Difference between revisions of "2000 AMC 10 Problems/Problem 16"
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<math>\mathrm{(A)}\ \frac{4\sqrt{5}}{3} \qquad\mathrm{(B)}\ \frac{5\sqrt{5}}{3} \qquad\mathrm{(C)}\ \frac{12\sqrt{5}}{7} \qquad\mathrm{(D)}\ 2\sqrt{5} \qquad\mathrm{(E)}\ \frac{5\sqrt{65}}{9}</math> | <math>\mathrm{(A)}\ \frac{4\sqrt{5}}{3} \qquad\mathrm{(B)}\ \frac{5\sqrt{5}}{3} \qquad\mathrm{(C)}\ \frac{12\sqrt{5}}{7} \qquad\mathrm{(D)}\ 2\sqrt{5} \qquad\mathrm{(E)}\ \frac{5\sqrt{65}}{9}</math> | ||
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+ | ==Solution 1== | ||
Let <math>l_1</math> be the line containing <math>A</math> and <math>B</math> and let <math>l_2</math> be the line containing <math>C</math> and <math>D</math>. If we set the bottom left point at <math>(0,0)</math>, then <math>A=(0,3)</math>, <math>B=(6,0)</math>, <math>C=(4,2)</math>, and <math>D=(2,0)</math>. | Let <math>l_1</math> be the line containing <math>A</math> and <math>B</math> and let <math>l_2</math> be the line containing <math>C</math> and <math>D</math>. If we set the bottom left point at <math>(0,0)</math>, then <math>A=(0,3)</math>, <math>B=(6,0)</math>, <math>C=(4,2)</math>, and <math>D=(2,0)</math>. | ||
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which is answer choice <math>\boxed{\text{B}}</math> | which is answer choice <math>\boxed{\text{B}}</math> | ||
− | + | ==Solution 2== | |
<asy> | <asy> | ||
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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. | 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== | |
<asy> | <asy> |
Latest revision as of 17:23, 19 October 2020
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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