $AB$, we obtain points $C_1^{\prime }=(3,5)$ and $C_3^{\prime }=(1,5)$ which yield the same areas (Fig. 19). Points $C_2$, $C_4$, $C_1^{\prime }$ and $C_3^{\prime }$ lie on the line $y=5$, whereby the one with the least x-coordinate, the point $C_3^{\prime }$, is the farthest from the line $AB$. Hence the largest area arises in the case $C=C_3=(6,2)$. The size of the rectangle surrounded by lines $y=2$, $y=4$, $x=2$ and $x=6$ is $2\times 4$. Removing three right triangles to extract the triangle $ABC$ (Fig. 20) enables to express the area of the triangle $ABC$ as $2\cdot 4-\frac{1}{2}\cdot 1\cdot 3-\frac{1}{2}\cdot 1\cdot 2-\frac{1}{2}\cdot 1\cdot 4=\frac{7}{2}$. As $\frac{7}{2}>3$, the largest possible area of the triangle $ABC$ is $\frac{7}{2}$.