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

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

Revision as of 20:28, 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 $1+BF=\sqrt{2}+\dfrac{1}{2}$.

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