Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2018 AIME II Problems/Problem 4

Revision as of 20:45, 5 February 2022 by Thisusernameistaken (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

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 $a + b$.

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$(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 $C_1 = (\frac{4}{3}, 2)$ where $C_1$ is the intersection of $CO$ and $AI$, and $C_2 = (\frac{5}{3}, 2)$ is the intersection of $RN$ and $CO$.

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

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 $EL$).

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_4&oldid=171003"