2021 AMC 10A Fall Problem 17

Attempt Problem 17 of the 2021 AMC 10A Fall below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 10A Fall solutions, or check the answer key.

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17.

An architect is building a structure that will place vertical pillars at the vertices of regular hexagon ABCDEF,ABCDEF, 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 A,A, B,B, and CC are 12,12, 9,9, and 1010 meters, respectively. What is the height, in meters, of the pillar at E?E?

99

636\sqrt{3}

838\sqrt{3}

1717

12312\sqrt{3}

Answer: D
Concepts:coordinate geometryregular polygonsystem of equations
Difficulty rating: 1660
Solution:

Put a regular hexagon in coordinates with A=(1,0),A=(-1,0), B=(12,32),B=(-\frac{1}{2},\frac{\sqrt3}{2}), C=(12,32),C=(\frac{1}{2},\frac{\sqrt3}{2}), and E=(12,32).E=(\frac{1}{2},-\frac{\sqrt3}{2}). Because the solar panel is flat, the height is an affine function h(x,y)=ux+vy+w.h(x,y)=ux+vy+w.

From h(A)=12,h(A)=12, h(B)=9,h(B)=9, and h(C)=10,h(C)=10, subtracting the last two equations gives u=1.u=1. Then u+w=12,-u+w=12, so w=13.w=13. Using h(B)=9h(B)=9 gives 12+32v+13=9,-\frac{1}{2}+\frac{\sqrt3}{2}v+13=9, so 3v=7.\sqrt3v=-7.

Therefore h(E)=1232v+13=12+72+13=17. \begin{aligned} h(E) &=\frac{1}{2}-\frac{\sqrt3}{2}v+13 \\ &=\frac{1}{2}+\frac{7}{2}+13 \\ &=17. \end{aligned}

Thus, D is the correct answer.

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