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

Difference between revisions of "2005 AMC 10A Problems/Problem 19"
(Improved formatting and explanation)
 
(17 intermediate revisions by 12 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== 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 <math>B</math> from the line on which the bases of the original squares were placed?
+
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?
  
 
<asy>
 
<asy>
Line 23: Line 23:
 
==Solution==
 
==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
+
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.
  
−
<math>BC-FC+1=\sqrt{2}-\frac{1}{2}+1=\sqrt{2}+\dfrac{1}{2}</math>
+
[[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>.
 +
 
 +
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
 +
 
 +
<cmath>BC-FC+1 = \sqrt{2}-\frac{1}{2}+1=\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}.</cmath>
 +
 
 +
(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==
 
==See Also==
−
*[[2005 AMC 10A Problems]]
 
  
−
*[[2005 AMC 10A Problems/Problem 18|Previous Problem]]
+
{{AMC10 box|year=2005|ab=A|num-b=18|num-a=20}}
  
−
*[[2005 AMC 10A Problems/Problem 20|Next Problem]]
+
[[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

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=2005_AMC_10A_Problems/Problem_19&oldid=253108"