Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 229

A circle with center O

A circle with center O and radius 1 cm rolls around the inside of a triangle whose sides are 6, 8, and 10 cm, always touching one or more of the sides as it rolls. How far does O travel in one complete circuit?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: A circle with center O

The answer I got was 3 + 4 + 5 = 12 cm. The path of O is a triangle whose sides are half the corresponding sides of the 6-8-10 triangle. Thank goodness for an easy question from you at last.

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: A circle with center O

Yes Jane, I too get 3 + 4 + 5 = 12 cm.
I drew it in inches with a perpendicular "L" square ruler.
Then after finding something round the right size, I drew it all
out and measured and got 11.95 inches (because I drew it 2.54 times too big).
After that I set out to use inverse tangents and half angles to the
center of each circle near the corners.  Then I tangent the half angles
to get the ratio between the radius of 1cm and the sides of the
right triangles of which there are two twin mirror image ones touching
in each corner. 
These sides of no travel from center of circle to corners are
1cm, 2cm and 3cm.   Sorry so confusing without a diagram.
Ask if want one.
So the 10cm hypotenuse's circle stops are stopped 2cm from the 6cm's side connecting corner, and is stopped at 3cm from the 8cm's side connecting corner.   
And the circle sits just squarely in the right angle, obviously, with its radius away from the corner in the parallel direction to the sides.
Also the inverse tangent of 4/3 and inverse tangent of 3/4 are
good starting points to the angles of the 6-8-10 triangle.
Recognition of similarity to the 3-4-5 right triangle is helpful, so
you just check 10^2 = 8^2 + 6^2 to start, if not sure.

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igloo myrtilles fourmis

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: A circle with center O

The path of O is the red triangle. This is similar to the big 6-8-10 triangle since their corresponding sides are parallel. (By the way, I’ve drawn the figure so that the vertical side of the big triangle is 6, the horizontal base 8 and the hypotenuse 10. I forgot to add these in while Painting the image.)

Now look at the top of the figure. We need to work out the length AC =  AB + BC = FD + BC. Using the fact that trangles BCD and EDF are similar to the 6-8-10 triangle, and BD = AF = FE = 1 cm. we should be able to find that FD = 5⁄4 and BC = 3⁄4. ∴ AC = 5⁄4 + 3⁄4 = 2.

And the red triangle is 1 cm above the base of the big triangle. Hence the length of the vertical side of the red triangle is 6 − 2 − 1 = 3 cm.

Since the red triangle is similar to the big triangle and we have found the length of one side of the red triangle, the lengths of the other sides should no longer be a mystery.