*Originally posted by phgao*

**1. If a + (1/(b+(1/c))) = 37/16
**

Then what does a + b + c = ?

2. What is the largest possible size of an angle of a triangle formed by joing the midpoints of three edges of a cube?

3. There are 10 couples at a party. Each pair of peo ...[text shortened]... ither her husband or another woman. How many handshakes are there?

1. Yeah, THUDandBLUNDER is right about defining what kind of answer you're looking for. I assumed you wanted positive integers, because they're nice and round, and the stipulation usually gives you the info you need to answer the question. Here's how I did it:

a + 1/(b + 1/c)) = (abc + a + c)/(bc + 1) = 37/16

I assumed that the numerators and denominators were the same, and equated them.

bc + 1 = 16

bc = 15

abc + a + c = 37

a(15) + a + c = 37

16a + c = 37

Then I tried values for "a". Letting a = 1 gave c = 21, which seemed odd for this type of question. However, letting a = 2 gave c = 5, and subsequently b = 3. Nice round numbers, so I was pretty sure I had your solution in mind.

a + b + c = 2 + 3 + 5 = 10.

2. I'm not sure, but I think the maximum angle is 90 degrees. You can create a right angle triangle by starting at any mid-point and moving to the mid-point opposite the first on the same face, and then connecting point 1 to point 3. Picking a different point for point 2 results in a longer triangle with the same base. If you imagine a right angle triangle laying on it's hypotenuse like so --> /\, stretching the triangle upwards will shrink the angle at the top (the 90 degree angle). So any elongation will result in a smaller angle. That's why I think the largest possible angle is 90 degrees.

3. All 10 men will shake hands with each other, which results in (10 choose 2) handshakes, and each woman will shake hands with 9 men (not her husband). So the total is:

(10 choose 2) + 10*9 = 45 + 90 = 135 handshakes.