2021 Fall AMC 10A Problems/Problem 17
Contents
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
Two-Dimensional Diagram
~ihatemath123 ~MRENTHUSIASM
Solution 1 (Height From the Center)
Since the pillar at has height and the pillar at has height and the solar panel is flat, the inclination from pillar to pillar would be . Call the center of the hexagon . Since , it follows that has a height of . Since the solar panel is flat, should be a straight line and therefore, has a height of .
~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 ~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)
Solution 6 (Vectors)
WLOG, let the side length of the hexagon be 6.
Establish a 3D coordinate system, in which . Let the coordinates of B and C be , , respectively. Then, the solar panel passes through .
The vector and . Computing using 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 .
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.