Quadrilateral Circular Reasoning (Take 2)

Say the quadrilateral has sides (in order) $a,b,c,d$ . From the question, either $b=8a$ and $d=2c$ , or $c=8a$ and $d=2b$ (other cases are just rotations and reflections of these).

In tangential quadrilaterals, the two pairs of opposite sides have the same sum. So $a+c=b+d=s$ , the semiperimeter.

If $b=8a$ and $d=2c$ , this is $a+c=8a+2c$

which has no solutions with all sides positive.

So $c=8a$ and $d=2b$ , and $a+8a=b+2b$

ie $b=3a$ . Now we have all the sides in terms of $a$ : $b=3a,\;c=8a,\;d=6a$

In cyclic quadrilaterals, the area is given by Bramhagupta's formula: $K=\sqrt{(s-a)(s-b)(s-c)(s-d)}$

Combining this with $a+c=b+d=s$ gives $K=\sqrt{abcd}$ in the case when the quadrilateral is tangential as well. Substituting for the sides as above, $300=\sqrt{a \cdot 3a \cdot 8a \cdot 6a}=12a^2$

so $a=5$ , $b=15$ , $c=40$ , $d=30$ and the perimeter is $\boxed{90}$ .

Let the sides of the quadrilateral be $a, b, c, d$ , where it can assumed without loss of generality that $a \gt b \gt c \gt d \gt 0$ . We know from the problem statement that

$a = 8 d , b = 2 c , a + d = b + c$

This system of equations has the following solution: $(a, b,c,d ) = t (8, 6, 3, 1 )$ , where $t$ is an arbitrary positive real.

The area of the cyclic quadrilateral is given by Brahmagupta's formula

$A = \sqrt{ (s - a)(s -b )(s - c)(s-d) }$

where $s$ is the semi-perimeter, $s = \frac{1}{2} (a + b + c + d )$ . Hence, $s = 9 t$ , and

$A = 300 = t^2 \sqrt{ (9 - 8)(9 - 6)(9- 3)(9 - 1) } = t^2 \sqrt{ (1)(3)(6)(8) } = 12 t^2$

Hence, $t = \sqrt{ \dfrac{300}{12} } = 5$

Therefore, $s = 9 t = 45$ and $P = 2 s =\boxed{ 90}$