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

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We can extend <math>BA</math> and <math>BC</math> to <math>G</math> and <math>H</math>, respectively, such that <math>AG = CH</math> and <math>E</math> lies on <math>\overline{GH}</math>:
 
We can extend <math>BA</math> and <math>BC</math> to <math>G</math> and <math>H</math>, respectively, such that <math>AG = CH</math> and <math>E</math> lies on <math>\overline{GH}</math>:
 
<asy>
 
<asy>
−
unitsize(1.5cm);
+
unitsize(1cm);
 
pair A = (-sqrt(3),1);
 
pair A = (-sqrt(3),1);
 
pair B = (0,2);
 
pair B = (0,2);

Revision as of 11:24, 28 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(1cm); 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

Solution 4

Denote by $h_X$ the height of any point $X$.

Denote by $M$ the midpoint of $A$ and $C$. Hence, $h_M = \frac{h_A + h_C}{1} = 11$.

Denote by $O$ the center of $ABCDEF$. Because $ABCDEF$ is a right hexagon, $O$ is the midpoint of $B$ and $E$. Hence, $h_O = \frac{h_E + h_B}{2} = \frac{h_E + 9}{2}$.

Because $ABCDEF$ is a right hexagon, $M$ is the midpoint of $B$ and $O$. Hence, $h_M = \frac{h_B + h_O}{2} = \frac{9 + h_O}{2}$.

Solving all these equations, we get $h_E = 17$.

Therefore, the answer is $\boxed{\textbf{(D) }17}$.

~Steven Chen (www.professorchenedu.com)

Solution 5 (Vectors)

WLOG, let the side length of the hexagon be 6.

Establish a 3D coordinate system, in which $A=(0,0,0)$. Let the coordinates of B and C be $(6,0,0)$, $(9,-3\sqrt{3},0)$, respectively. Then, the solar panel passes through $P=(0,0,12), Q=(6,0,9), R=(9,-3\sqrt{3},10)$.

The vector $\vec{PQ}=<6,0,-3>$ and $\vec{PR}=<9,-3\sqrt{3}, -2>$. Computing $\vec{PQ} \times \vec{PR}$ using the matrix

$\begin{bmatrix} i & j & k \\ 6 & 0 & -3 \\ 9 & -3\sqrt{3} & -2 \end{bmatrix}$

gives the result $-9\sqrt{3}i -15j -18\sqrt{3} k$. Therefore, a normal vector of the plane of the solar panel is $<-9\sqrt{3},-15,-18\sqrt{3}>$, and the equation of the plane is $-9\sqrt{3}x-15y-18\sqrt{3}z=k$. Substituting $(x,y,z)=(0,0,12)$, we find that $k=-216\sqrt{3}$

Since $E=(0,-6\sqrt{3})$, we substitute $(x,y)=(0,-6\sqrt{3})$ into $-9\sqrt{3}x-15y-18\sqrt{3}z=-216\sqrt{3}$, which gives $z=\boxed{\textbf{(D) }17}$.


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

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