All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes
Four Tangent Circles (Posted on 2003-10-20) Difficulty: 4 of 5
Four circles are tangent to each other and to a triangle that surrounds them, as shown in the figure. Each of the three circles in the corners have a radius of 1 cm. What is the radius of the circle in the middle?

No Solution Yet Submitted by Bryan    
Rating: 4.1429 (7 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution | Comment 1 of 13
Let's place this on a Cartesian graph, sitting on the x-axis, and centered on the y-axis.

R is the radius of the big (middle circle).

Let's introduce point A as the center of the top circle, point B as the center of the big (middle) circle, point C as the point of tangency between the two bottom circles, and point D as the center of the right, bottom circle.

Point C sits at (0,1) since it is on the y-axis, and is at the same height (y value) as the center of the two lower circles

Point D is at (1,1) since it is at the center of the the right, bottom circle

Draw right triangle BCD. The length of BD (the hypotenuse) is R+1. The length of CD is 1. Let the length of CB = x.
x² + 1² = (R+1)²
x = √(R² + 2R)

Therefore, point B sits at (0, 1 + √(R² + 2R) )

Point A is R + 1 (the sum of the two radii) higher than point B, therefore it is at (0, R + 2 + √(R² + 2R) ).
And of course, the length of AC is therefore R + 1 + √(R² + 2R).

Let's introduce point E, the point of tangency of the middle circle and the right side of the triangle.

Introduce line segment AD which connects the centers of two 1 unit circles. This line segment is parallel to the side of the right triangle (both circles have radius 1).

Introduce line segment BE which is perpendicular to the line segment AD and to the side of the triangle. Call the intersection of BE and AD, point F.

Note that length of BE = R, FE = 1, and BF = R-1.

Draw triangle ABF and DBF. Note that these two are similar right triangles (having hypotenuse = R+1, one leg = R-1, and therefore, the other leg must be 2√R).

Therefore, the length of AD = 2 * 2√R = 4√R.

This is the hypotenuse of right triangle ACD.

AC = has already been determined above: R + 1 + √(R² + 2R)
and CD = 1

So, let's set up the pythagorean theorem and solve for R:

( 4√R. )² = ( R + 1 + √(R² + 2R) )² + (1)²

This can be simplified to:
2R² - 12R + (2R+2)(√(R²+2R)) + 2 = 0

If we solve this, numerically for R, we get
R is approximately equal to 1.51067547264602

I probably made an error along the way, please check my work.
Edited on October 20, 2003, 6:24 pm
  Posted by SilverKnight on 2003-10-20 18:23:10
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information