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

Latest revision as of 19:28, 9 November 2025

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

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

Two-Dimensional Diagram

$[asy] /* Made by ihatemath123, Edited by MRENTHUSIASM */ path P; P = polygon(6); fill(P,yellow); draw(P); dot("$A,12$",dir(240),1.5*dir(240),linewidth(4.5)); dot("$B,9$",dir(300),1.5*dir(300),linewidth(4.5)); dot("$C,10$",dir(0),1.5*dir(0),linewidth(4.5)); dot("$D$",dir(60),1.5*dir(60),linewidth(4.5)); dot("$E$",dir(120),1.5*dir(120),linewidth(4.5)); dot("$F$",dir(180),1.5*dir(180),linewidth(4.5)); [/asy]$ ~ihatemath123 ~MRENTHUSIASM

Solution 1 (Height From the Center)

The pillar at $B$ has height $9$ and the pillar at $A$ has height $12.$ Since the solar panel is flat, the inclination from pillar $B$ to pillar $A$ is $3.$ Call the center of the hexagon $G.$ Since $\overrightarrow{CG}\parallel\overrightarrow{BA},$ it follows that the solar panel has height $13$ at $G.$ Since the solar panel is flat, the heights of the solar panel at $B,G,$ and $E$ are collinear. Therefore, the pillar at $E$ has height $9+4+4=\boxed{\textbf{(D) } 17}.$

~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$", H, S); draw(A--G, dashed); draw(C--H, dashed); dot(A^^B^^C^^D^^E^^F^^G^^H,linewidth(4.5)); [/asy]$ 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 as 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 $2\cdot3+2\cdot1=8.$

Together, 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 $z=\boxed{\textbf{(D) } 17}$ .

Solution 7 (3D Slopes)

Let the pillars be $AA', BB', \ldots, FF'$ . Since solar panel $A'B'C'D'E'F'$ is a hexagon, the line $B'E'$ hits the midpoint $M$ of $A'C'$ . So, the 3D slope (change in $x$ : change in $y$ : change in $z$ ) of $BE$ is same as $BD$ . If $a$ is side of the hexagonal solar panel, $B'M' = \frac{1}{2}a, B'E' = a+2\cdot \frac{1}{2}a = 2a$ . So, $B'M:B'E'$ = $1:4$ . Since the height of $M$ to ground $ABCDEF$ is $(10+12)/2 = 11$ , the rise (in z) from $B'$ to $M$ is 2 meaning the rise from $B'$ to $E'$ is $8$ . Thus, $EE' = 8+BB' = \boxed{\textbf{(D) } 17}$ .

~sml1809

Solution 8 (Midpoints, SIMPLE)

Set the midpoint of $\overline{AC}$ as $M$ : $[asy] unitsize(1cm); pair A = (-sqrt(3),1); pair B = (0,2); pair C = (sqrt(3),1); pair D = (sqrt(3),-1); pair E = (0,-2); pair F = (-sqrt(3),-1); pair M =(0,1.35); draw(A--B--C--D--E--F--cycle); label("$A, 12$", A, NW); label("$B, 9$", B, N); label("$C, 10$", C, NE); label("$D$", D, SE); label("$E$", E, S); label("$F$", F, SW); dot((0,1)); label("$M$", M); draw((-sqrt(3),1)--(sqrt(3),1),black); [/asy]$ We know that the height of $M$ is $11$ as it is the midpoint of $\overline{AC}$ , so the height is the average of $A$ and $C$ , which is $\frac{10 + 12}{2}= 11$ . Since $ABCDEF$ is a regular hexagon, $BE = 4\cdot BM$ . Because the increase in height is proportional to the length of the line segments, and the increase in height from $B$ to $M$ is $2$ , the increase in height from $B$ to $E$ is $2\cdot4=8.$ Adding to the height of $B$ , we get $8+9=\boxed{\textbf{(D) } 17}$ .

~iluvme

Solution 9 (Educated Guess)

Because the three points given are integers, it is likely that the answer is also an integer. This leaves us with $9$ or $17$ . Because both $A$ and $C$ are greater than $9$ and closer to $E$ than $B$ , we can assume that the height increases as the point gets closer to $E$ . Thus, we know the answer is greater than $9$ . The only choice that satisfies both these criteria is $\boxed{\textbf{(D) } 17}$ .

~iluvme

Video Solution (Under 2 min!)

https://youtu.be/lwEl3GxnysI

~Education, the Study of Everything

Video Solution by TheBeautyofMath

Inefficient, but it gets the job done https://youtu.be/SarZNOgo4DA

~IceMatrix

@@ Line 1: / Line 1: @@
 == Problem ==
-An architect is building a structure that will place vertical pillars at the vertices of regular hexagon <math>ABCDEF</math>, 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 <math>A</math>, <math>B</math>, and <math>C</math> are <math>12</math>, <math>9</math>, and <math>10</math> meters, respectively. What is the height, in meters, of the pillar at <math>E</math>?
+An architect is building a structure that will place vertical pillars at the vertices of regular hexagon <imath>ABCDEF</imath>, 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 <imath>A</imath>, <imath>B</imath>, and <imath>C</imath> are <imath>12</imath>, <imath>9</imath>, and <imath>10</imath> meters, respectively. What is the height, in meters, of the pillar at <imath>E</imath>?
-<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>
+<imath>\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}</imath>
 ==Diagrams==
@@ Line 61: / Line 61: @@
 ==Solution 1 (Height From the Center)==
-The pillar at <math>B</math> has height <math>9</math> and the pillar at <math>A</math> has height <math>12.</math> Since the solar panel is flat, the inclination from pillar <math>B</math> to pillar <math>A</math> is <math>3.</math> Call the center of the hexagon <math>G.</math> Since <math>\overrightarrow{CG}\parallel\overrightarrow{BA},</math> it follows that the solar panel has height <math>13</math> at <math>G.</math> Since the solar panel is flat, the heights of the solar panel at <math>B,G,</math> and <math>E</math> are collinear. Therefore, the pillar at <math>E</math> has height <math>9+4+4=\boxed{\textbf{(D) } 17}.</math>
+The pillar at <imath>B</imath> has height <imath>9</imath> and the pillar at <imath>A</imath> has height <imath>12.</imath> Since the solar panel is flat, the inclination from pillar <imath>B</imath> to pillar <imath>A</imath> is <imath>3.</imath> Call the center of the hexagon <imath>G.</imath> Since <imath>\overrightarrow{CG}\parallel\overrightarrow{BA},</imath> it follows that the solar panel has height <imath>13</imath> at <imath>G.</imath> Since the solar panel is flat, the heights of the solar panel at <imath>B,G,</imath> and <imath>E</imath> are collinear. Therefore, the pillar at <imath>E</imath> has height <imath>9+4+4=\boxed{\textbf{(D) } 17}.</imath>
 ~Arcticturn
@@ Line 67: / Line 67: @@
 ==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 <imath>D</imath> be <imath>x.</imath> Notice that the difference between the heights of pillar <imath>C</imath> and pillar <imath>D</imath> is equal to the difference between the heights of pillar <imath>A</imath> and pillar <imath>F.</imath> So, the height at <imath>F</imath> is <imath>x+2.</imath> Now, doing the same thing for pillar <imath>E</imath> we get the height is <imath>x+3.</imath> Therefore, we can see the difference between the heights at pillar <imath>C</imath> and pillar <imath>D</imath> is half the difference between the heights at <imath>B</imath> and <imath>E,</imath> so
 <cmath>\begin{align*}
 x+3-9&=2 \cdot (x-10) \\
@@ Line 73: / Line 73: @@
 x&=14.
 \end{align*}</cmath>
-The answer is <math>x+3=\boxed{\textbf{(D) } 17}.</math>
+The answer is <imath>x+3=\boxed{\textbf{(D) } 17}.</imath>
 ~kante314
 ==Solution 3 (Extend the Sides)==
-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 <imath>BA</imath> and <imath>BC</imath> to <imath>G</imath> and <imath>H</imath>, respectively, such that <imath>AG = CH</imath> and <imath>E</imath> lies on <imath>\overline{GH}</imath>:
 <asy>
 unitsize(1cm);
@@ Line 103: / Line 103: @@
 dot(A^^B^^C^^D^^E^^F^^G^^H,linewidth(4.5));
 </asy>
-Because of hexagon proportions, <math>\frac{BA}{AG} = \frac{1}{3}</math> and <math>\frac{BC}{CH} = \frac{1}{3}</math>. Let <math>g</math> be the height of <math>G</math>. Because <math>A</math>, <math>B</math> and <math>G</math> lie on the same line, <math>\frac{12-9}{g-12} = \frac{1}{3}</math>, so <math>g-12 = 9</math> and <math>g = 21</math>. Similarly, the height of <math>H</math> is <math>13</math>. <math>E</math> is the midpoint of <math>GH</math>, so we can take the average of these heights to get our answer, <math>\boxed{\textbf{(D) } 17}</math>.
+Because of hexagon proportions, <imath>\frac{BA}{AG} = \frac{1}{3}</imath> and <imath>\frac{BC}{CH} = \frac{1}{3}</imath>. Let <imath>g</imath> be the height of <imath>G</imath>. Because <imath>A</imath>, <imath>B</imath> and <imath>G</imath> lie on the same line, <imath>\frac{12-9}{g-12} = \frac{1}{3}</imath>, so <imath>g-12 = 9</imath> and <imath>g = 21</imath>. Similarly, the height of <imath>H</imath> is <imath>13</imath>. <imath>E</imath> is the midpoint of <imath>GH</imath>, so we can take the average of these heights to get our answer, <imath>\boxed{\textbf{(D) } 17}</imath>.
 ~ihatemath123
 == Solution 4 (Averages of Heights) ==
-Denote by <math>h_X</math> the height of any point <math>X</math>.
+Denote by <imath>h_X</imath> the height of any point <imath>X</imath>.
-Denote by <math>M</math> the midpoint of <math>A</math> and <math>C</math>.
+Denote by <imath>M</imath> the midpoint of <imath>A</imath> and <imath>C</imath>.
 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>.
+Denote by <imath>O</imath> the center of <imath>ABCDEF</imath>. Because <imath>ABCDEF</imath> is a regular hexagon, <imath>O</imath> is the midpoint of <imath>B</imath> and <imath>E</imath>.
 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>.
+Because <imath>ABCDEF</imath> is a regular hexagon, <imath>M</imath> is the midpoint of <imath>B</imath> and <imath>O</imath>.
 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>.
+Solving these equations, we get <imath>h_E = \boxed{\textbf{(D) } 17}</imath>.
 ~Steven Chen (www.professorchenedu.com)
 ==Solution 5 (Vectors)==
-In this solution, we define <b>rise</b> as the change of height (in meters) from the solar panel to the ground. It follows that the rise from <math>B</math> to <math>A</math> is <math>12-9=3,</math> and the rise from <math>B</math> to <math>C</math> is <math>10-9=1.</math> Note that <math>\vec{BE}=2\vec{BA}+2\vec{BC},</math> so the rise from <math>B</math> to <math>E</math> is <math>2\cdot3+2\cdot1=8.</math>
+In this solution, we define <b>rise</b> as the change of height (in meters) from the solar panel to the ground. It follows that the rise from <imath>B</imath> to <imath>A</imath> is <imath>12-9=3,</imath> and the rise from <imath>B</imath> to <imath>C</imath> is <imath>10-9=1.</imath> Note that <imath>\vec{BE}=2\vec{BA}+2\vec{BC},</imath> so the rise from <imath>B</imath> to <imath>E</imath> is <imath>2\cdot3+2\cdot1=8.</imath>
-Together, the height of the pillar at <math>E</math> is <math>9+8=\boxed{\textbf{(D) } 17}</math> meters.
+Together, the height of the pillar at <imath>E</imath> is <imath>9+8=\boxed{\textbf{(D) } 17}</imath> meters.
 ~MRENTHUSIASM
@@ Line 129: / Line 129: @@
 ==Solution 6 (Vectors)==
-WLOG, let the side length of the hexagon be <math>6</math>.
+WLOG, let the side length of the hexagon be <imath>6</imath>.
-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>.
+Establish a 3D coordinate system, in which <imath>A=(0,0,0)</imath>. Let the coordinates of <imath>B</imath> and <imath>C</imath> be <imath>(6,0,0)</imath>, <imath>\left(9,-3\sqrt{3},0\right)</imath>, respectively. Then, the solar panel passes through <imath>P=(0,0,12), Q=(6,0,9), R=\left(9,-3\sqrt{3},10\right)</imath>.
-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
+The vector <imath>\vec{PQ}=\langle 6,0,-3\rangle</imath> and <imath>\vec{PR}=\left\langle 9,-3\sqrt{3}, -2\right\rangle</imath>. Computing <imath>\vec{PQ} \times \vec{PR}</imath> by the matrix
 <cmath>
 \begin{bmatrix}
@@ Line 141: / Line 141: @@
 \end{bmatrix}
 </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 <imath>-9\sqrt{3}i -15j -18\sqrt{3} k</imath>. Therefore, a normal vector of the plane of the solar panel is <imath>\left\langle -9\sqrt{3},-15,-18\sqrt{3}\right\rangle</imath>, and the equation of the plane is <imath> -9\sqrt{3}x-15y-18\sqrt{3}z=k</imath>. Substituting <imath>(x,y,z)=(0,0,12)</imath>, we find that <imath>k=-216\sqrt{3}</imath>.
-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>.
+Since <imath>E=\left(0,-6\sqrt{3}\right)</imath>, we substitute <imath>(x,y)=\left(0,-6\sqrt{3}\right)</imath> into <imath>-9\sqrt{3}x-15y-18\sqrt{3}z=-216\sqrt{3}</imath>, which gives <imath>z=\boxed{\textbf{(D) } 17}</imath>.
 ==Solution 7 (3D Slopes)==
-Let the pillars be <math>AA', BB', \ldots, FF'</math>. Since solar panel <math>A'B'C'D'E'F'</math> is a hexagon, the line <math>B'E'</math> hits the midpoint <math>M</math> of <math>A'C'</math>. So, the 3D slope (change in <math>x</math> : change in <math>y</math> : change in <math>z</math>) of <math>BE</math> is same as <math>BD</math>. If <math>a</math> is side of the hexagonal solar panel, <cmath>B'M' = \frac{1}{2}a, B'E' = a+2\cdot \frac{1}{2}a = 2a</cmath>. So, <math>B'M:B'E'</math> = <math>1:4</math>. Since the height of <math>M</math> to ground <math>ABCDEF</math> is <math>(10+12)/2 = 11</math>, the rise (in z) from <math>B'</math> to <math>M</math> is 2 meaning the rise from <math>B'</math> to <math>E'</math> is <math>8</math>. Thus, <math>EE' = 8+BB' = \boxed{\textbf{(D) } 17}</math>.
+Let the pillars be <imath>AA', BB', \ldots, FF'</imath>. Since solar panel <imath>A'B'C'D'E'F'</imath> is a hexagon, the line <imath>B'E'</imath> hits the midpoint <imath>M</imath> of <imath>A'C'</imath>. So, the 3D slope (change in <imath>x</imath> : change in <imath>y</imath> : change in <imath>z</imath>) of <imath>BE</imath> is same as <imath>BD</imath>. If <imath>a</imath> is side of the hexagonal solar panel, <cmath>B'M' = \frac{1}{2}a, B'E' = a+2\cdot \frac{1}{2}a = 2a</cmath>. So, <imath>B'M:B'E'</imath> = <imath>1:4</imath>. Since the height of <imath>M</imath> to ground <imath>ABCDEF</imath> is <imath>(10+12)/2 = 11</imath>, the rise (in z) from <imath>B'</imath> to <imath>M</imath> is 2 meaning the rise from <imath>B'</imath> to <imath>E'</imath> is <imath>8</imath>. Thus, <imath>EE' = 8+BB' = \boxed{\textbf{(D) } 17}</imath>.
 ~sml1809
-==Solution 8 (Midpoints)==
+==Solution 8 (Midpoints, SIMPLE)==
-Set the midpoint of  <math>\overline{AC}</math> as <math>M</math>:
+Set the midpoint of  <imath>\overline{AC}</imath> as <imath>M</imath>:
 <asy>
 unitsize(1cm);
@@ Line 175: / Line 175: @@
 draw((-sqrt(3),1)--(sqrt(3),1),black);
 </asy>
-We know that the height of <math>M</math> is <math>11</math> as it is the midpoint of <math>\overline{AC}</math>, so the height is the average of <math>A</math> and <math>C</math>, which is <math>\frac{10 + 12}{2}= 11</math>. Since <math>ABCDEF</math> is a regular hexagon, <math>BE = 4\cdot BM</math>. Because the increase in height is proportional to the length of the line segments, and the increase in height from <math>B</math> to <math>M</math> is <math>2</math>, the increase in height from <math>B</math> to <math>E</math> is <math>2\cdot4=8.</math> Adding to the height of <math>B</math>, we get <math>8+9=\boxed{\textbf{(D) } 17}</math>.
+We know that the height of <imath>M</imath> is <imath>11</imath> as it is the midpoint of <imath>\overline{AC}</imath>, so the height is the average of <imath>A</imath> and <imath>C</imath>, which is <imath>\frac{10 + 12}{2}= 11</imath>. Since <imath>ABCDEF</imath> is a regular hexagon, <imath>BE = 4\cdot BM</imath>. Because the increase in height is proportional to the length of the line segments, and the increase in height from <imath>B</imath> to <imath>M</imath> is <imath>2</imath>, the increase in height from <imath>B</imath> to <imath>E</imath> is <imath>2\cdot4=8.</imath> Adding to the height of <imath>B</imath>, we get <imath>8+9=\boxed{\textbf{(D) } 17}</imath>.
 ~iluvme
@@ Line 181: / Line 181: @@
 ==Solution 9 (Educated Guess)==
-Because the three points given are integers, it is likely that the answer is also an integer. This leaves us with <math>9</math> or <math>17</math>. Because both <math>A</math> and <math>C</math> are greater than <math>9</math> and closer to <math>E</math> than <math>B</math>, we can assume that the height increases as the point gets closer to <math>E</math>. Thus, we know the answer is greater than <math>9</math>. The only choice that satisfies both these criteria is <math>\boxed{\textbf{(D) } 17}</math>.
+Because the three points given are integers, it is likely that the answer is also an integer. This leaves us with <imath>9</imath> or <imath>17</imath>. Because both <imath>A</imath> and <imath>C</imath> are greater than <imath>9</imath> and closer to <imath>E</imath> than <imath>B</imath>, we can assume that the height increases as the point gets closer to <imath>E</imath>. Thus, we know the answer is greater than <imath>9</imath>. The only choice that satisfies both these criteria is <imath>\boxed{\textbf{(D) } 17}</imath>.
 ~iluvme
+==Video Solution (Under 2 min!)==
+https://youtu.be/lwEl3GxnysI
+<i>~Education, the Study of Everything </i>
 ==Video Solution by TheBeautyofMath==

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