4 Colored Square II

Thanks to @Anton Wu in 4 Colored Square , we have a formula for computing the areas of the $4$ triangles constructed within a square, where $a<b<c$ stand for the areas of the triangles with square edges as their sides and $d$ is the area of the inscribed triangle (blue one in this case):

$d^2 = (a+b+c)^2 - 4ac$

From the question, we know that $d = 41$ and $96 = a+b+c+d$ . Thus, $a+b+c = 96-d = 96-41 =55$ .

Plugging in the values, we will obtain:

$41^2 = 55^2 - 4ac$

$55^2 - 41^2 = (55-41)(55+41) = 4ac$

$ac = 14\times 24 = 16\times 21 = 12\times 28$

Note that these are the only three possible combinations for $a,c$ because, for example, if $a = 8$ and $c = 4$ , $b = 55 -42 -8 = 5$ , which is not correlated with $a < b < c$ constraint.

Then for $(a,b,c)$ is $(14, 17, 24)$ , $b+c = 41$ , but the combination area must be less than $41$ (the blue area exceeds any combination). Thus, this is not the desired solution.

For $(a,b,c)$ is $(16, 18, 21)$ , $b+c = 39$ , so any combination will be less than $41$ .

Finally, for $(a,b,c)$ is $(12, 15, 28)$ , $b+c = 43$ , which exceeds the blue area and so is not satisfying the constraint.

Thus, $a = 16$ ; $b=18$ ; $c=21$ ; and $d=41$ . As a result, the smallest triangle’s area is $a = \boxed{16}$ .