Difference between revisions of "2021 Fall AMC 10A Problems/Problem 17"

Revision as of 16:42, 23 November 2021

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 $E$ ?

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

Diagram

$[asy] pair A = (-sqrt(3),1); pair B = (0,2); pair C = (sqrt(3),1); pair D = (sqrt(3),-1); pair E = (0,-2); pair F = (-sqrt(3),-1); draw(A--B--C--D--E--F--cycle); label("$A, 12$", A, NW); label("$B, 9$", B, N); label("$C, 10$", C, NE); label("$D$", D, SE); label("$E$", E, S); label("$F$", F, SW); [/asy]$

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

Solution 3 (Extend the lines)

We can extend $BA$ and $BC$ to $G$ and $H$ , respectively, such that $AG = CH$ and $E$ lies on $\overline{GH}$ : $[asy] unitsize(1.5cm); pair A = (-sqrt(3),1); pair B = (0,2); pair C = (sqrt(3),1); pair D = (sqrt(3),-1); pair E = (0,-2); pair F = (-sqrt(3),-1); draw(A--B--C--D--E--F--cycle); label("$A, 12$", A, NW); label("$B, 9$", B, N); label("$C, 10$", C, NE); label("$D$", D, SE); label("$E$", E, S); label("$F$", F, SW); pair G = (-4*sqrt(3),-2); pair H = (4*sqrt(3),-2); label("$G, 21$", G, W); label("$H, 13$", H, E); draw(A--G, dashed); draw(C--H, dashed); [/asy]$ Because of hexagon proportions, $\frac{BA}{AG} = \frac{1}{3}$ and $\frac{BC}{CH} = \frac{1}{3}$ . Let $g$ be the height of $G$ . Because $A$ , $B$ and $G$ lie on the same line, $\frac{12-9}{g-12} = \frac{1}{3}$ , so $g-12 = 9$ and $g = 21$ . Similarly, the height of $H$ is $13$ . $E$ is the midpoint of $GH$ , so we can take the average of these heights to get our answer, $\boxed{\textbf{(D) } 17}$ .

~ihatemath123

@@ Line 61: / Line 61: @@
 ~ihatemath123
+==See Also==
+{{AMC10 box|year=2021 Fall|ab=A|num-b=16|num-a=18}}
+{{MAA Notice}}

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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