WillBill

According to the Triangle Inequality, ABC is indeed a triangle, since 8 + 12 > 15.

Thus, the radius of the circle is the same as the circumradius of ABC, which can be found by:
\(R = \frac {abc} {4S}\)

Where S is the area of the triangle.

Using Heron's formula, we get:
\(S = \sqrt { \frac {35} {2} \left( \frac {35} {2} - 8 \right) \left( \frac {35} {2} - 15 \right) \left( \frac {35} {2} - 12 \right)}\)

\(S = \sqrt {\frac {35} {2} \left( \frac {19} {2} \right)\left( \frac {5} {2} \right)\left( \frac {11} {2} \right)}\)

\(S = \frac {5} {4} \sqrt {1463}\)

Therefore,

\(R = \frac {2520} {5\sqrt{1463}}\)

\(R = \frac {504\sqrt{1463}} {1463}\)

6 and 3 + 2sqrt(2)

\(\overline{AE}= \sqrt {\left( \frac {1} {2} \right)^2 + \left( \frac {5\sqrt{3}} {2} \right)^2} = \sqrt {\frac {1} {4} + \frac {75} {4}} = \frac {1} {2} \sqrt {76} = \boxed{\sqrt {19}}.\)

\(= \binom 43 \cdot 5! = 480.\)

\(= (a^2 - 3a + 3)(a^2 + 3a + 3)\)

Area:

\(= \sqrt {\frac {3a} {2} \left( \frac {3a} {2} - a \right)^3} = \sqrt {\frac {3a^4} {2^4}} = \frac {a^2} {4} \sqrt {3}\)

Circumradius:

\(= \frac {a^3} {\frac {a^2} {4} \sqrt {3}} = \frac {4a} {\sqrt {3}} = \frac {4\sqrt{3} a} {3}\)