Difference between revisions of "2000 AMC 10 Problems/Problem 16"
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− | <math>\ | + | <math>\textbf{(A)}\ \frac{4\sqrt{5}}{3} \qquad\textbf{(B)}\ \frac{5\sqrt{5}}{3} \qquad\textbf{(C)}\ \frac{12\sqrt{5}}{7} \qquad\textbf{(D)}\ 2\sqrt{5} \qquad\textbf{(E)}\ \frac{5\sqrt{65}}{9}</math> |
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==Solution 1== | ==Solution 1== |
Revision as of 10:18, 8 November 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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