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

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(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>.
+
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)from the side length of the square(s) (</math>1<math>), giving </math>\frac{1}{2}<math>.
  
−
Adding these up, we see that our answer is <math>\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}</math>.
+
Adding these up, we see that our answer is </math>\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}$.
  
 
- sdk652
 
- sdk652
 +
 +
Also, you can compute the distance from the top of the
  
 
==See Also==
 
==See Also==

Revision as of 00:13, 9 July 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$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}$.

- sdk652

Also, you can compute the distance from the top of the

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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