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

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(→‎Solution)
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which is answer choice <math>\boxed{\text{B}}</math>
 
which is answer choice <math>\boxed{\text{B}}</math>
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 +
===Solution 2===
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 +
<asy>
 +
path seg1, seg2;
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seg1=(6,0)--(0,3);
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seg2=(2,0)--(4,2);
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dot((0,0));
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dot((1,0));
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fill(circle((2,0),0.1),black);
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dot((3,0));
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dot((4,0));
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dot((5,0));
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fill(circle((6,0),0.1),black);
 +
dot((0,1));
 +
dot((1,1));
 +
dot((2,1));
 +
dot((3,1));
 +
dot((4,1));
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dot((5,1));
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dot((6,1));
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dot((0,2));
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dot((1,2));
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dot((2,2));
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dot((3,2));
 +
fill(circle((4,2),0.1),black);
 +
dot((5,2));
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dot((6,2));
 +
fill(circle((0,3),0.1),black);
 +
dot((1,3));
 +
dot((2,3));
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dot((3,3));
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dot((4,3));
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dot((5,3));
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dot((6,3));
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draw(seg1);
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draw(seg2);
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pair [] x=intersectionpoints(seg1,seg2);
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fill(circle(x[0],0.1),black);
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label("$A$",(0,3),NW);
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label("$B$",(6,0),SE);
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label("$C$",(4,2),NE);
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label("$D$",(2,0),S);
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label("$E$",x[0],N);
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label("$F$",(2.5,.5),E);
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draw((6,0)--(4,2));
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draw((0,3)--(2.5,.5));
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</asy>
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 +
Draw the perpendiculars from <math>A</math> and <math>B</math> to <math>CD</math>, respectively. As it turns out, <math>BC \perp CD</math>. Let <math>F</math> be the point on <math>CD</math> for which <math>AF\perp CD</math>.
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 +
<math>m\angle AFE=m\angle BCE=90^\circ</math>, and <math>m\angle AEF=m\angle CEB</math>, so by AA similarity, <cmath>\triangle AFE\sim \triangle BCE \Rightarrow \frac{AF}{AE}=\frac{BC}{BE}</cmath>
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 +
By the Pythagorean Theorem, we have <math>AB=\sqrt{3^2+6^2}=3\sqrt{5}</math>, <math>AF=\sqrt{2.5^2+2.5^2}=2.5\sqrt{2}</math>, and <math>BC=\sqrt{2^2+2^2}+2\sqrt{2}</math>. Let <math>AE=x</math>, so <math>BE=3\sqrt{5}-x</math>, then <cmath>\frac{2.5\sqrt{2}}{x}=\frac{2\sqrt{2}}{3\sqrt{5}-x}</cmath> <cmath>x=\frac{5\sqrt{5}}{3}</cmath>
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 +
This 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 07:33, 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 $\boxed{\text{B}}$

Solution 2

[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); label("$F$",(2.5,.5),E); draw((6,0)--(4,2)); draw((0,3)--(2.5,.5)); [/asy]

Draw the perpendiculars from $A$ and $B$ to $CD$, respectively. As it turns out, $BC \perp CD$. Let $F$ be the point on $CD$ for which $AF\perp CD$.

$m\angle AFE=m\angle BCE=90^\circ$, and $m\angle AEF=m\angle CEB$, so by AA similarity, \[\triangle AFE\sim \triangle BCE \Rightarrow \frac{AF}{AE}=\frac{BC}{BE}\]

By the Pythagorean Theorem, we have $AB=\sqrt{3^2+6^2}=3\sqrt{5}$, $AF=\sqrt{2.5^2+2.5^2}=2.5\sqrt{2}$, and $BC=\sqrt{2^2+2^2}+2\sqrt{2}$. Let $AE=x$, so $BE=3\sqrt{5}-x$, then \[\frac{2.5\sqrt{2}}{x}=\frac{2\sqrt{2}}{3\sqrt{5}-x}\] \[x=\frac{5\sqrt{5}}{3}\]

This 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
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