Hello, amerlaw1!

2. What is the distance between the incenter and circumcenter

of a triangle with sides of the length 12, 16 and 20? Code:

_ A *
: | *
: | *
: | *
: | *
: | *
: | * 20
: | * * * *
: | * * E
12 | * o
: |* r * *
: | * *
: * * * *
: F o - - - - o * *
: * r |O * *
: | | *
: r|* |r * *
: | * | * *
: | * | * *
- C * - - - * o * - - - - - - - - - - - - * B
r D
: - - - - - - - - 16 - - - - - - - - :

We have right triangle $\displaystyle ABC:\:AC = 12,\;BC = 16,\;AB = 20.$

Place the triangle on a coordinate system with $\displaystyle \,C$ at the Origin.

The incenter is $\displaystyle O(r,r).$

. . $\displaystyle OD = OE = OF = CD = CF = r.$

The circumcenter is the midpoint of $\displaystyle AB:\;P(8,6).$

. . (Not shown on the diagram.)

Note that: .$\displaystyle AF \,=\, 12-r$

Since $\displaystyle AE$ is also tangent to the circle: $\displaystyle AE \,=\, 12-r$

Note that: .$\displaystyle BD \,=\,16-r.$

Since $\displaystyle BE$ is also tangent to the circle: $\displaystyle BE \,=\,16-r$

Since $\displaystyle AB = 20$, we have: .$\displaystyle (12-r) + (16-r) \:=\:20 \quad\Rightarrow\quad r \,=\,4$

We have: .$\displaystyle \begin{Bmatrix}\text{Incenter:} & O(4,4) \\ \text{Circumcenter:} & P(8,6) \end{Bmatrix}$

Therefore: .$\displaystyle \iverline{OP} \;=\;\sqrt{(8-4)^2 + (6-5)^2} \;=\;\sqrt{16+4} \;=\;\sqrt{20} \;=\;2\sqrt{5}$