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Difference between revisions of "2005 AMC 10A Problems/Problem 19"

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Consider the rotated middle square shown in the figure. It will drop until length <math>DE</math> is 1 inch. Then, because <math>DEC</math> is a <math>45^{\circ}-45^{\circ}-90^{\circ}</math> triangle, <math>EC=\frac{\sqrt{2}}{2}</math>, and <math>FC=\frac{1}{2}</math>. We know that <math>BC=\sqrt{2}</math>, so the distance from <math>B</math> to the line is  
 
Consider the rotated middle square shown in the figure. It will drop until length <math>DE</math> is 1 inch. Then, because <math>DEC</math> is a <math>45^{\circ}-45^{\circ}-90^{\circ}</math> triangle, <math>EC=\frac{\sqrt{2}}{2}</math>, and <math>FC=\frac{1}{2}</math>. We know that <math>BC=\sqrt{2}</math>, so the distance from <math>B</math> to the line is  
  
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<math>BC-FC+1=\sqrt{2}-\frac{1}{2}+1=\sqrt{2}+\dfrac{1}{2}</math> <math>(D)</math>.
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<math>BC-FC+1=\sqrt{2}-\frac{1}{2}+1=\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}</math>.
  
 
[[File:AMC10200519Sol.png]]
 
[[File:AMC10200519Sol.png]]
  
  
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==Solution Add-On==
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===Note===
  
 
(Refer to Diagram Above)
 
(Refer to Diagram Above)
  
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After deducing that <math>BC=\sqrt{2}</math>, we can observe that the length from <math>C</math> to the baseline is <math>\frac{1}{2}</math>. This can be obtained by subtracting <math>FC</math> from the side length of the square(s), which is <math>1</math>.
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After deducing that <math>BC=\sqrt{2}</math>, we can compute the length from <math>C</math> to the baseline by subtracting <math>FC</math> (<math>1/2</math>) from the side length of the square(s) (<math>1</math>), giving <math>\frac{1}{2}</math>.
  
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Adding these up, we see that our answer is <math>(D)</math> <math>\sqrt{2}+\dfrac{1}{2}</math>.
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Adding these up, we see that our answer is <math>\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}</math>.
  
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sdk652
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- sdk652
  
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CHECK OUT Video Solution: https://youtu.be/frRxvmsUJQ4
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Also, you can compute the distance from B (the top) to the bottom
  
 
==See Also==
 
==See Also==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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[[Category:Area Ratio Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:58, 17 November 2024

Problem

Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point $B$ from the line on which the bases of the original squares were placed?

[asy] unitsize(1inch); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--((1/3) + 3*(1/2),0)); fill(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle, rgb(.7,.7,.7)); draw(((1/6),0)--((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6),(1/2))--cycle); draw(((1/6) + (1/2),0)--((1/6) + (1/2),(1/2))--((1/6) + 1,(1/2))--((1/6) + 1,0)--cycle); draw(((1/6) + 1,0)--((1/6) + 1,(1/2))--((1/6) + (3/2),(1/2))--((1/6) + (3/2),0)--cycle); draw((2,0)--(2 + (1/3) + (3/2),0)); draw(((2/3) + (3/2),0)--((2/3) + 2,0)--((2/3) + 2,(1/2))--((2/3) + (3/2),(1/2))--cycle); draw(((2/3) + (5/2),0)--((2/3) + (5/2),(1/2))--((2/3) + 3,(1/2))--((2/3) + 3,0)--cycle); label("$B$",((1/6) + (1/2),(1/2)),NW); label("$B$",((2/3) + 2 + (1/4),(29/30)),NNE); draw(((1/6) + (1/2),(1/2)+0.05)..(1,.8)..((2/3) + 2 + (1/4)-.05,(29/30)),EndArrow(HookHead,3)); fill(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle, rgb(.7,.7,.7)); draw(((2/3) + 2 + (1/4),(1/4))--((2/3) + (5/2) + (1/10),(1/2) + (1/9))--((2/3) + 2 + (1/4),(29/30))--((2/3) + 2 - (1/10),(1/2) + (1/9))--cycle);[/asy]

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \sqrt{2}+\frac{1}{2}\qquad\textbf{(E)}\ 2$

Solution

Consider the rotated middle square shown in the figure. It will drop until length $DE$ is 1 inch. Then, because $DEC$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, $EC=\frac{\sqrt{2}}{2}$, and $FC=\frac{1}{2}$. We know that $BC=\sqrt{2}$, so the distance from $B$ to the line is

$BC-FC+1=\sqrt{2}-\frac{1}{2}+1=\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}$.

AMC10200519Sol.png


Note

(Refer to Diagram Above)

After deducing that $BC=\sqrt{2}$, we can compute the length from $C$ to the baseline by subtracting $FC$ ($1/2$) from the side length of the square(s) ($1$), giving $\frac{1}{2}$.

Adding these up, we see that our answer is $\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}$.

- sdk652

Also, you can compute the distance from B (the top) to the bottom

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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

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