Difference between revisions of "2010 AIME II Problems/Problem 9"
(Created page with '== Problem 9 == Let <math>ABCDEF</math> be a regular hexagon. Let <math>G</math>, <math>H</math>, <math>I</math>, <math>J</math>, <math>K</math>, and <math>L</math> be the midpoi…') |
(→Solution) |
||
Line 54: | Line 54: | ||
</asy></center> | </asy></center> | ||
− | Let <math>M</math> be the intersection of <math>\ | + | Let <math>M</math> be the intersection of <math>\overline{AG}</math> and <math>\overline{BI}</math> |
− | and <math>N</math> be the intersection of <math>\ | + | and <math>N</math> be the intersection of <math>\overline{BI}</math> and <math>\overline{CJ}</math>. |
+ | |||
+ | and let <math>BC=2</math> | ||
+ | |||
+ | Note that <math>\angle BMH</math> is the vertical angle to an angle of regular hexagon, thus, it is <math>120^\circ</math>. | ||
+ | |||
+ | Because <math>\triangle ABH</math> and <math>\triangle BCI</math> are rotational images of one another, we get that <math>\angle{MBH}=\angle{HAB}</math> and hence <math>\triangle ABH \sim \triangle BMH \sim \triangle BCI</math>. | ||
+ | |||
+ | Using a simlar argument, <math>NI=MH</math>. | ||
+ | |||
+ | <math>MN=BI-NI-BM=BI-(BM+MH)</math> | ||
+ | |||
+ | Applying law of cosine on <math>\triangle BCI</math>, <math>BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))=\sqrt{7}</math> | ||
+ | |||
+ | <math>\frac{BC+CI}{BI}=\frac{3}{\sqrt{7}}=\frac{BM+MH}{BH}</math> | ||
+ | |||
+ | <math>BM+MH=\frac{3BH}{\sqrt{7}}=\frac{3}{\sqrt{7}}</math> | ||
+ | |||
+ | <math>MN=BI-(BM+MH)=\sqrt{7}-\frac{3}{\sqrt{7}}=\frac{4}{\sqrt{7}}</math> | ||
+ | |||
+ | <math>\frac{\text{Area of smaller hexagon}}{\text{Area of bigger hexagon}}=\left(\frac{MN}{BC}\right)^2=\left(\frac{2}{\sqrt{7}}\right)^2=\frac{4}{7}</math> | ||
+ | |||
+ | Thus, answer is <math>\boxed{011}</math> | ||
− | |||
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=8|num-a=10|n=II}} | {{AIME box|year=2010|num-b=8|num-a=10|n=II}} |
Revision as of 18:15, 3 April 2010
Problem 9
Let be a regular hexagon. Let , , , , , and be the midpoints of sides , , , , , and , respectively. The segments $\overbar{AH}$ (Error compiling LaTeX. Unknown error_msg), $\overbar{BI}$ (Error compiling LaTeX. Unknown error_msg), $\overbar{CJ}$ (Error compiling LaTeX. Unknown error_msg), $\overbar{DK}$ (Error compiling LaTeX. Unknown error_msg), $\overbar{EL}$ (Error compiling LaTeX. Unknown error_msg), and $\overbar{FG}$ (Error compiling LaTeX. Unknown error_msg) 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 .
and let
Note that is the vertical angle to an angle of regular hexagon, thus, it is .
Because and are rotational images of one another, we get that and hence .
Using a simlar argument, .
Applying law of cosine on , $BI=\sqrt{2^2+1^2-2(2)(1)(\cos(120^\circ))=\sqrt{7}$ (Error compiling LaTeX. Unknown error_msg)
Thus, 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 |