general-tiebreaker-solutions

# general-tiebreaker-solutions - Answer 30 SMT 2013 General...

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SMT 2013 General Tiebreaker Solutions February 2, 2013 1. Compute all real nonzero values of b such that bx 2 - bx - b = 0 has b as a solution. Answer: 1 ± 5 2 Solution: Plugging in x = b , we have b 3 - b 2 - b = b ( b 2 - b - 1) = 0. From the quadratic formula, we know that b 2 - b - 1 has 1 ± 5 2 as solutions. 2. Leo is at the bottom left corner of the grid. If he can only walk up and to the right, how many different paths can he take to the upper right corner? Answer: 20 Solution: 3. The lines y = 2 x + 1, y = - 1 2 x + 6, and y = - 2 x - 3 split the plane into multiple regions. One of the regions has finite area. Compute the area of that region.

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Unformatted text preview: Answer: 30 SMT 2013 General Tiebreaker Solutions February 2, 2013 Solution: The ﬁrst two lines have an intersection point of A = (2 , 5). The ﬁrst and third lines have an intersection point of B = (-1 ,-1). The last two lines have an intersection point of C = (-6 , 9). We observe that the ﬁrst and second line are perpendicular. Thus, the area of the triangle is AB · AC 2 = 3 √ 5 · 4 √ 5 2 = 30 ....
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