Difference between revisions of "2022 AMC 12B Problems/Problem 25"
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− | Begin by dividing the figure | + | Begin by dividing the figure as shown above. Clearly, the entire figure has 8-fold symmetry. Therefore, we can calculate the area of <math>ODBA</math> and multiply it by 8. We split <math>[ODBA]</math> into <math>[ODBC]+[ABC]</math>. |
Knowing the side length of the hexagon is <math>1</math>, we can use 30-60-90 triangles within the hexagon to find the total distance between opposite edges is <math>2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}.</math> Thus, <math>OD=\sqrt{3}-\frac{1}{2}</math> and <math>BC=\sqrt{3}-1.</math> Recognizing <math>DB=\frac{1}{2}</math> and <math>ODBC</math> is a trapezoid, <cmath>8\cdot[ODBC]=2\left(\sqrt{3}-\frac{1}{2}+\sqrt{3}-1\right)=4\sqrt{3}-3.</cmath> | Knowing the side length of the hexagon is <math>1</math>, we can use 30-60-90 triangles within the hexagon to find the total distance between opposite edges is <math>2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}.</math> Thus, <math>OD=\sqrt{3}-\frac{1}{2}</math> and <math>BC=\sqrt{3}-1.</math> Recognizing <math>DB=\frac{1}{2}</math> and <math>ODBC</math> is a trapezoid, <cmath>8\cdot[ODBC]=2\left(\sqrt{3}-\frac{1}{2}+\sqrt{3}-1\right)=4\sqrt{3}-3.</cmath> |
Revision as of 15:54, 21 November 2022
Contents
[hide]Problem
Four regular hexagons surround a square with side length 1, each one sharing an edge with the square, as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be written as , where , , and are integers and is not divisible by the square of any prime. What is ?
Solution 1
Begin by dividing the figure as shown above. Clearly, the entire figure has 8-fold symmetry. Therefore, we can calculate the area of and multiply it by 8. We split into .
Knowing the side length of the hexagon is , we can use 30-60-90 triangles within the hexagon to find the total distance between opposite edges is Thus, and Recognizing and is a trapezoid,
Next, we aim to find . By angle chasing, we find and We can use the law of sines to find :
We may not know what is by memory, but we can cleverly calculate it using a common trig identity:
With some simplification, we'll find . Now, we can easily calculate as
Thus, the area of the dodecagon is
Finally, we find
~Indiiiigo
Solution 2 (Coord bash)
Refer to the diagram above.
Let the origin be at the center of the square, be the intersection of the top and right hexagons, be the intersection of the top and left hexagons, and and be the top points in the diagram.
By symmetry, lies on the line . The equation of line is (due to it being one of the sides of the top hexagon). Thus, we can solve for the coordinates of by finding the intersection of the two lines:
This means that we can find the length , which is equal to . We will next find the area of trapezoid . The lengths of the bases are and , and the height is equal to the -coordinate of minus the -coordinate of . The height of the hexagon is and the bottom of the hexagon lies on the line . Thus, the -coordinate of is , and the height is . We can now find the area of the trapezoid:
The total area of the figure is the area of a square with side length plus four times the area of this trapezoid:
Our answer is .
~mathboy100
Solution 3
We calculate the area as the area of the red octagon minus the four purple congruent triangles: We first find the important angles in the figure. We note that 2 adjacent hexagons are rotated with respect to the other, so the angles between any sides is . In particular, as the purple triangles are isosceles, they have angles , and , and the octagon is equiangular (all its angles are ). Thus, we can draw a square around the octagon, and we note that the ``cut out" triangles are all isosceles right triangles.
Now, we calculate the side length of the square. Note that the hexagon has a height of , so the length of a side of the square is . In particular, the horizontal/vertical sides of the octagon have length , so the legs of the isosceles triangles are Thus, the area of the octagon is Now, we calculate the area of one of the four isosceles triangles. The base of the triangle is , so the area is Thus, the area of the dodecagon is Thus the answer is , or .
~cr. naman12
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last problem |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.