2021 Fall AMC 10A Problems/Problem 17

Problem

An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , 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 $A$ , $B$ , and $C$ are $12$ , $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$ ?

$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$

Solution

Since the pillar at $B$ has height $9$ and the pillar at $A$ has height $10$ and the solar panel is flat, the inclination from pillar $A$ to pillar $B$ would be $1$ . Call the center of the hexagon $G$ . Since $CG$ is parallel to $BA$ , $G$ has a height of $13$ . Since the solar panel is flat, $BGE$ should be a straight line and therefore, E has a height of $9+4+4$ = $\boxed {(D) 17}$ .

~Arcticturn

Solution 2

Let the height of the pillar at $D$ be $x.$ Notice that the difference between the heights of pillar $C$ and pillar $D$ is equal to the difference between the heights of pillar $A$ and pillar $F.$ So, the height at $F$ is $x+2.$ Now, doing the same thing for pillar $E$ we get the height is $x+3.$ Therefore, we can see the difference between the heights at pillar $C$ and pillar $D$ is half the difference between the heights at $B$ and $E,$ so $x+3-9=2 \cdot (x-10) \implies x-6=2 \cdot (x-10) \implies x=14 \implies x+3=\boxed{17}.$

- kante314

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2021 Fall AMC 10A Problems/Problem 17

Problem

Solution

Solution 2