A Math Question: the Crazed Carpenter
Yesterday I heard that a current union activist teacher, math teacher, was influenced to become a math teacher, in part, from reading this blog. Wow, even if that’s just a little true… Pretty cool. And then I was thinking I should write about math again. And today I was doing a problem with a friend on a car ride, and it was too hard, so we did this one again. So why not share it? Here goes:
A crazed carpenter perfectly attaches legs to the edge of the seat of a stool, except he pays no attention to where around the seat each leg goes. They could end up perfectly separated by 120º (and it stands), or they could be clustered together (and the stool would not stand). Any arrangement is possible. What is the probability that a 3-legged stool created by this crazed carpenter will stand?
and the legs attach to the edge of the seat
Credit to Bertie Taylor of the old Compuserve Science/Math Forum (or SIG?). Bertie called it the “Mad Carpenter” problem because in his dialect “mad” meant “crazy.” Alas, when I google Bertie I get nothing, and when I google “Mad Carpenter” I get a lovely bed and breakfast that offers no math problems whatsoever.
JD2718 
3/8
This is Sue V
Actually, that’s wrong. It’s 1/4.
Is “it stands” equivalent to “the center of the circle is on the interior of the triangle formed by the three legs”?
Yes, I think that works.
Ok so my first thought was to write down an integral, but maybe the following argument works instead: for each point A on the circle, let A’ be the diametrically opposite point. For each set of three points A, B, C on the circle, no two diametrically opposite, there are eight different triangles that can be formed by choosing one of {A, A’}, one of {B, B’}, and one of {C, C’} as vertices, and every non-right triangle with vertices on the circle belongs to exactly one of these groups of 8. Among the 8 triangles, there are always 2 that contain the center of the circle, and the excluded right triangles have measure 0 so don’t affect the final answer. So we get 1/4, like Sue V said.
My first instinct was the integral…
Love the alternate approach. Symmetry? Combinatorial?