Tatiana went to another Mu Alpha Theta competition this weekend. She took 9th place in the individual geometry competition! (Read on for one of the interesting problems she couldn't get.)

Unfortunately, she suffered an injury. You don't normally associate injuries with math; it's not a rough sport or anything. But it turns out Tatiana inherited her mother's brittle feet. She severely sprained her ankle on the way to the stage.

She's on crutches and ibuprofen.

**Update 2009-02-04:** I will no longer take my family to the CentraCare on University Blvd. After 48 hours without a reduction in swelling, we took her to our pediatrician, who discovered that the leg is actually fractured. Tatiana is now in a walking cast for 4 weeks.

Problem number 15 was interesting. Given a right triangle, such that the hypotenuse c = a + 1, what is b^{2}?

It's not really difficult. c^{2} = a^{2} + b^{2} so b^{2} = c^{2} - a^{2}. Find c^{2} with the classic FOIL method: (a + 1)^{2} = a^{2} + 2a +1. Subtract a^{2}, and you see that c = 2a + 1.

2a + 1 wasn't among the answers.

*But c + a was.*

You have to do an extra substitution step to realize that 2a + 1 is the same as a + c. After all, 2a + 1 = a + (a + 1). And earlier, we said that c = a + 1. So 2a + 1 = a + c = c + a.

I suppose you could also have realized that a 3,4,5 triangle could be said to satisfy c = a + 1 (if you swap the common definitions for a and b). Then b^{2} is 9. And 9 = 4 + 5.

If you wanted to do it the easy, sissy way.