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
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

m (→‎Solution 3)
 
(6 intermediate revisions by 6 users not shown)
Line 46: Line 46:
 
</asy>
 
</asy>
  
−
<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>\textbf{(A)}\ \frac{4\sqrt{5}}{3} \qquad\textbf{(B)}\ \frac{5\sqrt{5}}{3} \qquad\textbf{(C)}\ \frac{12\sqrt{5}}{7} \qquad\textbf{(D)}\ 2\sqrt{5} \qquad\textbf{(E)}\ \frac{5\sqrt{65}}{9}</math>
  
−
==Solution==
 
  
−
===Solution 1===
+
==Video Solution==
 +
https://youtu.be/oWxqYyW926I
 +
 
 +
==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>.  
  
−
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>
+
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>.
 
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>.
Line 69: Line 71:
 
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 124: Line 126:
  
 
This is answer choice <math>\boxed{\text{B}}</math>
 
This is answer choice <math>\boxed{\text{B}}</math>
−
 
−
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 176: Line 176:
  
 
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>
 
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>
 +
 +
== Solution 4 ==
 +
 +
<asy>
 +
// Coordinates
 +
pair A = (0,3), B = (6,0), C = (4,2), D = (2,0);
 +
path seg1 = B--A;
 +
path seg2 = D--C;
 +
pair[] intersectionPoints = intersectionpoints(seg1, seg2);
 +
pair E = intersectionPoints[0];
 +
 +
for (int i = 0; i <= 6; i = i + 1) {
 +
    for (int j = 0; j <= 3; j = j + 1) {
 +
        dot((i,j));
 +
    }
 +
}
 +
 +
// Draw
 +
draw(seg1);
 +
draw(seg2);
 +
 +
dot(E);
 +
 +
// Label
 +
label("$A$", A, NW);
 +
label("$B$", B, SE);
 +
label("$C$", C, dir(0));
 +
label("$D$", D, S);
 +
label("$E$", E, N);
 +
 +
// Add extras
 +
draw(C--(5,3), dashed);
 +
draw(D--B, dashed);
 +
draw((5,3)--A, dashed);
 +
</asy>
 +
 +
Extend line <math>\overline{DC}</math> as above.  This creates two similar triangles whose side lengths have the ratio <math>5:4</math>.  Therefore <math>AE=\frac{5}{9}AB</math>.  Using Pythagorean theorem to find <math>AB</math> gives us:
 +
 +
<cmath>AE=\frac{5}{9}AB=\frac{5}{9}\sqrt{3^2+6^2}=\frac{5}{9}\sqrt{45}=
 +
\boxed{\textbf{(B) }\frac{5\sqrt{5}}{3}}</cmath>
 +
 +
~ proloto
 +
 +
==Video Solution by Daily Dose of Math==
 +
 +
https://youtu.be/L1YNOIO-ZgY?si=oJWau9Y0J2p1mXnM
 +
 +
~Thesmartgreekmathdude
  
 
==See Also==
 
==See Also==
Line 181: Line 229:
 
{{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]]

Latest revision as of 08:38, 1 May 2025

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]

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


Video Solution

https://youtu.be/oWxqYyW926I

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 $\boxed{\text{B}}$

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 $AE = AF + FE = \sqrt{5} + \frac{2\sqrt{5}}{3} = \frac{5\sqrt{5}}{3} \Rightarrow \boxed{\text{B}}$

Solution 4

[asy] // Coordinates pair A = (0,3), B = (6,0), C = (4,2), D = (2,0); path seg1 = B--A; path seg2 = D--C; pair[] intersectionPoints = intersectionpoints(seg1, seg2); pair E = intersectionPoints[0]; for (int i = 0; i <= 6; i = i + 1) { for (int j = 0; j <= 3; j = j + 1) { dot((i,j)); } } // Draw draw(seg1); draw(seg2); dot(E); // Label label("$A$", A, NW); label("$B$", B, SE); label("$C$", C, dir(0)); label("$D$", D, S); label("$E$", E, N); // Add extras draw(C--(5,3), dashed); draw(D--B, dashed); draw((5,3)--A, dashed); [/asy]

Extend line $\overline{DC}$ as above. This creates two similar triangles whose side lengths have the ratio $5:4$. Therefore $AE=\frac{5}{9}AB$. Using Pythagorean theorem to find $AB$ gives us:

\[AE=\frac{5}{9}AB=\frac{5}{9}\sqrt{3^2+6^2}=\frac{5}{9}\sqrt{45}= \boxed{\textbf{(B) }\frac{5\sqrt{5}}{3}}\]

~ proloto

Video Solution by Daily Dose of Math

https://youtu.be/L1YNOIO-ZgY?si=oJWau9Y0J2p1mXnM

~Thesmartgreekmathdude

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_10_Problems/Problem_16&oldid=247735"