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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>.
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==Solution==
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We can draw <math>CORNELIA</math> and introduce some points.
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[[File:2018_AIME_II_Problem_4.png|200px]]
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The diagram is essentially a 3x3 grid where each of the 9 squares making up the grid have a side length of 1.
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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>.
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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>
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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>.
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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>
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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>
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<math>17 + 6 =</math> <math>\boxed{023}</math>
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==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 $CAROLINE$, $CA = RO = LI = NE =$ $\sqrt{2}$ and $AR = OL = IN = EC = 1$. The self-intersecting octagon $CORNELIA$ enclosed six non-overlapping triangular regions. Let $K$ be the area enclosed by $CORNELIA$, that is, the total area of the six triangular regions. Then $K =$ $\dfrac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find $a + b$.

Solution

We can draw $CORNELIA$ and introduce some points.

2018 AIME II Problem 4.png

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 $CORNELIA$, we need to find 4 times the area of $\bigtriangleup$$ACY$ and 2 times the area of $\bigtriangleup$$YZW$.

Using similar triangles $\bigtriangleup$$ARW$ and $\bigtriangleup$$YZW$, $YZ$ $=$ $\frac{1}{3}$. Therefore, the area of $\bigtriangleup$$YZW$ is $\frac{1}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}$ $=$ $\frac{1}{12}$

Since $YZ$ $=$ $\frac{1}{3}$ and $XY = ZQ$, $XY$ $=$ $\frac{1}{3}$ and $CY$ $=$ $\frac{4}{3}$.

Therefore, the area of $\bigtriangleup$$ACY$ is $\frac{4}{3}\cdot$ $1$ $\cdot$ $\frac{1}{2}$ $=$ $\frac{2}{3}$

Our final answer is $\frac{1}{12}$ $\cdot$ $2$ $+$ $\frac{2}{3}$ $\cdot$ $4$ $=$ $\frac{17}{6}$

$17 + 6 =$ $\boxed{023}$

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

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