I Took The AMC!

Recently, I have just been talking about love, love, and love. And I am boy gettin tired of it. So for today, I will switch my view to mathematics. Mostly because yesterday, I took the AMC 10 B, which I’d like to say I did poor on it. The beginning half I felt good, but the later half, there were some I left blank and some I just assumed it should be right. Man, the AMC is harder than it seems!

Here’s one question (no. 14) from that test I took:

Define a\clubsuit b=a^2b-ab^2. Which of the following describes the set of points (x,y) for which x\clubsuit y=y\clubsuit x.

\textbf{(A)}\ \text{a finite set of points}\\ \qquad\textbf{(B)}\ \text{one line}\\ \qquad\textbf{(C)}\ \text{two parallel li...

Here was my thought process, which eventually turned out to be wrong:

a\clubsuit b=a^2b-ab^2, so this also equals ab (a-b)Similarly, for x\clubsuit y=y\clubsuit x, it would be                       xy(x-y) = yx (y-x) thus x-y = y-x  thus 2x = 2y and thus x=y. That would satisfy answer B.

The actual way to solve it, unfortunately to my dismay:

x\clubsuit y = x^2y-xy^2 and y\clubsuit x = y^2x-yx^2. Therefore, we have the equation x^2y-xy^2 = y^2x-yx^2 Factoring out a -1 gives x^2y-xy^2 = -(x^2y-xy^2) Factoring both sides further, xy(x-y) = -xy(x-y). It follows that if x=0y=0, or (x-y)=0, both sides of the equation equal 0. By this, there are 3 lines (x=0y=0, or x=y) so the answer is (E) three lines.

Building Where I Took My AMC

Well, look’s like this question I missed. There were also a few other problems in which I missed due to miscalculations. For instance, one question I saw 5-3 as 5+3 and put my answer as 8, when it was supposed to be 2. Next time I take the AMC, I better be careful of these silly mistakes.

Updated 9:40 pm: By the way, refer to my previous mathematics posts to know more about the AMC.

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