Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 6"

Revision as of 19:41, 6 May 2007

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

$\displaystyle ABCD$ is a square of side length 2 and $FG$ is an arc of the circle with centre the midpoint $K$ of the side $\overline{AB}$ 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 $DF = CG = 2 - \sqrt{3} \Longrightarrow \mathrm{C}$ .

@@ Line 8: / Line 8: @@
 BC}</math>, respectively)
-Draw radii <math>\overline{KF},\ \overline{KG}</math>, both with length <math>2</math>. <math>\displaystyle AK = BK = 1</math>, so we form <math>30-60-90</math> [[right triangle]]s. <math>AF = BG = \sqrt{3}</math>, and so <math>DF = CG = 2 - \sqrt{3} \Rightarrow \mathrm{C}</math>.
+Draw radii <math>\overline{KF},\ \overline{KG}</math>, both with length <math>2</math>. <math>\displaystyle AK = BK = 1</math>, so we form <math>30-60-90</math> [[right triangle]]s. <math>AF = BG = \sqrt{3}</math>, and so <math>DF = CG = 2 - \sqrt{3} \Longrightarrow \mathrm{C}</math>.
 ==See also==

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

Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 6"

Revision as of 19:41, 6 May 2007

Solution

See also