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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 <math>45^{\circ}</math>, 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>
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unitsize(1inch);
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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);
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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);
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draw((2,0)--(2 + (1/3) + (3/2),0));
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draw(((2/3) + (3/2),0)--((2/3) + 2,0)--((2/3) + 2,(1/2))--((2/3) + (3/2),(1/2))--cycle);
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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));
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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>
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==Solution==
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The rotated middle square is lowered until it touches both the adjoining squares, so since the horizontal distance between those squares is <math>1</math> inch, the middle square will stop being lowered once the length <math>DE</math> in the diagram below is <math>1</math> inch. Now, since <math>DC</math> and <math>EC</math> respectively pass through the vertices <math>D</math> and <math>E</math> of the horizontal middle square, and intersect at right angles, they must be the diagonals of the horizontal middle square, so <math>\triangle DEC</math> is a <math>45^{\circ}-45^{\circ}-90^{\circ}</math> triangle.
  
 
[[File:AMC10200519Sol.png]]
 
[[File:AMC10200519Sol.png]]
 +
 +
It follows that when <math>DE = 1</math>, we have <math>DC = EC = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}</math>, and as the middle square was rotated by exactly <math>45^{\circ}</math> from its original horizontal position, the diagonal <math>BC</math> is now vertical, giving <math>BC \perp DE</math>. This means <math>\triangle DFC</math> and <math>\triangle EFC</math> are also <math>45^{\circ}-45^{\circ}-90^{\circ}</math> triangles, and hence <math>FC = \frac{\left(\frac{\sqrt{2}}{2}\right)}{\sqrt{2}} = \frac{1}{2}</math>.
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Lastly, since <math>BC</math> is the diagonal of a unit square, its length is <math>\sqrt{1^2+1^2} = \sqrt{2}</math> (by Pythagoras' theorem), so we deduce that the distance from <math>B</math> to the bottom horizontal line is
 +
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<cmath>BC-FC+1 = \sqrt{2}-\frac{1}{2}+1=\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}.</cmath>
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(An alternative for this last step is to compute the distance from <math>C</math> to the bottom horizontal line as <math>1-FC = 1-\frac{1}{2} = \frac{1}{2}</math>, so that the distance from <math>B</math> to the bottom horizontal line is, once again, <math>BC+\frac{1}{2} = \boxed{\textbf{(D) } \sqrt{2}+\frac{1}{2}}</math>.)
 +
 +
==See Also==
 +
 +
{{AMC10 box|year=2005|ab=A|num-b=18|num-a=20}}
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 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 01:43, 2 July 2025

Problem

Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated $45^{\circ}$, 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

The rotated middle square is lowered until it touches both the adjoining squares, so since the horizontal distance between those squares is $1$ inch, the middle square will stop being lowered once the length $DE$ in the diagram below is $1$ inch. Now, since $DC$ and $EC$ respectively pass through the vertices $D$ and $E$ of the horizontal middle square, and intersect at right angles, they must be the diagonals of the horizontal middle square, so $\triangle DEC$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle.

AMC10200519Sol.png

It follows that when $DE = 1$, we have $DC = EC = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$, and as the middle square was rotated by exactly $45^{\circ}$ from its original horizontal position, the diagonal $BC$ is now vertical, giving $BC \perp DE$. This means $\triangle DFC$ and $\triangle EFC$ are also $45^{\circ}-45^{\circ}-90^{\circ}$ triangles, and hence $FC = \frac{\left(\frac{\sqrt{2}}{2}\right)}{\sqrt{2}} = \frac{1}{2}$.

Lastly, since $BC$ is the diagonal of a unit square, its length is $\sqrt{1^2+1^2} = \sqrt{2}$ (by Pythagoras' theorem), so we deduce that the distance from $B$ to the bottom horizontal line is

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

(An alternative for this last step is to compute the distance from $C$ to the bottom horizontal line as $1-FC = 1-\frac{1}{2} = \frac{1}{2}$, so that the distance from $B$ to the bottom horizontal line is, once again, $BC+\frac{1}{2} = \boxed{\textbf{(D) } \sqrt{2}+\frac{1}{2}}$.)

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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