Difference between revisions of "2000 AMC 10 Problems/Problem 16"
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Latest revision as of 15:20, 19 April 2021
Problem
The diagram shows lattice points, each one unit from its nearest neighbors. Segment meets segment at . Find the length of segment .
Solution 1
Let be the line containing and and let be the line containing and . If we set the bottom left point at , then , , , and .
The line is given by the equation . The -intercept is , so . We are given two points on , hence we can compute the slope, to be , so is the line
Similarly, is given by . The slope in this case is , so . Plugging in the point gives us , so is the line .
At , the intersection point, both of the equations must be true, so
We have the coordinates of and , so we can use the distance formula here:
which is answer choice
Solution 2
Draw the perpendiculars from and to , respectively. As it turns out, . Let be the point on for which .
, and , so by AA similarity,
By the Pythagorean Theorem, we have , , and . Let , so , then
This is answer choice
Also, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B.
Solution 3
Drawing line and parallel line , we see that by AA similarity. Thus . Reciprocating, we know that so . Reciprocating again, we have . We know that , so by the Pythagorean Theorem, . Thus . Applying the Pythagorean Theorem again, we have . We finally have
See Also
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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