[superprismatic]

Begin with an isosceles triangle with sides 13, 13, and 10.

Put the 10-long side on the bottom and create an infinite

stack of circles inside this triangle as follows: Place the

largest inscribed circle in the triangle (it will be tangent

to the three sides of the triangle). Next, place the largest

circle you can fit in the triangle which sits atop the first

circle. Keep doing this to make an infinite stack of circles

with each circle tangent to sides of the triangle and to the

circles above or below it. What is the sum of the

circumferences of all the circles in the stack?

[Guest]

I believe the answer to be 20 pi

Draw the picture for the first inscribed circle and then use similar triangles rules to find that radius r. I found that to be 30/7.

Now in creating the second circle we can also use ratios and similar triangles. The second circle will be inscribed in triangle whose height is (10 - 2r). But the 2nd ratio(call it p)will be proportional. So

r/10 = p / (10 -2r) or p/r = (10-2r)/10 = 4/7

This ration holds true for each new circle.

To get the total circumference, then you just add it up.

2pi * sum(r + r1 + r2 ...).

This is just summing an infinite sequence.

2pi * 30/7 * (1/(1 - 4/7))

= 2pi *30/7*(7/3) = 20pi

[superprismatic]

I believe the answer to be 20 pi

Draw the picture for the first inscribed circle and then use similar triangles rules to find that radius r. I found that to be 30/7.

Now in creating the second circle we can also use ratios and similar triangles. The second circle will be inscribed in triangle whose height is (10 - 2r). But the 2nd ratio(call it p)will be proportional. So

r/10 = p / (10 -2r) or p/r = (10-2r)/10 = 4/7

This ration holds true for each new circle.

To get the total circumference, then you just add it up.

2pi * sum(r + r1 + r2 ...).

This is just summing an infinite sequence.

2pi * 30/7 * (1/(1 - 4/7))

= 2pi *30/7*(7/3) = 20pi

No, that's not correct. I haven't worked through your answer, but you must have made a mistake somewhere. Your method seems fine but I think the 30/7 is wrong.

[EventHorizon]

I'll hold myself back from giving the answer, but a significant clue to the simplest way to solve this is to realize that circumference = pi * diameter.

[Guest]

Overly simple approach

As eventHorizon pointed out, the answer is pi*diameter and as Treesfearme pointed out, you must sum the diameters of all the circles involved. Since each decreasing circle moves along a perpendicular from the base of the circle to the top vertex of the isosceles triangle. This sum of diameters must equal the height of the triangle. Simple Pythagorean Theorem application gives a height of 12, hence the total circumference of the all the circles is 12 pi.

Having demonstrated this solution, I must say Treesfearme did demonstrate excellent logic through his method, even though some error in process aquired the wron answer.

[EventHorizon]

Yup, that's the easy way I hinted at.

First, find the radius of the inscribed circle.

Method 1:

First, find the height of the triangle. Because it's a Pythagorean triple, it's simple. It is 12.

Next, break the triangle up into 6 pairs of congruent triangles by using lines from the vertices to the center of the circle and three lines which are perpendicular to each side and radii of the circle.

Choose one and solve for the length of the radius.

I'll use one of the two bottom triangles, since it is obvious the length of one of it's sides.

I'll calculate the angle at the bottom of the original triangle by using arctan(12/5) = 67.38013505.

Since the angle is bisected, I'll halve the angle and thats one angle of the triangle I chose.

The radius of the circle is then 5 * tan(67.38013505/2) = 3.33333333.

Method 2:

The radius of the inscribed circle of a triangle is equal to sqrt((s-a)(s-b)(s-c)/s), where a,b,c are the side lengths and s is the semiperimeter ((a+b+c)/2).

s = (10+13+13)/2 = 36/2 = 18.

so r = sqrt((18-10)(18-13)(18-13)/18) = sqrt(8*5*5/18) = 5* sqrt(4/9) = 5 * (2/3) = 10/3

Now that I have the radius of the inscribed circle, I can find the ratio to the radius of the next by finding the ratio of height of the original triangle (found in method 1 to be 12) to the height of the remaining triangle if you cut it off at the top of the inscribed circle.

The height of the remaining triangle is 12 - 2 * r = 12 - 2 * (10/3) = 36/3 - 20/3 = 16/3.

So the ratio is (16/3)/12 = 16/36 = 8/18 = 4/9.

The circumference of the first circle is 2 * pi * (10/3) = 20*pi/3.

Using the equation for the sum of a geometric series, sum = first/(1-ratio) = (20*pi/3) / (1-4/9) = (20*pi/3) / (5/9) = 9 * 20 * pi / (3 * 5) = 3 * 4 * pi = 12*pi.

So, like the easy way, the final answer is 12*pi.

The area of the first circle is 100*pi/9.

We know the ratio for the height of the triangles is 4/9. However, since area is a 2d concept, the ratio of the area of one circle to the next is the square of the ratio of lengths. So the ratio is 16/81.

Plugging numbers into the sum of a geometric series gives (100pi/9) / (1 - 16/81) = (100pi/9) / (65/81) = 81 * 100 * pi / (9 * 65) = 9 * 20 * pi / 13 = 180pi/13.

If you compare the equation above for the radius of the inscribed circle to Huron's formula for calculating the area of a triangle given it's side lengths you'll notice the following relation:

Area of triangle = radius of inscribed circle * semiperimeter.

[Guest]

i got 40pi/3