Difference between revisions of "2018 AIME II Problems/Problem 4"
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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>. | 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> | + | Using similar triangles <math>\bigtriangleup</math><math>ARW</math> and <math>\bigtriangleup</math><math>YZW</math>(We look at their heights), <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>. | 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>. |
Revision as of 23:29, 29 February 2020
Problem
In equiangular octagon ,
and
. The self-intersecting octagon
encloses 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
(We look at their heights),
. 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.