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 1 (Height From the Center))
(→‎Solution 9 (Educated Guess))
 
(49 intermediate revisions by 7 users not shown)
Line 27: Line 27:
 
F1 = (7.5*Cos(180),7.5*Sin(180),16);
 
F1 = (7.5*Cos(180),7.5*Sin(180),16);
 
draw(surface(A--B--C--D--E--F--cycle),yellow);
 
draw(surface(A--B--C--D--E--F--cycle),yellow);
−
draw(surface(A1--B1--C1--D1--E1--F1--cycle),green);
+
draw(surface(A1--B1--C1--D1--E1--F1--cycle),red);
−
draw(A--A1^^B--B1^^C--C1,red);
+
draw(A--A1^^B--B1^^C--C1,blue);
 
dot("$A$",A,A/3.75,linewidth(4.5));
 
dot("$A$",A,A/3.75,linewidth(4.5));
 
dot("$B$",B,B/3.75,linewidth(4.5));
 
dot("$B$",B,B/3.75,linewidth(4.5));
Line 36: Line 36:
 
dot("$F$",F,F/3.75,linewidth(4.5));
 
dot("$F$",F,F/3.75,linewidth(4.5));
 
dot(A1^^B1^^C1^^D1^^E1^^F1,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(F--F1^^F--A--B--C^^A1--B1--C1--D1--E1--F1--cycle);
−
draw(D--D1^^E--E1^^F--F1);
+
draw(C--D--E--F^^D--D1^^E--E1,dashed);
−
label("$12$",midpoint(A--A1),1.5*W,red);
+
label("$12$",midpoint(A--A1),1.5*W,blue);
−
label("$9$",midpoint(B--B1),1.5*W,red);
+
label("$9$",midpoint(B--B1),1.5*W,blue);
−
label("$10$",midpoint(C--C1),1.5*(1,0),red);
+
label("$10$",midpoint(C--C1),1.5*(1,0),blue);
 
</asy>
 
</asy>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
Line 48: Line 48:
 
/* Made by ihatemath123, Edited by MRENTHUSIASM */
 
/* Made by ihatemath123, Edited by MRENTHUSIASM */
 
path P;
 
path P;
−
P = rotate(30)*polygon(6);
+
P = polygon(6);
 +
fill(P,yellow);
 
draw(P);
 
draw(P);
−
label("$A,12$",dir(150),1.5*dir(150));
+
dot("$A,12$",dir(240),1.5*dir(240),linewidth(4.5));
−
label("$B,9$",dir(90),1.5*dir(90));
+
dot("$B,9$",dir(300),1.5*dir(300),linewidth(4.5));
−
label("$C,10$",dir(30),1.5*dir(30));
+
dot("$C,10$",dir(0),1.5*dir(0),linewidth(4.5));
−
label("$D$",dir(-30),1.5*dir(-30));
+
dot("$D$",dir(60),1.5*dir(60),linewidth(4.5));
−
label("$E$",dir(-90),1.5*dir(-90));
+
dot("$E$",dir(120),1.5*dir(120),linewidth(4.5));
−
label("$F$",dir(-150),1.5*dir(-150));
+
dot("$F$",dir(180),1.5*dir(180),linewidth(4.5));
−
dot(dir(150)^^dir(90)^^dir(30)^^dir(-30)^^dir(-90)^^dir(-150));
 
 
</asy>
 
</asy>
−
 
 
~ihatemath123 ~MRENTHUSIASM
 
~ihatemath123 ~MRENTHUSIASM
  
 
==Solution 1 (Height From the Center)==
 
==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>12</math> and the solar panel is flat, the inclination from pillar <math>B</math> to pillar <math>A</math> would be <math>3</math>. Call the center of the hexagon <math>G</math>. Since <math>\overrightarrow{CG}\parallel\overrightarrow{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>.
+
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
Line 102: 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 121: Line 121:
  
 
==Solution 5 (Vectors)==
 
==Solution 5 (Vectors)==
−
In this solution, we define <b>rise</b> to be 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 <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>
  
−
Finally, 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 <math>E</math> is <math>9+8=\boxed{\textbf{(D) } 17}</math> meters.
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
Line 144: Line 144:
  
 
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 <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, SIMPLE)==
 +
 +
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 00:52, 17 September 2024

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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_17&oldid=227379"