Difference between revisions of "2000 AMC 10 Problems/Problem 16"
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==Solution== | ==Solution== | ||
+ | |||
+ | ===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>. | ||
+ | |||
+ | The line <math>l_1</math> is given by the equation <math>y=m_1x+b_1</math>. The <math>y</math>-intercept is <math>A=(0,3)</math>, so <math>b_1=3</math>. We are given two points on <math>l_1</math>, hence we can compute the slope, <math>m_1</math> to be <math>\frac{0-3}{6-0}=\frac{-1}{2}</math>, so <math>l_1</math> is the line <math>y=\frac{-1}{2}x+3</math> | ||
+ | |||
+ | Similarly, <math>l_2</math> is given by <math>y=m_2x+b_2</math>. The slope in this case is <math>\frac{2-0}{4-2}=1</math>, so <math>y=x+b_2</math>. Plugging in the point <math>(2,0)</math> gives us <math>b_2=-2</math>, so <math>l_2</math> is the line <math>y=x-2</math>. | ||
+ | |||
+ | At <math>E</math>, the intersection point, both of the equations must be true, so | ||
+ | <cmath>\begin{align*} | ||
+ | y=x-2, y=\frac{-1}{2}x+3 &\Rightarrow x-2=\frac{-1}{2}x+3 \\ | ||
+ | &\Rightarrow x=\frac{10}{3} \\ | ||
+ | &\Rightarrow y=\frac{4}{3} \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | We have the coordinates of <math>A</math> and <math>E</math>, so we can use the distance formula here: <cmath>\sqrt{\left(\frac{10}{3}-0\right)^2+\left(\frac{4}{3}-3\right)^2}=\frac{5\sqrt{5}}{3}</cmath> | ||
+ | |||
+ | which is answer choice <math>\boxed{\text{B}}</math> | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=15|num-a=17}} | {{AMC10 box|year=2000|num-b=15|num-a=17}} |
Revision as of 02:12, 11 January 2009
Contents
Problem
The diagram shows lattice points, each one unit from its nearest neighbors. Segment
meets segment
at
. Find the length of segment
.
Solution
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
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 |