What is the radius of the circle inscribed in triangle $ABC$ if $AB = 12,$ $AC=14,$ $BC=20$? Express your answer in simplest radical form.
Using the Law of Cosines, $\cos B = \frac{20^2 - 12^2 - 14^2}{2 \cdot 12 \cdot 14} = -\frac{1}{7}.$ Because the triangle is not obtuse and $\cos B$ is negative, $\cos B = \cos (180^\circ - B) = -\frac{1}{7}.$ Then,
\[\sin B = \sqrt{1 - \left( -\frac{1}{7} \right)^2} = \frac{4 \sqrt{3}}{7}.\]Then from the Extended Law of Sines, the diameter of the circumcircle of triangle $ABC$ is
\[\frac{BC}{\sin B} = \frac{20}{\frac{4 \sqrt{3}}{7}} = \frac{35 \sqrt{3}}{2},\]so the circumradius of triangle $ABC$ is $R = \frac{BC}{2 \sin B} = \boxed{\frac{35 \sqrt{3}}{4}}.$
[asy]
unitsize(0.2 cm);
pair A, B, C;
B = (0,0);
C = (20,0);
A = 12*dir(acos(-1/7));
draw(A--B--C--cycle);
draw(rightanglemark(B,A,C,30));
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
[/asy]
To find the radius of the inscribed circle in triangle $ABC,$ we can use the formula:
$$r = \frac{A}{s},$$
where $r$ is the radius of the circle, $A$ is the area of the triangle, and $s$ is the semiperimeter of the triangle.
First, we need to find the area of triangle $ABC.$ We can use Heron's Formula:
$$A = \sqrt{s(s-a)(s-b)(s-c)},$$
where $a,$ $b,$ and $c$ are the side lengths of the triangle, and $s$ is the semiperimeter.
In this case, $a = 12,$ $b = 14,$ and $c = 20.$ So the semiperimeter is:
$$s = \frac{12 + 14 + 20}{2} = 23.$$
The area of triangle $ABC$ is then:
$$A = \sqrt{23(23-12)(23-14)(23-20)} = \sqrt{23(11)(9)(3)} = \sqrt{3^2 \cdot 11 \cdot 9 \cdot 23} = 3\sqrt{11 \cdot 9 \cdot 23}.$$
Now, we can find the radius of the inscribed circle:
$$r = \frac{A}{s} = \frac{3\sqrt{11 \cdot 9 \cdot 23}}{23} = \frac{3\sqrt{11\cdot 3 \cdot 3 \cdot 23}}{23} = \frac{9\sqrt{33}}{23}.$$
Therefore, the radius of the circle inscribed in triangle $ABC$ is $\boxed{\frac{9\sqrt{33}}{23}}.$
To find the radius of the circle inscribed in triangle $ABC,$ we can make use of the formula:
\[ r = \frac{A}{s}, \]
where $r$ is the radius of the inscribed circle, $A$ is the area of the triangle, and $s$ is the semiperimeter of the triangle.
First, we need to find the area of triangle $ABC.$ We can use Heron's formula:
\[ A = \sqrt{s(s-a)(s-b)(s-c)}, \]
where $a,$ $b,$ and $c$ are the lengths of the sides of the triangle.
In this case, $a=BC=20,$ $b=AC=14,$ and $c=AB=12.$ We can compute the semiperimeter as $s=\frac{20+14+12}{2}=23.$
Now, we can find the area of triangle $ABC$ using Heron's formula:
\[ A = \sqrt{23(23-20)(23-14)(23-12)} = \sqrt{23(3)(9)(11)} = 3\sqrt{759}. \]
Finally, we can substitute the values of $A$ and $s$ into the formula for the radius of the inscribed circle:
\[ r = \frac{3\sqrt{759}}{23} = \boxed{\frac{3\sqrt{759}}{23}}. \]
Therefore, the radius of the circle inscribed in triangle $ABC$ is $\frac{3\sqrt{759}}{23}$ in simplest radical form.