### Small Geometry Puzzle

Feb. 13th, 2015 11:11 pm**ch3cooh**

Challenge: write up your initial instinct, then try to prove it and let me know what happens. :D

I really like this puzzle. And I'm sure that there's a simpler proof than the one I'm using right now.

Puzzle:

Solution:

The orange is

Proof:

Step 1: The Orange circle is exactly 1/4 of the big circle. Proof by picture:(Draw another equlateral triangle inscribed in the circle, with a 180 degree rotation relative to the original triangle. These two triangles overlap in a regular hexagon that can be split into 6 triangles like a pie. Each of these triangles is the same size as any of the 6 points of the star now inscribed in the circle. So it is clear that the radius of the orange circle is half the radius of the large circle. Since A = pi*r^2, this means that the area of the orange circle is 1/4 the area of the whole circle.

Step 2: Blue area is exactly 1/3 of the remaining area in the large circle that is not part of the orange circle. Proof by picture:The remaining area has three semetrical components.

Step 3: Since the orange circle is 1/4 of the large circle, the remaining interior of the large circle is 3/4 of the area. The blue area is 1/3 of that, and 1/3 of 3/4 is 1/4. So both the orange region and the combined two blue regions are 1/4 the area of the whole circle.

Boom! Done!

Using this to draw a very pretty infinite summation picture. :)

1/4 + 1/16 + 1/64 + ... 1/4^n as n approaches infinity = 1/3

I really like this puzzle. And I'm sure that there's a simpler proof than the one I'm using right now.

Puzzle:

Solution:

The orange is

**exactly the same**area as the sum of the two blue areas.Proof:

Step 1: The Orange circle is exactly 1/4 of the big circle. Proof by picture:(Draw another equlateral triangle inscribed in the circle, with a 180 degree rotation relative to the original triangle. These two triangles overlap in a regular hexagon that can be split into 6 triangles like a pie. Each of these triangles is the same size as any of the 6 points of the star now inscribed in the circle. So it is clear that the radius of the orange circle is half the radius of the large circle. Since A = pi*r^2, this means that the area of the orange circle is 1/4 the area of the whole circle.

Step 2: Blue area is exactly 1/3 of the remaining area in the large circle that is not part of the orange circle. Proof by picture:The remaining area has three semetrical components.

Step 3: Since the orange circle is 1/4 of the large circle, the remaining interior of the large circle is 3/4 of the area. The blue area is 1/3 of that, and 1/3 of 3/4 is 1/4. So both the orange region and the combined two blue regions are 1/4 the area of the whole circle.

Boom! Done!

Using this to draw a very pretty infinite summation picture. :)

1/4 + 1/16 + 1/64 + ... 1/4^n as n approaches infinity = 1/3