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Difference between revisions of "2005 AMC 10A Problems/Problem 19"
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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>.
 
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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[[File:AMC10200519Sol.png]]

Revision as of 21:29, 30 January 2011

(D) Consider the rotated middle square shown in the figure. It will drop until length $DE$ is 1 inch. Thus $FC=DF=FE=\dfrac{1}{2}$ and $BC=\sqrt{2}$. Hence, $BF=BC-FC=\sqrt{2}-\dfrac{1}{2}$.

Then you have to add the 1 from the height of the other two squares, so you get $1+BF=\sqrt{2}+\dfrac{1}{2}$.

AMC10200519Sol.png

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