SOLUTION: in rectangle ABCD,AB=8 and BC=20.let P be a point on AD such that angle BPC=90 degree.if r1,r2,r3 are radii of the incircles of the triangles APB,BPC and CPD, what is the value of

Algebra ->  Triangles -> SOLUTION: in rectangle ABCD,AB=8 and BC=20.let P be a point on AD such that angle BPC=90 degree.if r1,r2,r3 are radii of the incircles of the triangles APB,BPC and CPD, what is the value of       Log On


   

Question 1120565: in rectangle ABCD,AB=8 and BC=20.let P be a point on AD such that angle BPC=90 degree.if r1,r2,r3 are radii of the incircles of the triangles APB,BPC and CPD, what is the value of r1+ r2+r3 ?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The figure looks like this:


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:

8%2Fx+=+%2820-x%29%2F8
20x-x%5E2+=+64
x%5E2-20x%2B64+=+0
%28x-4%29%28x-16%29+=+0

So x is 4.

The Pythagorean Theorem then gives us 4*sqrt(5) as the length of BP.

r1, the inradius of triangle PAB, can be found using the formula

triangle area = one-half perimeter times inradius:


16+=+%286%2B2%2Asqrt%285%29%29%28r1%29
8+=+%283%2Bsqrt%285%29%29%28r1%29


To find r2 and r3, we can use the similarity of the three triangles. r2 is r1*sqrt(5); r3 is r1*2:

r2 = 6%2Asqrt%285%29-10
r3 = 12-4%2Asqrt%285%29

Then r1+r2+r3 = %286-2%2Asqrt%285%29%29%2B%286%2Asqrt%285%29-10%29%2B%2812-4%2Asqrt%285%29%29+=+8

Final answer: r1+r2+r3 = 8.

Answer by ikleyn(52817) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let me offer different solution. 




For right triangles, there is the simple and straightforward formula for the radius of the inscribed circle

    r = %28a+%2B+b+-+c%29%2F2,

where "a" and "b" are the legs lengths and "c" is the hypotenuse length.

    For the proof of this formula see the lesson 
        Solved problems on tangent lines released from a point outside a circle
    in this site.

Using this formula, you get

    r1 = %281%2F2%29.(|AB| + |AP| - |BP|)    (1)  for the triangle ABP;

    r2 = %281%2F2%29.(|BP| + |PC| - |BC|)    (2)  for the triangle PBC;  and  

    r3 = %281%2F2%29.(|PD| + |DC| - |PC|)    (3)  for the triangle PDC.


Now add the formulas (1), (2) and (3).  You will get

r1 + r2 + r3 = %281%2F2%29.(|AB| + |AP| - |BP| + |BP| + |PC| - |BC| + |PD| + |DC| - |PC|) = 

                   the terms -|BP|  and  |BP|,  |PC|  and  -|PC|  cancel each other, and you get

             = %281%2F2%29.(|AB| + |AP|  - |BC| + |PD| + |DC|) = %281%2F2%29.(8 + 20 - 20 + 8) = 8.

Answer.   r1 + r2 + r3 = 8.