Kite Riddle

Since the diagonals of a kite intersect each other at right angles, the point $G$ in this case also acts a midpoint of $EF$ , and $GC$ is a bisector of an isosceles $EFC$ . With $EG = a$ , $GC = b$ , and $CE = c$ for some pairwise co-prime $a,b,c$ , these lengths are, in fact, Pythagorean triples. Then, according to Euclid's formula, we can parametize these lengths, for some co-prime $m>n$ as followed:

$a = m^2 - n^2 = (m+n)(m-n)$

$b = 2mn$

$c = m^2 + n^2$

We can then calculate the area of the triangle $EFC$ = $ab = (2mn)(m+n)(m-n)$

Then we know that $\dfrac{BD}{2} = \dfrac{b}{2} = mn$ . Thus, the area of the triangle $ABC$ equals to that of $EFC$ because both colored regions have the same area.

Hence, area of $ABC = \dfrac{mn}{2}(AC) = \dfrac{mn}{2}(8+b) = mn(4 + mn)$ .

Therefore, $mn(4+mn) = 2(mn)(m+n)(m-n)$

$4+mn = 2(m+n)(m-n)$

$8 + 2mn = AC = 4a$

$b = 4a - 8$

Now if we recreate the figure to be the right triangle $ABC$ such that the colored regions have the same area as shown below, we'll let $h$ be the height $AB$ :

By similar triangle ratios, the maxmum $h = \dfrac{a(8+b)}{b}$ , and by the area calculation, $h = \dfrac{2ab}{8+b}$ .

Hence, $h^2 = 2a^2$ . $h = a\sqrt{2}$ .

Now we can set inequality: $a\sqrt{2} > 2a - 4 > a$ .

For $2a-4 > a$ , we'll obtain $a > 4$ .

Then for $a\sqrt{2} > 2a - 4$ , we will obtain $\dfrac{4}{2-\sqrt{2}} \approx 6.8 > a$ .

Thus, $a$ can only be $5$ or $6$ , but if $a = 6$ , then $b = 4\times 6 - 8 = 16$ , which is not pairwise co-prime and not Pythagorean triples.

As a result, $a= 5$ , and $b = 4\times 5 - 8 =12$ .

Finally, the area of the kite $ABCD$ = $6\times 20 = \boxed{120}$ .

Note : We can alternatively calculate the length $AG = c - a = 13 - 5 = 8$ .