Difference between revisions of "2010 AIME II Problems/Problem 9"
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== Problem == | == Problem == | ||
− | Let <math>ABCDEF</math> be a [[regular polygon|regular]] [[hexagon]]. Let <math>G</math>, <math>H</math>, <math>I</math>, <math>J</math>, <math>K</math>, and <math>L</math> be the [[midpoint]]s of sides <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EF</math>, and <math>AF</math>, respectively. The [[segment]]s <math>\ | + | Let <math>ABCDEF</math> be a [[regular polygon|regular]] [[hexagon]]. Let <math>G</math>, <math>H</math>, <math>I</math>, <math>J</math>, <math>K</math>, and <math>L</math> be the [[midpoint]]s of sides <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EF</math>, and <math>AF</math>, respectively. The [[segment]]s <math>\overline{AH}</math>, <math>\overline{BI}</math>, <math>\overline{CJ}</math>, <math>\overline{DK}</math>, <math>\overline{EL}</math>, and <math>\overline{FG}</math> bound a smaller regular hexagon. Let the [[ratio]] of the area of the smaller hexagon to the area of <math>ABCDEF</math> be expressed as a fraction <math>\frac {m}{n}</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>. |
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Revision as of 20:22, 13 March 2015
Problem
Let be a regular hexagon. Let , , , , , and be the midpoints of sides , , , , , and , respectively. The segments , , , , , and bound a smaller regular hexagon. Let the ratio of the area of the smaller hexagon to the area of be expressed as a fraction where and are relatively prime positive integers. Find .
Solution
Let be the intersection of and
and be the intersection of and .
Let be the center.
Solution 1
Let (without loss of generality).
Note that is the vertical angle to an angle of regular hexagon, and so has degree .
Because and are rotational images of one another, we get that and hence .
Using a similar argument, , and
Applying the Law of cosines on ,
Thus, the answer is 4 + 7 = .
Solution 2
We can use coordinates. Let be at with at ,
then is at ,
is at ,
is at ,
Line has the slope of and the equation of
Line has the slope of and the equation
Let's solve the system of equation to find
Finally,
Thus, the answer is .
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.