Talk:2013 AMC 10B

Revision as of 00:34, 26 February 2013 by Yongh96 (talk | contribs) (Solution to problem 19: new section)
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I think there is a simpler solution for number 16.

PE = 1.5, PD = 2, and DE = 2.5. This is a right triangle because of the Pythagorean Theorem (1.5^2 + 2^2 = 2.5^2). Since PED is a right triangle, ACP, CPD, and APE must be right triangles as well.

Since we know CE and AD are medians, they bust be bisected into 2:1 ratios. Therefore, CP = 3 (because PE * 2 = 3) and AP = 4 (because PD * 2 = 4).

Now you have AP = 4, CP = 3, PD = 2, and PE = 1.5 Apply A = 1/2bh to all of these triangles and add them up. You get 13.5

Solution to problem 19

The currently uploaded solution is correct until the part where it says $x=\sqrt{3}\/b/2$.

But since c=b+x but c is less than b, x has to be lower than zero, $x=-\sqrt{3}\/b/2$.

The root is $\frac{-b}{2a}$, equal to $\frac{-b}{2b+\sqrt{3}b}$. This is $\frac{-1}{2+\sqrt{3}}$, and multiplying by the conjugate gives the answer $-2+\sqrt{3}$.