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> | + | <math>BC-FC+1=\sqrt{2}-\frac{1}{2}+1=\sqrt{2}+\dfrac{1}{2}</math> <math>(D)</math>. |
[[File:AMC10200519Sol.png]] | [[File:AMC10200519Sol.png]] |
Revision as of 11:53, 26 November 2015
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
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See Also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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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