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2007 Cyprus MO/Lyceum/Problem 6

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$\displaystyle ABCD$ is a square of side length 2 and $FG$ is an arc of the circle with centre the midpoint $K$ of the side $AB$ and radius 2. The length of the segments $\displaystyle FD=GC=x$ is

$\mathrm{(A) \ } \frac 14\qquad \mathrm{(B) \ } \frac{\sqrt{2}}2\qquad \mathrm{(C) \ } 2-\sqrt{3}\qquad \mathrm{(D) \ } \sqrt{3} - 1\qquad \mathrm{(E) \ } \sqrt{2} - 1$

Solution

($F,\ G$ are on $\overline{AD},\ \overline{ BC}$, respectively)

Draw radii $\overline{KF},\ \overline{KG}$, both with length $2$. $\displaystyle AK = BK = 1$, so we form $30-60-90$ right triangles. $AF = BG = \sqrt{3}$, and so $DF = CG = 2 - \sqrt{3} \Longrightarrow \mathrm{C}$.

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 5
Followed by
Problem 7
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