Star and cross dissections

[bonanova]

Construct a regular 5-pointed star, and cut it into acute triangles. Or sketch the star and draw lines. The triangles need not be congruent. What is the smallest number of cuts (lines) needed to accomplish this?

A Greek cross is the union of five squares: one each above, below and either side of a central square. Ignoring the lines joining the squares and taking only the outside perimeter, or by constructing the shape, divide a Greek cross into the smallest number of acute triangles. How many?

[Prof. Templeton]

If the triangles must be acute, then the center square has to bump out on all four side into the four squares on top, bottom, left and right. That will form an octagon in the center that can be divided into 8 acute triangles. The now "less then square" squares at the four cardinals can be divided into three acute triangles each. (4 x 3) + 8 = 20 triangles total. I'm not sure if this is minimum, but it seems the most logical.

[bonanova]

If the triangles must be acute, then the center square has to bump out on all four side into the four squares on top, bottom, left and right. That will form an octagon in the center that can be divided into 8 acute triangles. The now "less then square" squares at the four cardinals can be divided into three acute triangles each. (4 x 3) + 8 = 20 triangles total. I'm not sure if this is minimum, but it seems the most logical.

That's it. Nice explanation.

And the star?

[Prof. Templeton]

After making some doodles at lunch-time

If we just cut off all the points we are left with five triangles and a pentagon to divide, which isn't ideal. So, we just cut off two points leaving something like this

That's cuts No.1 and No.2

We can't follow the lines of the pentagon because those are 108 degree angles. We need to cut from the middle of the two bottom points "A" to the sides forming an angle less then 90 but more then the original 72 degrees. In fact angles CAB and DAE would need to be greater then 81 degrees to make triangle ABE acute.

We have 6 acute triangles after making 5 cuts.

Edited May 18, 2012 by Prof. Templeton

[bonanova]

Bravo.

[Morningstar]

Is that the minimum?

[bonanova]

Is that the minimum?

It is, unless you can show a dissection with fewer.

You can't. It is the fewest number of triangles.