Hrvatska 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 $S=\frac{(a+b)h}{2}$. So: $$72=\frac{(a+b)\cdot 6}{2}⟹a+b=\frac{72\times 2}{6}=24$$ But $a+b=4\sqrt{10}$. So $4\sqrt{10}=24⟹\sqrt{10}=6$, 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, $4\sqrt{10}=24⟹\sqrt{10}=6$, 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: $$c=\sqrt{x^2+h^2},d=\sqrt{y^2+h^2}$$ 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 $a+b=24$, $a-b=s⟹a=\frac{24+s}{2},b=\frac{24-s}{2}$.

The area is also: $$S=\frac{(a+b)h}{2}=\frac{24\times 6}{2}=72$$ 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 $x+y=s⟹x=y=\frac{s}{2}$.

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

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

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: $$R=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$ where $a,b,c,d$ are the sides, and $s=\frac{a+b+c+d}{2}$.

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

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: $$R=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$$ 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: $$R=\frac{1}{4}\sqrt{\frac{4,147,200}{5184}}$$

Now, $4,147,200÷5184=800$

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

Answer:

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