2000 AMC 10 Problem 16

Below is the professionally curated solution for Problem 16 of the 2000 AMC 10, 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.

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Concepts:coordinate geometry distance formula system of equations

Difficulty rating: 1690

16.

The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E.$ Find the length of segment $AE.$

$\dfrac{4\sqrt5}{3}$

$\dfrac{5\sqrt5}{3}$

$\dfrac{12\sqrt5}{7}$

$2\sqrt5$

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

Solution:

Place the points at $A = (0, 3),$ $B = (6, 0),$ $C = (4, 2),$ $D = (2, 0).$

Line $AB$ is $x + 2y = 6$ and line $CD$ is $x - y = 2.$ Solving simultaneously gives $E = \left(\dfrac{10}{3}, \dfrac{4}{3}\right).$

Then $AE = \sqrt{\left(\dfrac{10}{3}\right)^2 + \left(\dfrac{4}{3} - 3\right)^2} = \sqrt{\dfrac{100}{9} + \dfrac{25}{9}} = \dfrac{5\sqrt5}{3}.$

Thus, the correct answer is B.

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