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>
  
==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
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<cmath>\begin{align*}
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y=x-2, y=\frac{-1}{2}x+3 &\Rightarrow x-2=\frac{-1}{2}x+3 \\
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&\Rightarrow x=\frac{10}{3} \\
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&\Rightarrow y=\frac{4}{3} \\
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\end{align*}</cmath>
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 +
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>
 +
 
 +
==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));
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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));
 +
dot((6,2));
 +
fill(circle((0,3),0.1),black);
 +
dot((1,3));
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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));
 +
</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>
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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.
 +
 
 +
==Solution 3==
 +
 
 +
<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);
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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("$F$",(2,2),NE);
 +
draw((2,2)--(4,2));
 +
draw((6,0)--(2,0));
 +
</asy>
 +
 
 +
Drawing line <math>\overline{BD}</math> and parallel line <math>\overline{CF}</math>, we see that <math>\triangle FCE \sim \triangle BDE</math> by AA similarity. Thus <math>\frac{FE}{EB} = \frac{FC}{DB} = \frac{2}{4} = \frac{1}{2}</math>. Reciprocating, we know that <math>\frac{EB}{FE} = 2</math> so <math>\frac{EB+FE}{FE} = 2+1 \Rightarrow \frac{FB}{FE} = 3</math>. Reciprocating again, we have <math>\frac{FE}{FB} = \frac{1}{3} \Rightarrow FE = \frac{1}{3}FB</math>. We know that <math>FD = 2</math>, so by the Pythagorean Theorem, <math>FB = \sqrt{2^{2} + 4^{2}} = 2\sqrt{5}</math>. Thus <math>FE = \frac{1}{3}FB = \frac{2\sqrt{5}}{3}</math>. Applying the Pythagorean Theorem again, we have <math>AF = \sqrt{1^{2}+2^{2}} = \sqrt{5}</math>. We finally have <math>AE = AF + FE = \sqrt{5} + \frac{2\sqrt{5}}{3} = \frac{5\sqrt{5}}{3} \Rightarrow \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}}
 +
{{MAA Notice}}
 +
 +
[[Category:Introductory Geometry Problems]]

Revision as of 15:20, 19 April 2021

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 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 $\boxed{\text{B}}$

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] 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("$F$",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]

Drawing line $\overline{BD}$ and parallel line $\overline{CF}$, we see that $\triangle FCE \sim \triangle BDE$ by AA similarity. Thus $\frac{FE}{EB} = \frac{FC}{DB} = \frac{2}{4} = \frac{1}{2}$. Reciprocating, we know that $\frac{EB}{FE} = 2$ so $\frac{EB+FE}{FE} = 2+1 \Rightarrow \frac{FB}{FE} = 3$. Reciprocating again, we have $\frac{FE}{FB} = \frac{1}{3} \Rightarrow FE = \frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \sqrt{2^{2} + 4^{2}} = 2\sqrt{5}$. Thus $FE = \frac{1}{3}FB = \frac{2\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \sqrt{1^{2}+2^{2}} = \sqrt{5}$. We finally have $AE = AF + FE = \sqrt{5} + \frac{2\sqrt{5}}{3} = \frac{5\sqrt{5}}{3} \Rightarrow \boxed{\text{B}}$

See Also

2000 AMC 10 (ProblemsAnswer KeyResources)
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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