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Difference between revisions of "2005 AMC 10A Problems/Problem 19"
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'''(D)''' Consider the rotated middle square shown in the figure. It will drop until length <math>DE</math> is 1 inch. Thus <math>FC=DF=FE=\dfrac{1}{2}</math> and <math>BC=\sqrt{2}</math>. Hence, <math>BF=BC-FC=\sqrt{2}-\dfrac{1}{2}</math>.
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== Problem ==
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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 <math>B</math> from the line on which the bases of the original squares were placed?
  
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Then you have to add the 1 from the height of the other two squares, so you get <math>1+BF=\sqrt{2}+\dfrac{1}{2}</math>.
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<asy>
 +
unitsize(1inch);
 +
defaultpen(linewidth(.8pt)+fontsize(8pt));
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draw((0,0)--((1/3) + 3*(1/2),0));
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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);
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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);
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label("$B$",((1/6) + (1/2),(1/2)),NW);
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label("$B$",((2/3) + 2 + (1/4),(29/30)),NNE);
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draw(((1/6) + (1/2),(1/2)+0.05)..(1,.8)..((2/3) + 2 + (1/4)-.05,(29/30)),EndArrow(HookHead,3));
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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>
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 +
<math> \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 </math>
 +
 
 +
==Solution==
 +
 
 +
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
 +
 
 +
<math>BC-FC+1=\sqrt{2}-\frac{1}{2}+1=\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}</math>.
 +
 
 +
[[File:AMC10200519Sol.png]]
 +
 
 +
 
 +
===Note===
 +
 
 +
(Refer to Diagram Above)
 +
 
 +
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>.
 +
 
 +
Adding these up, we see that our answer is <math>\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}</math>.
 +
 
 +
- sdk652
 +
 
 +
==See Also==
 +
 
 +
{{AMC10 box|year=2005|ab=A|num-b=18|num-a=20}}
 +
 
 +
[[Category:Introductory Geometry Problems]]
 +
[[Category:Area Ratio Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:46, 14 December 2021

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 observe that the length from $C$ to the baseline is $\frac{1}{2}$. This can be obtained by subtracting $FC$ from the side length of the square(s), which is $1$.

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

- sdk652

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

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