Difference between revisions of "2021 Fall AMC 10A Problems/Problem 17"
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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=== | ||
<asy> | <asy> | ||
− | + | /* Made by MRENTHUSIASM */ | |
− | + | size(225); | |
− | + | import graph3; | |
− | + | import solids; | |
− | + | ||
− | + | currentprojection=orthographic((0.1,-0.7,0.2)); | |
− | draw(A--B--C--D--E--F-- | + | triple A, B, C, D, E, F, A1, B1, C1, D1, E1, F1; |
− | label("$A, 12$", | + | 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> | </asy> | ||
+ | ~ihatemath123 ~MRENTHUSIASM | ||
− | ==Solution== | + | ==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> | |
~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) \ | + | <cmath>\begin{align*} |
+ | x+3-9&=2 \cdot (x-10) \\ | ||
+ | x-6&=2 \cdot (x-10) \\ | ||
+ | x&=14. | ||
+ | \end{align*}</cmath> | ||
+ | The answer is <math>x+3=\boxed{\textbf{(D) } 17}.</math> | ||
− | + | ~kante314 | |
==Solution 3 (Extend the Sides)== | ==Solution 3 (Extend the Sides)== | ||
Line 57: | Line 101: | ||
draw(A--G, dashed); | draw(A--G, dashed); | ||
draw(C--H, dashed); | draw(C--H, dashed); | ||
+ | dot(A^^B^^C^^D^^E^^F^^G^^H,linewidth(4.5)); | ||
</asy> | </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, <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>. | ||
Line 62: | Line 107: | ||
~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, < | + | 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>. | ||
− | + | ~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> | ||
− | + | Together, the height of the pillar at <math>E</math> is <math>9+8=\boxed{\textbf{(D) } 17}</math> meters. | |
− | + | ~MRENTHUSIASM | |
− | + | ==Solution 6 (Vectors)== | |
− | + | 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>\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>. | |
− | <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> by the matrix |
+ | <cmath> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
i & j & k \\ | i & j & k \\ | ||
Line 94: | Line 140: | ||
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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ~sml1809 | ||
+ | |||
+ | ==Solution 8 (Midpoints)== | ||
+ | |||
+ | Set the midpoint of <math>\overline{AC}</math> as <math>M</math>: | ||
+ | <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 <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>. | ||
+ | |||
+ | ~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 <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>. | ||
+ | |||
+ | ~iluvme | ||
+ | |||
+ | |||
+ | ==Video Solution (Under 2 min!)== | ||
+ | https://youtu.be/lwEl3GxnysI | ||
− | + | <i>~Education, the Study of Everything </i> | |
− | + | ==Video Solution by TheBeautyofMath== | |
+ | Inefficient, but it gets the job done | ||
+ | https://youtu.be/SarZNOgo4DA | ||
+ | ~IceMatrix | ||
==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}} |
Latest revision as of 00:13, 17 July 2023
Contents
- 1 Problem
- 2 Diagrams
- 3 Solution 1 (Height From the Center)
- 4 Solution 2 (Height From Each Vertex)
- 5 Solution 3 (Extend the Sides)
- 6 Solution 4 (Averages of Heights)
- 7 Solution 5 (Vectors)
- 8 Solution 6 (Vectors)
- 9 Solution 7 (3D Slopes)
- 10 Solution 8 (Midpoints)
- 11 Solution 9 (Educated Guess)
- 12 Video Solution (Under 2 min!)
- 13 Video Solution by TheBeautyofMath
- 14 See Also
Problem
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon , 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
,
, and
are
,
, and
meters, respectively. What is the height, in meters, of the pillar at
?
Diagrams
Three-Dimensional Diagram
~MRENTHUSIASM
Two-Dimensional Diagram
~ihatemath123 ~MRENTHUSIASM
Solution 1 (Height From the Center)
The pillar at has height
and the pillar at
has height
Since the solar panel is flat, the inclination from pillar
to pillar
is
Call the center of the hexagon
Since
it follows that the solar panel has height
at
Since the solar panel is flat, the heights of the solar panel at
and
are collinear. Therefore, the pillar at
has height
~Arcticturn
Solution 2 (Height From Each Vertex)
Let the height of the pillar at be
Notice that the difference between the heights of pillar
and pillar
is equal to the difference between the heights of pillar
and pillar
So, the height at
is
Now, doing the same thing for pillar
we get the height is
Therefore, we can see the difference between the heights at pillar
and pillar
is half the difference between the heights at
and
so
The answer is
~kante314
Solution 3 (Extend the Sides)
We can extend and
to
and
, respectively, such that
and
lies on
:
Because of hexagon proportions,
and
. Let
be the height of
. Because
,
and
lie on the same line,
, so
and
. Similarly, the height of
is
.
is the midpoint of
, so we can take the average of these heights to get our answer,
.
~ihatemath123
Solution 4 (Averages of Heights)
Denote by the height of any point
.
Denote by the midpoint of
and
.
Hence,
Denote by
the center of
. Because
is a regular hexagon,
is the midpoint of
and
.
Hence,
Because
is a regular hexagon,
is the midpoint of
and
.
Hence,
Solving these equations, we get
.
~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 to
is
and the rise from
to
is
Note that
so the rise from
to
is
Together, the height of the pillar at is
meters.
~MRENTHUSIASM
Solution 6 (Vectors)
WLOG, let the side length of the hexagon be .
Establish a 3D coordinate system, in which . Let the coordinates of
and
be
,
, respectively. Then, the solar panel passes through
.
The vector and
. Computing
by the matrix
gives the result
. Therefore, a normal vector of the plane of the solar panel is
, and the equation of the plane is
. Substituting
, we find that
.
Since , we substitute
into
, which gives
.
Solution 7 (3D Slopes)
Let the pillars be . Since solar panel
is a hexagon, the line
hits the midpoint
of
. So, the 3D slope (change in
: change in
: change in
) of
is same as
. If
is side of the hexagonal solar panel,
. So,
=
. Since the height of
to ground
is
, the rise (in z) from
to
is 2 meaning the rise from
to
is
. Thus,
.
~sml1809
Solution 8 (Midpoints)
Set the midpoint of as
:
We know that the height of
is
as it is the midpoint of
, so the height is the average of
and
, which is
. Since
is a regular hexagon,
. Because the increase in height is proportional to the length of the line segments, and the increase in height from
to
is
, the increase in height from
to
is
Adding to the height of
, we get
.
~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 or
. Because both
and
are greater than
and closer to
than
, we can assume that the height increases as the point gets closer to
. Thus, we know the answer is greater than
. The only choice that satisfies both these criteria is
.
~iluvme
Video Solution (Under 2 min!)
~Education, the Study of Everything
Video Solution by TheBeautyofMath
Inefficient, but it gets the job done https://youtu.be/SarZNOgo4DA
~IceMatrix
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.