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

(Added solution 3)
(See Also)
(4 intermediate revisions by 3 users not shown)
Line 48: Line 48:
 
<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===
+
 
 +
==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>.  
 
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>.  
Line 69: Line 69:
 
which is answer choice <math>\boxed{\text{B}}</math>
 
which is answer choice <math>\boxed{\text{B}}</math>
  
===Solution 2===
+
==Solution 2==
  
 
<asy>
 
<asy>
Line 127: Line 127:
 
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===
+
==Solution 3==
  
 
<asy>
 
<asy>
Line 171: Line 171:
 
label("$E$",x[0],N);
 
label("$E$",x[0],N);
 
label("$F$",(2,2),NE);
 
label("$F$",(2,2),NE);
draw((6,2)--(4,2));
+
draw((2,2)--(4,2));
draw((6,0)--(2,.0));
+
draw((6,0)--(2,0));
 
</asy>
 
</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>
+
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==
Line 181: Line 181:
 
{{AMC10 box|year=2000|num-b=15|num-a=17}}
 
{{AMC10 box|year=2000|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png