Question 1120565
<br>
The figure looks like this:
{{{drawing(400,200,-2,22,-2,10,
line(0,0,20,0),line(20,0,20,8),line(20,8,0,8),line(0,8,0,0),line(0,8,4,0),line(4,0,20,8),locate(-1,-.5,A),locate(-1,8.5,B),locate(20.5,8.5,C),locate(20.5,-.5,D),locate(4,-.5,P),
locate(-1,4,"8"),locate(10,9,"20"),locate(1.5,1,x),locate(12,1,"20-x")
)}}}<br>
Triangles PAB, BPC, and CDP are all similar right triangles.  We can find the value of x using the similarity of triangles PAB and CDP:<br>
{{{8/x = (20-x)/8}}}
{{{20x-x^2 = 64}}}
{{{x^2-20x+64 = 0}}}
{{{(x-4)(x-16) = 0}}}<br>
So x is 4.<br>
The Pythagorean Theorem then gives us 4*sqrt(5) as the length of BP.<br>
r1, the inradius of triangle PAB, can be found using the formula<br>
triangle area = one-half perimeter times inradius:<br>
{{{A = (1/2)(p)(r1)}}<br>
where A is the area of the triangle and p is its perimeter.<br>
{{{16 = (1/2)(8+4+4*sqrt(5))(r1)}}}
{{{16 = (6+2*sqrt(5))(r1)}}}
{{{8 = (3+sqrt(5))(r1)}}}
{{{r1 = 8/(3+sqrt(5)) = 8(3-sqrt(5))/((3+sqrt(5))(3-sqrt(5))) = 8(3-sqrt(5))/4 = 6-2*sqrt(5)}}}<br>
To find r2 and r3, we can use the similarity of the three triangles.  r2 is r1*sqrt(5); r3 is r1*2:<br>
r2 = {{{6*sqrt(5)-10}}}
r3 = {{{12-4*sqrt(5)}}}<br>
Then r1+r2+r3 = {{{(6-2*sqrt(5))+(6*sqrt(5)-10)+(12-4*sqrt(5)) = 8}}}<br>
Final answer: r1+r2+r3 = 8.