2010 AMC 10B Problems/Problem 16

The radius of circle is $\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}$

Half the diagonal of the square is $\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2}$ . We can see that the circle passes outside the square, but the square is NOT completely contained in the circle

Therefore the picture will look something like this:

Then we proceed to find: 4 * (area of the sector marked off by the two radii drawn - area of the triangle with side on the square and the two radii).

To find this, we do the following:

First we realize that the radius perpendicular to the side of the square between the extra lines marking off the sector splits the chord in half. Let this half-length be $a$ . Power of a point states that if two chords (AB and CD) intersect at X, then AX $\cdot$ BX $=$ CX $\cdot$ DX. Applying power of a point to this situation, $a^2=(\frac{\sqrt{3}}{3}-\frac{1}{2})(\frac{\sqrt{3}}{3}+\frac{1}{2})$ . (We know the center of a square to be halfway in each direction, if you know what I mean by direction).

Solving, $a= \frac{\sqrt{3}}{6}$ . The significance? This means the chord is equal to the radius of a circle, so the sector has angle $60^{\circ}$ . Since this is a sixth of the circle, the sector has area $\frac{\pi}{6}\cdot \frac{\sqrt{3}}{3}^2=\frac{\pi}{18}$ .

Now we turn to the triangle.

Since it is equilateral, we know it has area equal to $\frac{\sqrt{3}}{4}$ times the square of its sidelength. Therefore, our triangle has area $\frac{\frac{\sqrt{3}}{3}^2\sqrt{3}}{4}=\frac{\sqrt{3}}{12}$ .

Putting it together, we get the answer to be $4( \frac{\pi}{18}-\frac{\sqrt{3}{12} )=$ (Error compiling LaTeX. Unknown error_msg)

$\boxed{\frac{2\pi}{9}-\frac{\sqrt{3}}{3} (B)}$

Page

Toolbox

Search

2010 AMC 10B Problems/Problem 16

See also