# Difference between revisions of "2010 AMC 10B Problems/Problem 16"

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

− | Putting it together, we get the answer to be <math>4 | + | Putting it together, we get the answer to be <math>4( \frac{\pi}{18}-\frac{\sqrt{3}{12} )=</math> |

− | <cmath>\frac{2\pi}{9}-\frac{\sqrt{3}}{3} (B)</cmath> | + | <cmath>\boxed{\frac{2\pi}{9}-\frac{\sqrt{3}}{3} (B)}</cmath> |

== See also == | == See also == | ||

[[Area of an equilateral triangle]] | [[Area of an equilateral triangle]] |

## Revision as of 17:28, 12 June 2011

The radius of circle is

Half the diagonal of the square is . 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 . Power of a point states that if two chords (AB and CD) intersect at X, then AXBXCXDX. Applying power of a point to this situation, . (We know the center of a square to be halfway in each direction, if you know what I mean by direction).

Solving, . The significance? This means the chord is equal to the radius of a circle, so the sector has angle . Since this is a sixth of the circle, the sector has area .

Now we turn to the triangle.

Since it is equilateral, we know it has area equal to times the square of its sidelength. Therefore, our triangle has area .

Putting it together, we get the answer to be $4( \frac{\pi}{18}-\frac{\sqrt{3}{12} )=$ (Error compiling LaTeX. ! File ended while scanning use of \frac .)