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

m (→‎Solution 2)
(Added solution 3)
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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.
 
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((6,2)--(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 by AA similarity. Thus </math>\frac{FE}{EB} = \frac{FC}{DB} = \frac{2}{4} = \frac{1}{2}<math></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 = </math>\sqrt{5} + \frac{2\sqrt{5}}{3}<math> = \frac{5\sqrt{5}}{3} \Rightarrow \boxed{\text{B}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 17:35, 29 October 2017

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 $\boxed{\text{B}}$

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((6,2)--(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}$$ (Error compiling LaTeX. Unknown error_msg). 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 (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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