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2018/19 AHSMC Part 1 Problem 11

Problem 11 of 16

A set SS of five different real numbers is given. When each pair of these numbers are added, you get the ten sums −4,−3,−2,−1,0,1,4,5,6,8-4,-3,-2,-1,0,1,4,5,6,8. The ratio of the largest number in SS to the smallest number in SS is

Answer choices

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Solution

Let x1≤x2≤x3≤x4≤x5x_1 \le x_2 \le x_3 \le x_4 \le x_5 be the numbers. Then x1+x2=−4x_1 + x_2 = -4, x1+x3=−3x_1 + x_3 = -3, x4+x5=8x_4 + x_5 = 8, x3+x5=6x_3 + x_5 = 6. On the other hand 4(x1+x2+x3+x4+x5)=−4−3−2−1+0+1+4+5+6+8=14  ⟺  x1+x2+x3+x4+x5=72. 4(x_1 + x_2 + x_3 + x_4 + x_5) = -4-3-2-1+0+1+4+5+6+8 = 14 \iff x_1 + x_2 + x_3 + x_4 + x_5 = \frac{7}{2}. Hence x3=−12x_3 = -\dfrac{1}{2} and also x1=−52x_1 = -\dfrac{5}{2}, x2=−32x_2 = -\dfrac{3}{2}, x4=32x_4 = \dfrac{3}{2}, x5=132x_5 = \dfrac{13}{2} and thus x5x1=−135\dfrac{x_5}{x_1} = -\dfrac{13}{5}. The answer is (B).

University of Alberta