Difference between revisions of "2023 AMC 10A Problems/Problem 24"
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− | Notice that when sliding the smaller hexagon along the edge, we see that the contact edge withe the smaller hexagon "in front" of it is <math>60^{\circ}</math>, thus meaning the hexagon "in front" is pushed at a speed <math>\sin{60^{\circ}}</math> times the actual speed of the hexagon. We can preform a similar analysis on the hexagon that is being pushed and get that the speed at which that hexagon is moving is <math>\frac{1}{\sin{60^{\circ}}}</math> times the speed it pushed by. As we can see, the two factors cancel out and by the same argument, every small hexagon can move at the same speed while mantaining an edge of contact with the two adjacent hexagons | + | Notice that when sliding the smaller hexagon along the edge, we see that the contact edge withe the smaller hexagon "in front" of it is <math>60^{\circ}</math>, thus meaning the hexagon "in front" is pushed at a speed <math>\sin{60^{\circ}}</math> times the actual speed of the hexagon. We can preform a similar analysis on the hexagon that is being pushed and get that the speed at which that hexagon is moving is <math>\frac{1}{\sin{60^{\circ}}}</math> times the speed it pushed by. As we can see, the two factors cancel out and by the same argument, every small hexagon can move at the same speed while mantaining an edge of contact with the two adjacent hexagons. |
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===Note=== | ===Note=== |
Revision as of 22:55, 9 November 2023
Contents
[hide]Problem
Six regular hexagonal blocks of side length 1 unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is unit. What is the area of the region inside the frame not occupied by the blocks?
Solution 1
Examining the red isosceles trapezoid with and as two bases, we know that the side lengths are from triangle.
We can conclude that the big hexagon has side length 3.
Thus the target area is: area of big hexagon - 6 * area of small hexagon.
~Technodoggo
Solution 2 (Not rigorous)
Note that one can "slide' the small hexagons along their respective edges, and either by sliding them to the center or to the corners, and thus getting that the side length of the larger hexagon is 3. The rest proceeds the same as solution 1.
Solution 2.1 (Clarification)
Notice that when sliding the smaller hexagon along the edge, we see that the contact edge withe the smaller hexagon "in front" of it is , thus meaning the hexagon "in front" is pushed at a speed times the actual speed of the hexagon. We can preform a similar analysis on the hexagon that is being pushed and get that the speed at which that hexagon is moving is times the speed it pushed by. As we can see, the two factors cancel out and by the same argument, every small hexagon can move at the same speed while mantaining an edge of contact with the two adjacent hexagons.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Note
The number is irrelevant to solve the problem.
Video Solution 1 by OmegaLearn
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 10 Problems and Solutions |
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