AIME 2018 II · 第 4 题

Solution

We can draw $CORNELIA$ 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 $CORNELIA$, 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 $\frac{1}{12}$ $\cdot$ $2$ $+$ $\frac{2}{3}$ $\cdot$ $4$ $=$ $\frac{17}{6}$

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

Solution 2

$CAROLINE$ is essentially a plus sign with side length 1 with a few diagonals, which motivates us to coordinate bash. We let $N = (1, 0)$ and $E = (0, 1)$. To find $CORNELIA$'s self intersections, we take

$$CO = y = 2, AI = y = -3x + 6, RN = y = 3x - 3$$ And plug them in to get $C_1 = \left(\frac{4}{3}, 2 \right)$ where $C_1$ is the intersection of $CO$ and $AI$, and $C_2 = \left(\frac{5}{3}, 2 \right)$ is the intersection of $RN$ and $CO$.

We also track the intersection of $AI$ and $RN$ to get $\left(\frac{3}{2}, \frac{3}{2} \right)$.

By vertical symmetry, the other 2 points of intersection should have the same x-coordinates. We can then proceed with Solution 1 to calculate the area of the triangle (compare the $y$-coordinates of $A,R,I,N$ and $CO$ and $EL$).