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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{...

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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