Difference between revisions of "2009 AMC 8 Problems/Problem 9"
| (7 intermediate revisions by 6 users not shown) | |||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | Construct a square on one side of an equilateral triangle. | + | Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have? |
<asy> | <asy> | ||
| Line 14: | Line 14: | ||
==Solution== | ==Solution== | ||
Of the six shapes used to create the polygon, the triangle and octagon are adjacent to the others on one side, and the others are adjacent on two sides. In the triangle and octagon <math>3+8-2(1)=9</math> sides are on the outside of the final polygon. In the other shapes <math>4+5+6+7-4(2) = 14</math> sides are on the outside. The resulting polygon has <math>9+14 = \boxed{\textbf{(B)}\ 23}</math> sides. | Of the six shapes used to create the polygon, the triangle and octagon are adjacent to the others on one side, and the others are adjacent on two sides. In the triangle and octagon <math>3+8-2(1)=9</math> sides are on the outside of the final polygon. In the other shapes <math>4+5+6+7-4(2) = 14</math> sides are on the outside. The resulting polygon has <math>9+14 = \boxed{\textbf{(B)}\ 23}</math> sides. | ||
| + | |||
| + | ==Solution 2== | ||
| + | We can quickly see a pattern if we draw out the other shapes. Every shape will have two of its sides taken out except the triangle and octagon. We can then make the expression <math>2+2+3+4+5+7</math> which is <math>\boxed{\textbf{(B)}\ 23}</math> | ||
| + | |||
| + | ~Trex226 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/eG8w13-3S8M | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=8|num-a=10}} | {{AMC8 box|year=2009|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 07:24, 30 May 2023
Problem
Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?
![[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]](http://latex.artofproblemsolving.com/d/4/a/d4a192b1c509fea9e4f0b63b29b58c056ff0a1ab.png)
Solution
Of the six shapes used to create the polygon, the triangle and octagon are adjacent to the others on one side, and the others are adjacent on two sides. In the triangle and octagon 
sides are on the outside of the final polygon. In the other shapes 
sides are on the outside. The resulting polygon has 
sides.
Solution 2
We can quickly see a pattern if we draw out the other shapes. Every shape will have two of its sides taken out except the triangle and octagon. We can then make the expression 
which is 
~Trex226
Video Solution
~savannahsolver
See Also
| 2009 AMC 8 (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
