MHB What is the radius of the circumcircle for cyclic pentagon $PQRST$?

[anemone]

Here is this week's POTW:

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Cyclic pentagon $PQRST$ has side lengths $PQ=QR=5,\,RS=ST=12$ and $PT=14$.

Determine the radius of its circumcircle.

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[anemone]

Congratulations to greg1313 for his correct solution, which you can find below::)

Spoiler

Let $r$ be the circumradius. Let $O$ be the centre of the circle. Let $PR$ = $a$.

By the law of cosines, $\angle{QOR}=\arccos\left(\dfrac{2r^2-25}{2r^2}\right)$ and $\angle{POR}$ is $2\arccos\left(\dfrac{2r^2-25}{2r^2}\right)$.

Construct a segment from $Q$ Through $O$ to meet the circle at $X$.

By the inscribed angle theorem, $\angle{PXR}=\arccos\left(\dfrac{2r^2-25}{2r^2}\right)$.

As $PQRX$ is cyclic, $\angle{PQR}=180^\circ-\arccos\left(\dfrac{2r^2-25}{2r^2}\right)$ and

$a^2=50+25\cdot\dfrac{2r^2-25}{r^2}$. Solving for $r^2$, we have $r^2=\dfrac{625}{100-a^2}$

From Wolfram Mathworld we have for a cyclic quadrilateral

$r=\dfrac14\sqrt{\dfrac{(ac+bd)(ad+bc)(ab+cd)}{(s-a)(s-b)(s-c)(s-d)}}\quad(1)$

where $s$ is the semiperimeter and $a,b,c,d$ are the sides of the quadrilateral.

Plugging in and simplifying we get

$625=\dfrac{144(14a+144)(10-a)}{38-a}\Rightarrow a=\dfrac{86}{9}$

Plugging this $a$ into $(1)$ we arrive at $r=\dfrac{225}{4\sqrt{44}}=\dfrac{225\sqrt{11}}{88}$.