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Problem 6 Easy

The line y = \frac { 1 } { 3 } x + 1 divides the square region defined by 0 \leqslant x \leqslant 2 and 0 \leqslant y \leqslant 2 into an upper region and a lower region. The line x=a divides the lower region into two regions of equal area. Then a can be written as \sqrt { s } -t, where s and t are positive integers. What is s+t?

  • A.

    18

  • B.

    19

  • C.

    20

  • D.

    21

  • E.

    22

Answer:C
Solution

Points: E(0,1), F(2, \frac{5}{3}), G(a,0), H(a, \frac{1}{3}a+1), A(0,0), B(2,0).

Given S_{AGHE} = \frac{1}{2} S_{ABFE}:

\frac{1}{2} \cdot a \cdot \left(\frac{1}{3}a+1+1\right) = \frac{1}{2} \cdot \frac{1}{2} \cdot 2 \times \left(1 + \frac{5}{3}\right)

This gives a^2 + 6a - 8 = 0, so a = \pm \sqrt{17} - 3.

Since a > 0, we have a = \sqrt{17} - 3.

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