## Sunday, March 15, 2015

### Problem 1099 Four Circles, Common External Tangent, Common Internal Tangent, Radius, Metric Relations, Sangaku

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it. 1. http://s2.postimg.org/3z5fe947d/pro_1099.png

Draw lines per attached sketch
Let CA1 cut DB1 at A . We have O1 and O2 are incircle and excircle of triangle ADC
We have DE= CF … ( property of incircle and excicle of a triangle ADC)
DE= r2.tan(D/2)
CF= r1/tan(C/2)
DE=CF => r2= r1/(tan(D/2). tan(C/2)) ….. (1)
In right triangle O1HO4 we have HO4 ^2= (r1+r4)^2 – (r1-r4)^2 => HO4=2.sqrt(r1.r4)
So tan(D/2)= (r1-r4)/(2.sqrt(r1.r4))
Similarly tan(C/2) = (r1-r3)/(2.sqrt(r1.r3))
Replace value of tan(C/2) and tan(D/2) in (1) we have
r2= 4. r1^2.sqrt(r3-r4)/((r1-r3).(r1-r4))

2. There is a typo error in the last line of my comment. The correction will be
r2= 4. r1^2.sqrt(r3.r4)/((r1-r3).(r1-r4))

3. Purely geometry proof

Let A1A2 be tangent to circle O3 at A3 and let CD be tangent to circles O1,
O2, O3, O4 at C1,C2,C3,C4. Let BD be tangent to circle O4 at B4.

Further let A1C=CC1= a, A3C = C3C =p, C1C2 = b, C2C4 = c, C4D = d so that
B4D = d and so B1B4 = b+c and DB2 =c+d

Now O1A1C and O2A2C are similar triangles so r1r2 = a(a+b)…..(1)

Similarly r1r2 = (b+c+d)(c+d) …..(2)

From (1) and (2) a=c+d …(3)

From similar triangles a/p =r1/r3 and using Pythagoras (a-p)^2 = (r1+r3)^2
– (r1-r3)^2 = 4r1r3 , hence solving and simplifying,

a=2r1sqrt(r1r3)/(r1-r3)….(4) and similarly from (3)

C1D = b+c+d = a+b = 2r1sqrt(r1r4)/(r1-r4)…(5)

From (4) and (5), a(a+b) = 4r1^3.sqrt(r3r4)/{(r1-r3)(r1-r4)}

Hence from (1) r2 = a(a+b)/r1 = 4r1^2.sqrt(r3r4)/{(r1-r3)(r1-r4)}

Sumith Peiris
Moratuwa
Sri Lanka