2000 AMC 10 Problem 16

Attempt Problem 16 of the 2000 AMC 10 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 10 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

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.

453\dfrac{4\sqrt5}{3}

553\dfrac{5\sqrt5}{3}

1257\dfrac{12\sqrt5}{7}

252\sqrt5

5659\dfrac{5\sqrt{65}}{9}

Answer: B
Concepts:coordinate geometrydistance formulasystem of equations
Difficulty rating: 1690
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 xy=2.x - y = 2. Solving simultaneously gives E=(103,43).E = \left(\dfrac{10}{3}, \dfrac{4}{3}\right).

Then AE=(103)2+(433)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.

← Problem 15#15
Full Exam

Problem 16 in Other Years