Dammit not again
Nov. 1st, 2005 06:43 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
logicbuttonSo I'm walking from the apartment to campus today, and what do I see on the patio lying up against the gate but a DEAD BIRD. It's only been a month since I had to walk past that other one on the sidewalk six times in one day. I swear, they are everywhere. If the avian flu thingy has come to Decatur, that's it for me; I'm staying home for the rest of the semester. Or maybe I'll come live in the math center. I practically do already, and at least I'll be able to go to class.
On another note, one of the problems I got back in math seminar today was marked with a 7/10 and this:
Diana, can we talk about this mistake in class? It is quite possibly the most intelligent one ever made. In fact, I haven't yet found the hole, except that the "standard" method yields the "right" answer of [answer] which is a hair above [my answer]. I don't want to embarrass you, but I think it would be a good discussion topic.
So I'm ridiculously flattered, in a way. The thing is, this is easily the nicest compliment I've ever gotten on a wrong answer, but it's actually not the first time I've been told that my way of getting one was clever. Clearly, my talent lies in proving things that aren't true. For my next trick, I will be demonstrating that 0 = 1.
I finally figured out my mistake in class, incidentally. Apparently I was assuming that the way to maximize the rectangle with side lengths x and (y-x) was to make it a square, which only works when y is a constant and not dependent on x. Ha ha! Silly me!
Here's the problem:
Find the smallest number M such that for any two squares of combined area 1, there exists a rectangle of area M such that the two squares can be packed into the rectangle (with no overlap of the interior of the squares).
My answer was 6/5. The real answer is (1 + root(2))/2.
On another note, one of the problems I got back in math seminar today was marked with a 7/10 and this:
Diana, can we talk about this mistake in class? It is quite possibly the most intelligent one ever made. In fact, I haven't yet found the hole, except that the "standard" method yields the "right" answer of [answer] which is a hair above [my answer]. I don't want to embarrass you, but I think it would be a good discussion topic.
So I'm ridiculously flattered, in a way. The thing is, this is easily the nicest compliment I've ever gotten on a wrong answer, but it's actually not the first time I've been told that my way of getting one was clever. Clearly, my talent lies in proving things that aren't true. For my next trick, I will be demonstrating that 0 = 1.
I finally figured out my mistake in class, incidentally. Apparently I was assuming that the way to maximize the rectangle with side lengths x and (y-x) was to make it a square, which only works when y is a constant and not dependent on x. Ha ha! Silly me!
Here's the problem:
Find the smallest number M such that for any two squares of combined area 1, there exists a rectangle of area M such that the two squares can be packed into the rectangle (with no overlap of the interior of the squares).
My answer was 6/5. The real answer is (1 + root(2))/2.
