I'm a senior year maths student and I stumbled upon a question from a maths competition from a previous year. I seem to be on the cusp of solving it but I am unable to solve for the radius (to give me the answer).
The question reads as follows: A rectangle has sides of length 5 and 12 units. A diagonal is drawn and then the largest possible circle is drawn in each of the two triangles. What is the distance between the centres of these two circles?
Image below for reference:
What I have attempted so far is connecting the points of tangency for each circle for their respective centres and labelled them $r$. From this, I was able to label sides $5-r$ and $12-r$. I noticed that the diagonal and both widths of the rectangle were lines of tangency that met in the top left and bottom right corners, and therefore those parts of the diagonal from the corner to the point of tangency were also $5 - r$. From this, I could label the middle part of the diagonal $3 + 2r$. Drawing the line I had to solve for, I broke this middle part into equal sections of $3/2 + r$. From there I found out I could use Pythagoras' theorem to calculate the hypotenuse which was half of the length of the line I was trying to find. I ended up calculating this to be $\sqrt{8r^2+12r+9}$.
The only problem is I am unsure of how to solve for $r$. Help is much appreciated.
Hint
The inradius of a right-angled triangle is given by $\frac{1}{2}(a+b-c)$, where $a,b$ are the legs and $c$ is the hypotenuse. A proof of this is here on the line after the word 'Proof'.
Alternatively, you can observe the following diagram:
The area of the triangle is $\frac{1}{2} (5)(12)$. However, the area of the triangle is also $\frac{1}{2} (5r + 12r + 13r)$ by adding the areas of $\Delta CDA, \Delta ADB, \Delta BDC$ together. Therefore:
$$\frac{1}{2} (5r + 12r + 13r) = \frac{1}{2}(5)( 12) \Rightarrow 30r=60 \Rightarrow r=2$$
A generalisation of this for any triangle gives the fact that $A = rs \Rightarrow r = A/s$, where $s$ is the semiperimeter $\frac{a+b+c}{2}$.