Difference between revisions of "2011 AMC 10B Problems/Problem 20"
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Rhombus <math>ABCD</math> has side length <math>2</math> and <math>\angle B = 120</math>°. Region <math>R</math> consists of all points inside the rhombus that are closer to vertex <math>B</math> than any of the other three vertices. What is the area of <math>R</math>? | Rhombus <math>ABCD</math> has side length <math>2</math> and <math>\angle B = 120</math>°. Region <math>R</math> consists of all points inside the rhombus that are closer to vertex <math>B</math> than any of the other three vertices. What is the area of <math>R</math>? | ||
| − | <math> \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{ | + | <math> \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2</math> |
== Solution == | == Solution == | ||
Revision as of 03:40, 27 December 2011
- The following problem is from both the 2011 AMC 12B #16 and 2011 AMC 10B #20, so both problems redirect to this page.
Problem
Rhombus 
has side length 
and 
°. Region 
consists of all points inside the rhombus that are closer to vertex 
than any of the other three vertices. What is the area of ?
Solution
Suppose that 
is a point in the rhombus and let 
be the perpendicular bisector of 
. Then 
if and only if is on the same side of as . The line divides the plane into two half-planes; let 
be the half-plane containing . Let us define similarly 
and 
. Then is equal to 
. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:
![[asy] unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,lightgray); draw(A--B--C--D--cycle); draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G); label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW); label("$E$",E,N);label("$F$",F,SW);label("$G$",G,SW);label("$H$",H,S);label("$I$",I,NE); label("$2$",(D--C),SW); [/asy]](http://latex.artofproblemsolving.com/c/2/a/c2a7f37d7dc652d27ff712c7c2c95862ed92c123.png)
Since 
and 
are equilateral, contains 
, 
contains 
and 
, and 
contains . Then 
with 
and 
so ![$[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}$](http://latex.artofproblemsolving.com/9/9/5/99522386e5bdd18a46622bba455d07407e52c51b.png)
and has area 
.
See Also
| 2011 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 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 | |
| All AMC 12 Problems and Solutions | |
