2005 AMC 10A Problems/Problem 19

(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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2005 AMC 10A Problems/Problem 19