2021 Fall AMC 10A Problems/Problem 17
Contents
- 1 Problem
- 2 Diagrams
- 3 Solution 1 (Height From the Center)
- 4 Solution 2 (Height From Each Vertex)
- 5 Solution 3 (Extend the Sides)
- 6 Solution 4 (Averages of Heights)
- 7 Solution 5 (Vectors)
- 8 Solution 6 (Vectors)
- 9 Solution 7 (3D Slopes)
- 10 Solution 8 (Midpoints)
- 11 Solution 9 (Educated Guess)
- 12 Video Solution (Under 2 min!)
- 13 Video Solution by TheBeautyofMath
- 14 See Also
Problem
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at , , and are , , and meters, respectively. What is the height, in meters, of the pillar at ?
Diagrams
Three-Dimensional Diagram
~MRENTHUSIASM
Two-Dimensional Diagram
~ihatemath123 ~MRENTHUSIASM
Solution 1 (Height From the Center)
The pillar at has height and the pillar at has height Since the solar panel is flat, the inclination from pillar to pillar is Call the center of the hexagon Since it follows that the solar panel has height at Since the solar panel is flat, the heights of the solar panel at and are collinear. Therefore, the pillar at has height
~Arcticturn
Solution 2 (Height From Each Vertex)
Let the height of the pillar at be Notice that the difference between the heights of pillar and pillar is equal to the difference between the heights of pillar and pillar So, the height at is Now, doing the same thing for pillar we get the height is Therefore, we can see the difference between the heights at pillar and pillar is half the difference between the heights at and so The answer is
~kante314
Solution 3 (Extend the Sides)
We can extend and to and , respectively, such that and lies on : Because of hexagon proportions, and . Let be the height of . Because , and lie on the same line, , so and . Similarly, the height of is . is the midpoint of , so we can take the average of these heights to get our answer, .
~ihatemath123
Solution 4 (Averages of Heights)
Denote by the height of any point .
Denote by the midpoint of and . Hence, Denote by the center of . Because is a regular hexagon, is the midpoint of and . Hence, Because is a regular hexagon, is the midpoint of and . Hence, Solving these equations, we get .
~Steven Chen (www.professorchenedu.com)
Solution 5 (Vectors)
In this solution, we define rise as the change of height (in meters) from the solar panel to the ground. It follows that the rise from to is and the rise from to is Note that so the rise from to is
Together, the height of the pillar at is meters.
~MRENTHUSIASM
Solution 6 (Vectors)
WLOG, let the side length of the hexagon be .
Establish a 3D coordinate system, in which . Let the coordinates of and be , , respectively. Then, the solar panel passes through .
The vector and . Computing by the matrix gives the result . Therefore, a normal vector of the plane of the solar panel is , and the equation of the plane is . Substituting , we find that .
Since , we substitute into , which gives .
Solution 7 (3D Slopes)
Let the pillars be . Since solar panel is a hexagon, the line hits the midpoint of . So, the 3D slope (change in : change in : change in ) of is same as . If is side of the hexagonal solar panel, . So, = . Since the height of to ground is , the rise (in z) from to is 2 meaning the rise from to is . Thus, .
~sml1809
Solution 8 (Midpoints)
Set the midpoint of as : We know that the height of is as it is the midpoint of , so the height is the average of and , which is . Since is a regular hexagon, . Because the increase in height is proportional to the length of the line segments, and the increase in height from to is , the increase in height from to is Adding to the height of , we get .
~iluvme
Solution 9 (Educated Guess)
Because the three points given are integers, it is likely that the answer is also an integer. This leaves us with or . Because both and are greater than and closer to than , we can assume that the height increases as the point gets closer to . Thus, we know the answer is greater than . The only choice that satisfies both these criteria is .
~iluvme
Video Solution (Under 2 min!)
~Education, the Study of Everything
Video Solution by TheBeautyofMath
Inefficient, but it gets the job done https://youtu.be/SarZNOgo4DA
~IceMatrix
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.