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2000 AMC 10 Problem 16

Problem 16 of 25IntermediateAlgebraGeometry

The diagram shows 2828 lattice points, each one unit from its nearest neighbors. Segment ABAB meets segment CDCD at E.E. Find the length of segment AE.AE.

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Solution

Place the points at A=(0,3),A = (0, 3), B=(6,0),B = (6, 0), C=(4,2),C = (4, 2), D=(2,0).D = (2, 0). Line ABAB is x+2y=6x + 2y = 6 and line CDCD is x−y=2.x - y = 2. Solving simultaneously gives E=(103,43).E = \left(\dfrac{10}{3}, \dfrac{4}{3}\right). Then AE=(103)2+(43−3)2=1009+259=553. \begin{aligned} AE &= \sqrt{\left(\dfrac{10}{3}\right)^2 + \left(\dfrac{4}{3} - 3\right)^2} \\ &= \sqrt{\dfrac{100}{9} + \dfrac{25}{9}} \\ &= \dfrac{5\sqrt5}{3}. \end{aligned} Thus, the correct answer is B.
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Tagged: coordinate geometry · distance formula · system of equations

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