2007 Cyprus MO/Lyceum/Problem 6

$\displaystyle ABCD$ 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 $DF = CG = 2 - \sqrt{3} \Longrightarrow \mathrm{C}$ .

2007 Cyprus MO, Lyceum (Problems)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Page

Toolbox

Search

2007 Cyprus MO/Lyceum/Problem 6

Solution

See also