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

m (→‎Solution 3 (Extend the lines))
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<math>\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}</math>
 
<math>\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}</math>
  
−
==Diagram==
+
==Diagrams==
 +
===Three-Dimensional Diagram===
 +
 
 +
===Two-Dimensional Diagram===
 
<asy>
 
<asy>
 
pair A = (-sqrt(3),1);
 
pair A = (-sqrt(3),1);
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</asy>
 
</asy>
  
−
==Solution==
+
==Solution 1 (Height From the Center)==
−
Since the pillar at <math>B</math> has height <math>9</math> and the pillar at <math>A</math> has height <math>10</math> and the solar panel is flat, the inclination from pillar <math>A</math> to pillar <math>B</math> would be <math>1</math>. Call the center of the hexagon <math>G</math>. Since <math>CG</math> is parallel to <math>BA</math>, <math>G</math> has a height of <math>13</math>. Since the solar panel is flat, <math>BGE</math> should be a straight line and therefore, E has a height of <math>9+4+4</math> = <math>\boxed {(D) 17}</math>.
+
Since the pillar at <math>B</math> has height <math>9</math> and the pillar at <math>A</math> has height <math>10</math> and the solar panel is flat, the inclination from pillar <math>A</math> to pillar <math>B</math> would be <math>1</math>. Call the center of the hexagon <math>G</math>. Since <math>\overline{CG}\parallel\overline{BA}</math>, it follows that <math>G</math> has a height of <math>13</math>. Since the solar panel is flat, <math>BGE</math> should be a straight line and therefore, <math>E</math> has a height of <math>9+4+4=\boxed{\textbf{(D) } 17}</math>.
  
 
~Arcticturn
 
~Arcticturn
  
−
==Solution 2==
+
==Solution 2 (Height From Each Vertex)==
  
−
Let the height of the pillar at <math>D</math> be <math>x.</math> Notice that the difference between the heights of pillar <math>C</math> and pillar <math>D</math> is equal to the difference between the heights of pillar <math>A</math> and pillar <math>F.</math> So, the height at <math>F</math> is <math>x+2.</math> Now, doing the same thing for pillar <math>E</math> we get the height is <math>x+3.</math> Therefore, we can see the difference between the heights at pillar <math>C</math> and pillar <math>D</math> is half the difference between the heights at <math>B</math> and <math>E,</math> so
+
Let the height of the pillar at <math>D</math> be <math>x.</math> Notice that the difference between the heights of pillar <math>C</math> and pillar <math>D</math> is equal to the difference between the heights of pillar <math>A</math> and pillar <math>F.</math> So, the height at <math>F</math> is <math>x+2.</math> Now, doing the same thing for pillar <math>E</math> we get the height is <math>x+3.</math> Therefore, we can see the difference between the heights at pillar <math>C</math> and pillar <math>D</math> is half the difference between the heights at <math>B</math> and <math>E,</math> so <cmath>x+3-9=2 \cdot (x-10) \implies x-6=2 \cdot (x-10) \implies x=14 \implies x+3=\boxed{\textbf{(D) } 17}.</cmath>
−
<cmath>x+3-9=2 \cdot (x-10) \implies x-6=2 \cdot (x-10) \implies x=14 \implies x+3=\boxed{17}.</cmath>
+
~kante314
−
 
 
−
- kante314
 
  
 
==Solution 3 (Extend the Sides)==
 
==Solution 3 (Extend the Sides)==
Line 62: Line 63:
 
~ihatemath123
 
~ihatemath123
  
−
== Solution 4 ==
+
== Solution 4 (Averages of Heights) ==
 
Denote by <math>h_X</math> the height of any point <math>X</math>.
 
Denote by <math>h_X</math> the height of any point <math>X</math>.
  
 
Denote by <math>M</math> the midpoint of <math>A</math> and <math>C</math>.
 
Denote by <math>M</math> the midpoint of <math>A</math> and <math>C</math>.
−
Hence, <math>h_M = \frac{h_A + h_C}{1} = 11</math>.
+
Hence, <cmath>h_M = \frac{h_A + h_C}{2} = 11.</cmath>
 +
Denote by <math>O</math> the center of <math>ABCDEF</math>. Because <math>ABCDEF</math> is a regular hexagon, <math>O</math> is the midpoint of <math>B</math> and <math>E</math>.
 +
Hence, <cmath>h_O = \frac{h_E + h_B}{2} = \frac{h_E + 9}{2}.</cmath>
 +
Because <math>ABCDEF</math> is a regular hexagon, <math>M</math> is the midpoint of <math>B</math> and <math>O</math>.
 +
Hence, <cmath>h_M = \frac{h_B + h_O}{2} = \frac{9 + h_O}{2}.</cmath>
 +
Solving these equations, we get <math>h_E = \boxed{\textbf{(D) } 17}</math>.
  
−
Denote by <math>O</math> the center of <math>ABCDEF</math>. Because <math>ABCDEF</math> is a right hexagon, <math>O</math> is the midpoint of <math>B</math> and <math>E</math>.
+
~Steven Chen (www.professorchenedu.com)
−
Hence, <math>h_O = \frac{h_E + h_B}{2} = \frac{h_E + 9}{2}</math>.
 
−
 
 
−
Because <math>ABCDEF</math> is a right hexagon, <math>M</math> is the midpoint of <math>B</math> and <math>O</math>.
 
−
Hence, <math>h_M = \frac{h_B + h_O}{2} = \frac{9 + h_O}{2}</math>.
 
−
 
 
−
Solving all these equations, we get <math>h_E = 17</math>.
 
−
 
 
−
Therefore, the answer is <math>\boxed{\textbf{(D) }17}</math>.
 
  
−
~Steven Chen (www.professorchenedu.com)
+
==Solution 5 (Vectors)==
  
−
==Solution 5 (Vectors) ==
+
==Solution 6 (Vectors)==
  
 
WLOG, let the side length of the hexagon be 6.  
 
WLOG, let the side length of the hexagon be 6.  
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Establish a 3D coordinate system, in which <math>A=(0,0,0)</math>. Let the coordinates of B and C be <math>(6,0,0)</math>, <math>(9,-3\sqrt{3},0)</math>, respectively. Then, the solar panel passes through <math>P=(0,0,12), Q=(6,0,9), R=(9,-3\sqrt{3},10)</math>.  
 
Establish a 3D coordinate system, in which <math>A=(0,0,0)</math>. Let the coordinates of B and C be <math>(6,0,0)</math>, <math>(9,-3\sqrt{3},0)</math>, respectively. Then, the solar panel passes through <math>P=(0,0,12), Q=(6,0,9), R=(9,-3\sqrt{3},10)</math>.  
  
−
The vector <math>\vec{PQ}=<6,0,-3></math> and <math>\vec{PR}=<9,-3\sqrt{3}, -2></math>. Computing <math>\vec{PQ} \times \vec{PR}</math> using the matrix
+
The vector <math>\vec{PQ}=\langle 6,0,-3\rangle</math> and <math>\vec{PR}=\left\langle 9,-3\sqrt{3}, -2\right\rangle</math>. Computing <math>\vec{PQ} \times \vec{PR}</math> using the matrix
−
 
+
<cmath>
−
<math>
 
 
\begin{bmatrix}
 
\begin{bmatrix}
 
i & j & k \\
 
i & j & k \\
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9 & -3\sqrt{3} & -2
 
9 & -3\sqrt{3} & -2
 
\end{bmatrix}
 
\end{bmatrix}
−
</math>
+
</cmath>
−
 
+
gives the result <math>-9\sqrt{3}i -15j -18\sqrt{3} k</math>. Therefore, a normal vector of the plane of the solar panel is <math>\left\langle -9\sqrt{3},-15,-18\sqrt{3}\right\rangle</math>, and the equation of the plane is <math> -9\sqrt{3}x-15y-18\sqrt{3}z=k</math>. Substituting <math>(x,y,z)=(0,0,12)</math>, we find that <math>k=-216\sqrt{3}</math>
−
gives the result <math>-9\sqrt{3}i -15j -18\sqrt{3} k</math>. Therefore, a normal vector of the plane of the solar panel is <math><-9\sqrt{3},-15,-18\sqrt{3}></math>, and the equation of the plane is <math> -9\sqrt{3}x-15y-18\sqrt{3}z=k</math>. Substituting <math>(x,y,z)=(0,0,12)</math>, we find that <math>k=-216\sqrt{3}</math>
 
−
 
 
−
Since <math>E=(0,-6\sqrt{3})</math>, we substitute <math>(x,y)=(0,-6\sqrt{3})</math> into <math>-9\sqrt{3}x-15y-18\sqrt{3}z=-216\sqrt{3}</math>, which gives <math>z=\boxed{\textbf{(D) }17}</math>.
 
  
 +
Since <math>E=(0,-6\sqrt{3})</math>, we substitute <math>(x,y)=(0,-6\sqrt{3})</math> into <math>-9\sqrt{3}x-15y-18\sqrt{3}z=-216\sqrt{3}</math>, which gives <math>z=\boxed{\textbf{(D) } 17}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=16|num-a=18}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:57, 30 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 $E$?

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

Diagrams

Three-Dimensional Diagram

Two-Dimensional 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 1 (Height From the Center)

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 $\overline{CG}\parallel\overline{BA}$, it follows that $G$ has a height of $13$. Since the solar panel is flat, $BGE$ should be a straight line and therefore, $E$ has a height of $9+4+4=\boxed{\textbf{(D) } 17}$.

~Arcticturn

Solution 2 (Height From Each Vertex)

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 \[x+3-9=2 \cdot (x-10) \implies x-6=2 \cdot (x-10) \implies x=14 \implies x+3=\boxed{\textbf{(D) } 17}.\] ~kante314

Solution 3 (Extend the Sides)

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, S); label("$H, 13$", H, S); 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 (Averages of Heights)

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

Denote by $M$ the midpoint of $A$ and $C$. Hence, \[h_M = \frac{h_A + h_C}{2} = 11.\] Denote by $O$ the center of $ABCDEF$. Because $ABCDEF$ is a regular 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 regular hexagon, $M$ is the midpoint of $B$ and $O$. Hence, \[h_M = \frac{h_B + h_O}{2} = \frac{9 + h_O}{2}.\] Solving these equations, we get $h_E = \boxed{\textbf{(D) } 17}$.

~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 $A=(0,0,0)$. 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 $\vec{PQ}=\langle 6,0,-3\rangle$ and $\vec{PR}=\left\langle 9,-3\sqrt{3}, -2\right\rangle$. 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 $\left\langle -9\sqrt{3},-15,-18\sqrt{3}\right\rangle$, 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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