Difference between revisions of "2018 AIME II Problems/Problem 4"
(Created page with "==Problem== In equiangular octagon <math>CAROLINE</math>, <math>CA = RO = LI = NE =</math> <math>\sqrt{2}</math> and <math>AR = OL = IN = EC = 1</math>. The self-intersecting...") |
(If anyone is good with Asymptote, please change the diagrams because I am not good with Asymptote at all) |
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In equiangular octagon <math>CAROLINE</math>, <math>CA = RO = LI = NE =</math> <math>\sqrt{2}</math> and <math>AR = OL = IN = EC = 1</math>. The self-intersecting octagon <math>CORNELIA</math> enclosed six non-overlapping triangular regions. Let <math>K</math> be the area enclosed by <math>CORNELIA</math>, that is, the total area of the six triangular regions. Then <math>K =</math> <math>\dfrac{a}{b}</math>, where <math>a</math> and <math>b</math> are relatively prime positive integers. Find <math>a + b</math>. | In equiangular octagon <math>CAROLINE</math>, <math>CA = RO = LI = NE =</math> <math>\sqrt{2}</math> and <math>AR = OL = IN = EC = 1</math>. The self-intersecting octagon <math>CORNELIA</math> enclosed six non-overlapping triangular regions. Let <math>K</math> be the area enclosed by <math>CORNELIA</math>, that is, the total area of the six triangular regions. Then <math>K =</math> <math>\dfrac{a}{b}</math>, where <math>a</math> and <math>b</math> are relatively prime positive integers. Find <math>a + b</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | We can draw <math>CORNELIA</math> and introduce some points. | ||
+ | |||
+ | [[File:2018_AIME_II_Problem_4.png|200px]] | ||
+ | |||
+ | The diagram is essentially a 3x3 grid where each of the 9 squares making up the grid have a side length of 1. | ||
+ | |||
+ | In order to find the area of <math>CORNELIA</math>, we need to find 4 times the area of <math>\bigtriangleup</math><math>ACY</math> and 2 times the area of <math>\bigtriangleup</math><math>YZW</math>. | ||
+ | |||
+ | Using similar triangles <math>\bigtriangleup</math><math>ARW</math> and <math>\bigtriangleup</math><math>YZW</math>, <math>YZ</math> <math>=</math> <math>\frac{1}{3}</math>. Therefore, the area of <math>\bigtriangleup</math><math>YZW</math> is <math>\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}</math> <math>=</math> <math>\frac{1}{12}</math> | ||
+ | |||
+ | Since <math>YZ</math> <math>=</math> <math>\frac{1}{3}</math> and <math>XY = ZQ</math>, <math>XY</math> <math>=</math> <math>\frac{1}{3}</math> and <math>CY</math> <math>=</math> <math>\frac{4}{3}</math>. | ||
+ | |||
+ | Therefore, the area of <math>\bigtriangleup</math><math>ACY</math> is <math>\frac{4}{3}\cdot</math> <math>1</math> <math>\cdot</math> <math>\frac{1}{2}</math> <math>=</math> <math>\frac{2}{3}</math> | ||
+ | |||
+ | Our final answer is <math>\frac{1}{12}</math> <math>\cdot</math> <math>2</math> <math>+</math> <math>\frac{2}{3}</math> <math>\cdot</math> <math>4</math> <math>=</math> <math>\frac{17}{6}</math> | ||
+ | |||
+ | <math>17 + 6 =</math> <math>\boxed{023}</math> | ||
+ | |||
+ | ==See Also== | ||
{{AIME box|year=2018|n=II|num-b=3|num-a=5}} | {{AIME box|year=2018|n=II|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 09:49, 25 March 2018
Problem
In equiangular octagon ,
and
. The self-intersecting octagon
enclosed six non-overlapping triangular regions. Let
be the area enclosed by
, that is, the total area of the six triangular regions. Then
, where
and
are relatively prime positive integers. Find
.
Solution
We can draw and introduce some points.
The diagram is essentially a 3x3 grid where each of the 9 squares making up the grid have a side length of 1.
In order to find the area of , we need to find 4 times the area of
and 2 times the area of
.
Using similar triangles and
,
. Therefore, the area of
is
Since
and
,
and
.
Therefore, the area of is
Our final answer is
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.