Four Right Triangles in One

Relevant wiki: Pythagorean Triples

Suppose the diagonals intersect each other at point E.

And we know that AC : BD = 8 : 9, so suppose AC = 8k and BD = 9k for some constant k.

Suppose AE = x and BE = y. Then we will obtain the system of equations below:

$x^2 + y^2 = 252$ _(1)

$y^2 + (8k - x)^2 = 392$ _(2)

$(8k - x)^2 + (9k - y)^2 = 602$ _(3)

$(9k - y)^2 + x^2 = 522$ _(4)

Subtracting equation (2) with (1);

$(8k - x)^2 - x^2 = 392 - 252 = 896$

$64k^2 - 16kx = 896$

$64k^2 - 896 = 16kx$

$x = 4k - \dfrac{56}{k}$

Now subtracting equation (4) with (1);

$(9k - y)^2 - y^2 = 522 - 252 = 2079$

$81k^2 - 18ky = 2079$

$81k^2 - 2079 = 18ky$

$y = \dfrac{9k}{2} - \dfrac{231}{2k}$

Substituting k terms for x and y in equation (1):

$(4k - 56/k)^2 + (\dfrac{9k}{2} - \dfrac{231}{2k} )^2 = 625$

Multiplying by 4k2 both sides;

$(8k^2 - 112)^2 + (9k^2 - 231)^2 = 2500k^2$

$64k^4 - 1792k^2 + 12544 + 81k^4 - 4158k^2 + 53361 = 2500k^2$

$145k^4 - 8450k^2 + 65905 = 0$

$29k^4 - 1690k^2 + 13181 = 0$

$(k^2 - 49)(29k^2 - 269) = 0$

Only k = 7 will result in diagonals of integer lengths.

Therefore, $x = 4×7 - \dfrac{56}{7} = 20$ .

And $y = \dfrac{63}{2} - \dfrac{231}{14} = \dfrac{63-33}{2} = 15$ .

Then we will get $CE^2 = 39^2 - 15^2 = 36^2. CE = 36$ .

Finally, $DE^2 = 60^2 - 36^2 = 48^2. DE = 48$ .

As a result, $AC = 20 + 36 = 56$ , and $BD = 15 + 48 = 63$ . (Note that 56 : 63 = 8 : 9)

Therefore, the area of the quadrilateral ABCD = $(\dfrac{1}{2})(56×63) = 1764$

Moreover, let AE = a, BE = b, CE = c, and DE = d for some numbers a, b, c, d.

We will obtain the system of equations below:

$a^2 + b^2 = 25^2$

$b^2 + c^2 = 39^2$

$c^2 + d^2 = 60^2$

$d^2 + a^2 = 52^2$

Observe that by adding first and third equations: $a^2 + b^2 + c^2 + d^2 = 25^2 + 60^2$

And by adding second and fourth: $a^2 + b^2 + c^2 + d^2 = 39^2 + 52^2$

Thus, $39^2 + 52^2 = 25^2 + 60^2$ .

This can be related to British Flag theorem , which deals with summation of squares of distances from a point in a rectangle to its four vertices.

The only possible triplets with $25$ as the hypotenuse are $(7, 24, 25)$ and $(15, 20, 25)$ . This can be checked using the Euclid's formula of triples.

We need to choose $OB$ , the common side of $\triangle AOB$ and $\triangle BOC$ , from the above triples such that $OC$ is an integer.

Now,

$39^2 - 7^2 = (39 + 7)(39 - 7) = 46×32$

$39^2 - 24^2 = (39 + 24)(39 - 24) = 63×15$

$39^2 - \textcolor{grey}{15}^2 = (39 + 15)(39 - 15) = \textcolor{#624F41}{54×24}$

$39^2 - 20^2 = (39 + 20)(39 - 20) = 59×19$

Out of these only the third case yields a perfect square.

So, $OB = \textcolor{grey}{15}$ units and $OC = \textcolor{#624F41}{36}$

The other sides are:

$OA = 20, OD = 48$ units

$\therefore$ Area $=\dfrac{D_1D_2}{2} = \dfrac{56×63}{2} = \boxed{1764}$ unit²

Let BE=a and EC=b and AE=c.

a square+b square=39 square.

a square+c square=25 square.

b square-c square=896.

(b-c)(b+c)=896

writing all factors:

1 x 896

2 x 448

4 x 224

7 X 128

8 x 112

14 x 64

16 x 56

28 x 32

b-c < b+c and 2b<78

so only 16 X 56 and 28 X 32 are valid

For 16 x 56,

b-c=16 and b+c=56

2b=72

b=36 and c=20 so a=15 and EB=48

checking 56/63=AC/BD satisfies the condition.

so .5[15x20+15x36+36x48+20x48]=1764.