Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2 (Height From Each Vertex))
m (→‎Solution 6 (Vectors))
Line 129: Line 129:
 
==Solution 6 (Vectors)==
 
==Solution 6 (Vectors)==
  
−
WLOG, let the side length of the hexagon be 6.  
+
WLOG, let the side length of the hexagon be <math>6</math>.  
  
−
Establish a 3D coordinate system, in which <math>A=(0,0,0)</math>. Let the coordinates of <math>B</math> and <math>C</math> 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 <math>B</math> and <math>C</math> be <math>(6,0,0)</math>, <math>\left(9,-3\sqrt{3},0\right)</math>, respectively. Then, the solar panel passes through <math>P=(0,0,12), Q=(6,0,9), R=\left(9,-3\sqrt{3},10\right)</math>.  
  
−
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
+
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> by the matrix
 
<cmath>
 
<cmath>
 
\begin{bmatrix}
 
\begin{bmatrix}
Line 141: Line 141:
 
\end{bmatrix}
 
\end{bmatrix}
 
</cmath>
 
</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>\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>.
  
−
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=\left(0,-6\sqrt{3}\right)</math>, we substitute <math>(x,y)=\left(0,-6\sqrt{3}\right)</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 14:11, 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

[asy] /* Made by MRENTHUSIASM */ size(225); import graph3; import solids; currentprojection=orthographic((0.1,-0.7,0.2)); triple A, B, C, D, E, F, A1, B1, C1, D1, E1, F1; A = (7.5*Cos(240),7.5*Sin(240),0); B = (7.5*Cos(300),7.5*Sin(300),0); C = (7.5*Cos(0),7.5*Sin(0),0); D = (7.5*Cos(60),7.5*Sin(60),0); E = (7.5*Cos(120),7.5*Sin(120),0); F = (7.5*Cos(180),7.5*Sin(180),0); A1 = (7.5*Cos(240),7.5*Sin(240),12); B1 = (7.5*Cos(300),7.5*Sin(300),9); C1 = (7.5*Cos(0),7.5*Sin(0),10); D1 = (7.5*Cos(60),7.5*Sin(60),14); E1 = (7.5*Cos(120),7.5*Sin(120),17); F1 = (7.5*Cos(180),7.5*Sin(180),16); draw(surface(A--B--C--D--E--F--cycle),yellow); draw(surface(A1--B1--C1--D1--E1--F1--cycle),green); draw(A--A1^^B--B1^^C--C1,red); dot("$A$",A,A/3.75,linewidth(4.5)); dot("$B$",B,B/3.75,linewidth(4.5)); dot("$C$",C,C/3.75,linewidth(4.5)); dot("$D$",D,D/3.75,linewidth(4.5)); dot("$E$",E,E/3.75,linewidth(4.5)); dot("$F$",F,F/3.75,linewidth(4.5)); dot(A1^^B1^^C1^^D1^^E1^^F1,linewidth(4.5)); draw(A--B--C--D--E--F--cycle^^A1--B1--C1--D1--E1--F1--cycle); draw(D--D1^^E--E1^^F--F1); label("$12$",midpoint(A--A1),1.5*W,red); label("$9$",midpoint(B--B1),1.5*W,red); label("$10$",midpoint(C--C1),1.5*(1,0),red); [/asy] ~MRENTHUSIASM

Two-Dimensional Diagram

[asy] /* Made by ihatemath123, Edited by MRENTHUSIASM */ path P; P = rotate(30)*polygon(6); draw(P); label("$A,12$",dir(150),1.5*dir(150)); label("$B,9$",dir(90),1.5*dir(90)); label("$C,10$",dir(30),1.5*dir(30)); label("$D$",dir(-30),1.5*dir(-30)); label("$E$",dir(-90),1.5*dir(-90)); label("$F$",dir(-150),1.5*dir(-150)); dot(dir(150)^^dir(90)^^dir(30)^^dir(-30)^^dir(-90)^^dir(-150)); [/asy]

~ihatemath123 ~MRENTHUSIASM

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 \begin{align*} x+3-9&=2 \cdot (x-10) \\ x-6&=2 \cdot (x-10) \\ x&=14. \end{align*} The answer is $x+3=\boxed{\textbf{(D) } 17}.$

~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)

In this solution, we define rise to be the change of height (in meters) from the solar panel to the ground. It follows that the rise from $B$ to $A$ is $12-9=3,$ and the rise from $B$ to $C$ is $10-9=1.$ Note that $\vec{BE}=2\vec{BA}+2\vec{BC},$ so the rise from $B$ to $E$ is $2\cdot3+2\cdot1=8.$

Finally, the height of the pillar at $E$ is $9+8=\boxed{\textbf{(D) } 17}$ meters.

~MRENTHUSIASM

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 $(6,0,0)$, $\left(9,-3\sqrt{3},0\right)$, respectively. Then, the solar panel passes through $P=(0,0,12), Q=(6,0,9), R=\left(9,-3\sqrt{3},10\right)$.

The vector $\vec{PQ}=\langle 6,0,-3\rangle$ and $\vec{PR}=\left\langle 9,-3\sqrt{3}, -2\right\rangle$. Computing $\vec{PQ} \times \vec{PR}$ by 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 $E=\left(0,-6\sqrt{3}\right)$, we substitute $(x,y)=\left(0,-6\sqrt{3}\right)$ 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_17&oldid=167119"