2018 AIME II Problems/Problem 4


In equiangular octagon $CAROLINE$, $CA = RO = LI = NE =$ $\sqrt{2}$ and $AR = OL = IN = EC = 1$. The self-intersecting octagon $CORNELIA$ encloses 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 $a + b$.


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$(We look at their heights), $YZ$ $=$ $\frac{1}{3}$. 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 (ProblemsAnswer KeyResources)
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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