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 | ||
− | <math>BC-FC+1=\sqrt{2}-\frac{1}{2}+1=\sqrt{2}+\dfrac{1}{2} | + | <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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(Refer to Diagram Above) | (Refer to Diagram Above) | ||
− | After deducing that <math>BC=\sqrt{2}</math>, we can | + | 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>. |
− | Adding these up, we see that our answer is <math>(D) | + | Adding these up, we see that our answer is <math>\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}</math>. |
− | sdk652 | + | - sdk652 |
+ | |||
+ | Also, you can compute the distance from the top of the | ||
==See Also== | ==See Also== |
Latest revision as of 01:14, 9 July 2024
Contents
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 from the line on which the bases of the original squares were placed?
Solution
Consider the rotated middle square shown in the figure. It will drop until length is 1 inch. Then, because
is a
triangle,
, and
. We know that
, so the distance from
to the line is
.
Note
(Refer to Diagram Above)
After deducing that , we can compute the length from
to the baseline by subtracting
(
) from the side length of the square(s) (
), giving
.
Adding these up, we see that our answer is .
- 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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