Difference between revisions of "2014 AMC 10B Problems/Problem 13"

(Solution)
(Problem)
Line 4: Line 4:
  
 
(diagram needed)
 
(diagram needed)
 +
 +
<asy>
 +
draw((0,0)--(-5,8.66025404)--(0, 17.3205081)--(10, 17.3205081)--(15,8.66025404)--(10, 0)--(0, 0));
 +
draw((30,0)--(25,8.66025404)--(30, 17.3205081)--(40, 17.3205081)--(45, 8.66025404)--(40, 0)--(30, 0));
 +
draw((30,0)--(25,-8.66025404)--(30, -17.3205081)--(40, -17.3205081)--(45, -8.66025404)--(40, 0)--(30, 0));
 +
draw((0,0)--(-5, -8.66025404)--(0, -17.3205081)--(10, -17.3205081)--(15, -8.66025404)--(10, 0)--(0, 0));
 +
label("A", (0,0), W);
 +
label("B", (30, 17.3205081), NE);
 +
label("C", (30, -17.3205081), SE);
 +
draw((0,0)--(30, 17.3205081)--(30, -17.3205081)--(0, 0));
 +
</asy>
  
 
==Solution==
 
==Solution==

Revision as of 19:49, 20 February 2014

Problem

Six regular hexagons surround a regular hexagon of side length $1$ as shown. What is the area of $\triangle{ABC}$?

(diagram needed)

[asy] draw((0,0)--(-5,8.66025404)--(0, 17.3205081)--(10, 17.3205081)--(15,8.66025404)--(10, 0)--(0, 0)); draw((30,0)--(25,8.66025404)--(30, 17.3205081)--(40, 17.3205081)--(45, 8.66025404)--(40, 0)--(30, 0)); draw((30,0)--(25,-8.66025404)--(30, -17.3205081)--(40, -17.3205081)--(45, -8.66025404)--(40, 0)--(30, 0)); draw((0,0)--(-5, -8.66025404)--(0, -17.3205081)--(10, -17.3205081)--(15, -8.66025404)--(10, 0)--(0, 0)); label("A", (0,0), W); label("B", (30, 17.3205081), NE); label("C", (30, -17.3205081), SE); draw((0,0)--(30, 17.3205081)--(30, -17.3205081)--(0, 0)); [/asy]

Solution

We note that the $6$ triangular sections in $\triangle{ABC}$ can be put together to form a hexagon congruent to each of the seven other hexagons. By the formula for the area of the hexagon, we get the area for each hexagon as $\dfrac{6\sqrt{3}}{4}$. The area of $\triangle{ABC}$, which is equivalent to two of these hexagons together, is $\boxed{\textbf{(B)}  3\sqrt{3}}$.

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png