## Sunday, October 25, 2015

### Geometry Problem 1157: Four Circles, Mutually Tangent, 90 Degrees, Radius, Metric Relations

Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below for more details. 1. From rt. triangle, O1O2O3, (4+1)²+(1+r3)²=(4+r3)² or r3 = 5/3 and thus r4 = 4+5/3+1=20/3.

1. (To Ajit)
Can u pls explain why r4=r1+r2+r3 ?

2. Got easily r3 = 5/3

But r4 still eludes me

3. If we have three touching circle of radii a, b & c then the radius of the external circle that touches all three may be obtained from, Rext = abc/(ab+bc+ca -2√abc(a+b+c)), (The relevant problem # from this collection which gives this formula now eludes me)
In this case, a=4,b=1 & c=5/3. Apply the formula above to obtain, r4 =20/3. The radius of the internal circle which touches the inside three circles may be obtained from: Rint = abc/(ab+bc+ca +2√abc(a+b+c)),

4. Can u share the basic steps of the proof ie for the formulae of Rext and Rint Ajit ?

Rgds

Sumith

1. Sumith,
If you find any proof simpler than this, you can write to me at: ajitathle@gmail.com

5. From right trO1O2O3, r3=5/3. Let <O1O2O4=x. Using cosine rule in trs O1O2O4 & O3O2O4, and using the fact sin^2x+cos^2x=1 we get a quadratic in r4 which yeilds r4=20/3.

6. For some reason as yet unknown to me < O1O4O3 = 90

Then using Pythagoras for Tr. O1O4O3 we get a quadratic from which r4 = 20/3

Or easier we can use Ptolemy on cyclic quadrilateral O1O4O3O2

But the 1st step is still beyond me.

Ajit / Antonio - any ideas?