Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 AMC 10 Problems/Problem 16"

(→‎Problem)
(→‎Solution)
Line 49: Line 49:
  
 
==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

Problem

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 $AE$.

[asy] path seg1, seg2; seg1=(6,0)--(0,3); seg2=(2,0)--(4,2); dot((0,0)); dot((1,0)); fill(circle((2,0),0.1),black); dot((3,0)); dot((4,0)); dot((5,0)); fill(circle((6,0),0.1),black); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); fill(circle((4,2),0.1),black); dot((5,2)); dot((6,2)); fill(circle((0,3),0.1),black); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); draw(seg1); draw(seg2); pair [] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label("$A$",(0,3),NW); label("$B$",(6,0),SE); label("$C$",(4,2),NE); label("$D$",(2,0),S); label("$E$",x[0],N); [/asy]

$\mathrm{(A)}\ \frac{4\sqrt{5}}{3} \qquad\mathrm{(B)}\ \frac{5\sqrt{5}}{3} \qquad\mathrm{(C)}\ \frac{12\sqrt{5}}{7} \qquad\mathrm{(D)}\ 2\sqrt{5} \qquad\mathrm{(E)}\ \frac{5\sqrt{65}}{9}$

Solution

Solution 1

Let $l_1$ be the line containing $A$ and $B$ and let $l_2$ be the line containing $C$ and $D$. If we set the bottom left point at $(0,0)$, then $A=(0,3)$, $B=(6,0)$, $C=(4,2)$, and $D=(2,0)$.

The line $l_1$ is given by the equation $y=m_1x+b_1$. The $y$-intercept is $A=(0,3)$, so $b_1=3$. We are given two points on $l_1$, hence we can compute the slope, $m_1$ to be $\frac{0-3}{6-0}=\frac{-1}{2}$, so $l_1$ is the line $y=\frac{-1}{2}x+3$

Similarly, $l_2$ is given by $y=m_2x+b_2$. The slope in this case is $\frac{2-0}{4-2}=1$, so $y=x+b_2$. Plugging in the point $(2,0)$ gives us $b_2=-2$, so $l_2$ is the line $y=x-2$.

At $E$, the intersection point, both of the equations must be true, so \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*}

We have the coordinates of $A$ and $E$, so we can use the distance formula here: \[\sqrt{\left(\frac{10}{3}-0\right)^2+\left(\frac{4}{3}-3\right)^2}=\frac{5\sqrt{5}}{3}\]

which is answer choice $\boxed{\text{B}}$

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_10_Problems/Problem_16&oldid=29444"