, 2011

Solution:

Let the trapezoid have bases $a$ and $b$ ($a > b$), and legs $c$ and $d$. Since a circle can be circumscribed about the trapezoid, it must be tangential, so $a + b = c + d$.

Given:
- $c + d = 4\sqrt{10}$
- $a + b = 4\sqrt{10}$
- Height $h = 6$
- Area $S = 72$

The area of a trapezoid is $S = \dfrac{(a + b)h}{2}$.
So:
$$72 = \dfrac{(a + b) \cdot 6}{2} \implies a + b = \dfrac{72 \times 2}{6} = 24$$
But $a + b = 4\sqrt{10}$. So $4\sqrt{10} = 24 \implies \sqrt{10} = 6$, which is not true. Therefore, the sum $a + b = 24$ and $c + d = 4\sqrt{10}$.

But since the circle can be circumscribed, $a + b = c + d$, so $a + b = c + d = 24$.

Therefore, $4\sqrt{10} = 24 \implies \sqrt{10} = 6$, which is not true. So the problem must mean that the sum of the legs is $4\sqrt{10}$, and the sum of the bases is $24$.

Let the bases be $a$ and $b$, with $a > b$, and the legs be $c$ and $d$.

Let us denote the bases as $a$ and $b$, and the legs as $c$ and $d$.

Let us drop perpendiculars from the endpoints of the shorter base to the longer base. Let the projections from the endpoints of the shorter base to the longer base be $x$ and $y$, so $x + b + y = a$.

By the Pythagorean theorem:
$$c = \sqrt{x^2 + h^2}, \quad d = \sqrt{y^2 + h^2}$$
So:
$$c + d = \sqrt{x^2 + 36} + \sqrt{y^2 + 36} = 4\sqrt{10}$$
Also, $x + y = a - b$.

But $a + b = 24$, so $a - b = x + y$.

Let $s = x + y = a - b$.

Let us try to find $a$ and $b$.

Let $a + b = 24$, $a - b = s \implies a = \dfrac{24 + s}{2}, \ b = \dfrac{24 - s}{2}$.

The area is also:
$$S = \dfrac{(a + b)h}{2} = \dfrac{24 \times 6}{2} = 72$$
Which matches the given area.

Now, $c + d = 4\sqrt{10}$.

Let us try to find $x$ and $y$ such that $\sqrt{x^2 + 36} + \sqrt{y^2 + 36} = 4\sqrt{10}$ and $x + y = s$.

Let us try $x = y$ for simplicity.

Then $x + y = s \implies x = y = \dfrac{s}{2}$.

Then:
$$2\sqrt{x^2 + 36} = 4\sqrt{10} \implies \sqrt{x^2 + 36} = 2\sqrt{10} \implies x^2 + 36 = 4 \times 10 = 40 \implies x^2 = 4 \implies x = 2$$
So $x = y = 2$, $s = 4$.

Therefore,
$$a = \dfrac{24 + 4}{2} = 14, \quad b = \dfrac{24 - 4}{2} = 10$$

So the bases are $14$ and $10$, the legs are both $\sqrt{2^2 + 36} = \sqrt{40} = 2\sqrt{10}$, so $c + d = 4\sqrt{10}$, as required.

Now, the trapezoid is isosceles, so the circle can be circumscribed.

The radius $R$ of the circle circumscribed about a tangential quadrilateral (which is also cyclic) is given by:
$$R = \dfrac{1}{4} \sqrt{\dfrac{(ab + cd)(ac + bd)(ad + bc)}{(s - a)(s - b)(s - c)(s - d)}}$$
where $a, b, c, d$ are the sides, and $s = \dfrac{a + b + c + d}{2}$.

Let us label the sides in order: $a = 14$, $b = 2\sqrt{10}$, $c = 10$, $d = 2\sqrt{10}$.

But in the isosceles trapezoid, the order is $AB = 14$, $BC = 2\sqrt{10}$, $CD = 10$, $DA = 2\sqrt{10}$.

So $a = 14$, $b = 2\sqrt{10}$, $c = 10$, $d = 2\sqrt{10}$.

Then $s = \dfrac{14 + 10 + 2\sqrt{10} + 2\sqrt{10}}{2} = \dfrac{24 + 4\sqrt{10}}{2} = 12 + 2\sqrt{10}$.

Now, the area $S$ of a cyclic quadrilateral is:
$$S = \sqrt{(s - a)(s - b)(s - c)(s - d)}$$
But $S = 72$.

So:
$$72 = \sqrt{(s - a)(s - b)(s - c)(s - d)}$$
Let us compute each term:

$s - a = (12 + 2\sqrt{10}) - 14 = -2 + 2\sqrt{10}$

$s - b = (12 + 2\sqrt{10}) - 2\sqrt{10} = 12$

$s - c = (12 + 2\sqrt{10}) - 10 = 2 + 2\sqrt{10}$

$s - d = (12 + 2\sqrt{10}) - 2\sqrt{10} = 12$

So:
$$(s - a)(s - b)(s - c)(s - d) = [-2 + 2\sqrt{10}] \times 12 \times [2 + 2\sqrt{10}] \times 12$$
$= 12 \times 12 \times [-2 + 2\sqrt{10}] \times [2 + 2\sqrt{10}]$

Now,
$[-2 + 2\sqrt{10}] \times [2 + 2\sqrt{10}] = (-2)(2) + (-2)(2\sqrt{10}) + (2\sqrt{10})(2) + (2\sqrt{10})(2\sqrt{10})$
$= -4 - 4\sqrt{10} + 4\sqrt{10} + 4 \times 10$
$= -4 + 40 = 36$

So the product is $12 \times 12 \times 36 = 144 \times 36 = 5184$

So $S = \sqrt{5184} = 72$, as required.

Now, the radius $R$ of the circumscribed circle about a cyclic quadrilateral is:
$$R = \dfrac{1}{4} \sqrt{\dfrac{(ab + cd)(ac + bd)(ad + bc)}{(s - a)(s - b)(s - c)(s - d)}}$$
Let us compute each term:

$ab + cd = (14)(2\sqrt{10}) + (10)(2\sqrt{10}) = 28\sqrt{10} + 20\sqrt{10} = 48\sqrt{10}$

$ac + bd = (14)(10) + (2\sqrt{10})(2\sqrt{10}) = 140 + 4 \times 10 = 140 + 40 = 180$

$ad + bc = (14)(2\sqrt{10}) + (10)(2\sqrt{10}) = 28\sqrt{10} + 20\sqrt{10} = 48\sqrt{10}$

So the numerator is:
$(ab + cd)(ac + bd)(ad + bc) = [48\sqrt{10}] \times 180 \times [48\sqrt{10}]$

$= (48\sqrt{10})^2 \times 180 = 2304 \times 10 \times 180 = 23040 \times 180 = 4,147,200$

The denominator is $(s - a)(s - b)(s - c)(s - d) = 5184$

So:
$$R = \dfrac{1}{4} \sqrt{\dfrac{4,147,200}{5184}}$$

Now, $4,147,200 \div 5184 = 800$

So:
$$R = \dfrac{1}{4} \sqrt{800} = \dfrac{1}{4} \times 20\sqrt{2} = 5\sqrt{2}$$

Answer:

The radius of the circumscribed circle is $5\sqrt{2}$.