Geometry Problem. Post your solution in the comments box below.

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

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## Wednesday, October 28, 2015

### Geometry Problem 1158: Four Tangent Circles, Tangent Chord, Radius, Metric Relations

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ReplyDeleteLet x= O4M

Since O2O3E is a right triangle => O3E^2= 9^2-1^2= 80

Since O4EO3 is a right triangle => O3O4=Sqrt((x+4)^2+80)

We have O4C=O4D => x+10= sqrt((x+4)^2+80) +4

Solve this equation we get x= 15

In right triangle CBF we have MB^2= MC. MF= 10 .x 40= 400

So MB=20 and AB=40

Let O2O4 meet AB at C the point of tangency and mid point of AB. Let O1D be an altitude of Tr. O1O2O4 and let O1D = h

ReplyDeleteWe need to use Pythagoras thrice

h^2 = 9^2 -1 = 80 .... (1) from Tr. O1O2D

If the radius of O4 = r then

(r-4)^2 = h^2 + (r-6)^2 from which upon simplifying and using (1) r = 25 (Tr. O1O4D)

Now use Pythagoras on Tr. ACO4

(AB/2)^2 = r^2 -(r-10)^2 = 20r-100 = 400

So AB = 40

Sumith Peiris

Moratuwa

Sri Lanka