The Hidden Ice Cream Cone

Applying the converse of the Power of a Point theorem shows that $ABDC$ is cyclic. (That's why it looks a bit like an ice cream with its cone when you add the circle.)

Also note that by SAS similarity $PCD$ ~ $PBA$ . So $AB = 2CD$ .

Now by Ptolemy's Theorem, $7\times 4 + 2CD^2 = 190$ and it follows that $CD =9 \implies \boxed{AB=18}$ as required.

First, we will use the Law of Cosines twice in order to find expressions for $\overline{AD}$ and $\overline{BC}$ .

Let $\overline{AD} = x$ , and $\overline{BC} = y$ . We can write the following:

$x^{2} = 6^{2} + 12^{2} - 2*6*12*cos(P)$

$x^{2} = 6^{2}*(5 - 4*cos(P))$

$x = 6*\sqrt{5 - 4*cos(P)}$

And also:

$y^{2} = 5^{2} + 10^{2} - 2*5*10*cos(P)$

$y^{2} = 5^{2}*(5 - 4*cos(P))$

$y = 5*\sqrt{5 - 4*cos(P)}$

Then, taking the product $xy$ yields $30*(5 - 4*cos(P))$ . But this value is known to be 190, thus:

$30*(5 - 4*cos(P)) = 190$ , which implies $cos(P) = -\frac {1}{3}$ .

Now we have everything we need to solve the problem.

Let $\overline{AB} = z$ . Applying the Law of Cosines one last time will yield the following:

$z^{2} = 10^{2} + 12^{2} - 2*10*12*cos(P)$

$z^{2} = 100 + 144 + 80 = 324$

$z = 18$ which is the answer sought.

Solution using Co-ordinate geometry.