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 "2013 AMC 10B Problems/Problem 22"
 
(32 intermediate revisions by 20 users not shown)
Line 3: Line 3:
 
The regular octagon <math>ABCDEFGH</math> has its center at <math>J</math>.  Each of the vertices and the center are to be associated with one of the digits <math>1</math> through <math>9</math>, with each digit used once, in such a way that the sums of the numbers on the lines <math>AJE</math>, <math>BJF</math>, <math>CJG</math>, and <math>DJH</math> are all equal.  In how many ways can this be done?
 
The regular octagon <math>ABCDEFGH</math> has its center at <math>J</math>.  Each of the vertices and the center are to be associated with one of the digits <math>1</math> through <math>9</math>, with each digit used once, in such a way that the sums of the numbers on the lines <math>AJE</math>, <math>BJF</math>, <math>CJG</math>, and <math>DJH</math> are all equal.  In how many ways can this be done?
  
−
<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 </math>
+
<math> \textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456 </math>
  
−
==Solution==
+
<asy>
 +
pair A,B,C,D,E,F,G,H,J;
 +
A=(20,20(2+sqrt(2)));
 +
B=(20(1+sqrt(2)),20(2+sqrt(2)));
 +
C=(20(2+sqrt(2)),20(1+sqrt(2)));
 +
D=(20(2+sqrt(2)),20);
 +
E=(20(1+sqrt(2)),0);
 +
F=(20,0);
 +
G=(0,20);
 +
H=(0,20(1+sqrt(2)));
 +
J=(10(2+sqrt(2)),10(2+sqrt(2)));
 +
draw(A--B);
 +
draw(B--C);
 +
draw(C--D);
 +
draw(D--E);
 +
draw(E--F);
 +
draw(F--G);
 +
draw(G--H);
 +
draw(H--A);
 +
dot(A);
 +
dot(B);
 +
dot(C);
 +
dot(D);
 +
dot(E);
 +
dot(F);
 +
dot(G);
 +
dot(H);
 +
dot(J);
 +
label("A",A,NNW);
 +
label("B",B,NNE);
 +
label("C",C,ENE);
 +
label("D",D,ESE);
 +
label("E",E,SSE);
 +
label("F",F,SSW);
 +
label("G",G,WSW);
 +
label("H",H,WNW);
 +
label("J",J,SE);
 +
</asy>
  
−
First of all, note that <math>J</math> must be <math>1</math>, <math>5</math>, or <math>9</math> to preserve symmetry. We also notice that <math>A+E = B+F = C+G = D+H</math>. 
+
==Solution 1==
 +
First of all, note that <math>J</math> must be <math>1</math>, <math>5</math>, or <math>9</math> to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:
  
−
WLOG assume that <math>J = 1</math>.  Thus the pairs of vertices must be <math>9</math> and <math>2</math>, <math>8</math> and <math>3</math>, <math>7</math> and <math>4</math>, and <math>6</math> and <math>5</math>.  There are <math>4! = 24</math> ways to assign these to the vertices.  Furthermore, there are <math>2^{4} = 16</math> ways to switch them (i.e. do <math>2</math> <math>9</math> instead of <math>9</math> <math>2</math>).   
+
<asy>
 +
pair A,B,C,D,E,F,G,H,J;
 +
A=(20,20(2+sqrt(2)));
 +
B=(20(1+sqrt(2)),20(2+sqrt(2)));
 +
C=(20(2+sqrt(2)),20(1+sqrt(2)));
 +
D=(20(2+sqrt(2)),20);
 +
E=(20(1+sqrt(2)),0);
 +
F=(20,0);
 +
G=(0,20);
 +
H=(0,20(1+sqrt(2)));
 +
J=(10(2+sqrt(2)),10(2+sqrt(2)));
 +
draw(A--B);
 +
draw(B--C);
 +
draw(C--D);
 +
draw(D--E);
 +
draw(E--F);
 +
draw(F--G);
 +
draw(G--H);
 +
draw(H--A);
 +
dot(A);
 +
dot(B);
 +
dot(C);
 +
dot(D);
 +
dot(E);
 +
dot(F);
 +
dot(G);
 +
dot(H);
 +
dot(J);
 +
label("A",A,NNW);
 +
label("B",B,NNE);
 +
label("C",C,ENE);
 +
label("D",D,ESE);
 +
label("E",E,SSE);
 +
label("F",F,SSW);
 +
label("G",G,WSW);
 +
label("H",H,WNW);
 +
label("J $(1, 5, 9)$",J,SE);
 +
</asy>
 +
 
 +
We also notice that <math>A+E = B+F = C+G = D+H</math>. 
 +
 
 +
WLOG, assume that <math>J = 1</math>.  Thus the pairs of vertices must be <math>9</math> and <math>2</math>, <math>8</math> and <math>3</math>, <math>7</math> and <math>4</math>, and <math>6</math> and <math>5</math>.  There are <math>4! = 24</math> ways to assign these to the vertices.  Furthermore, there are <math>2^{4} = 16</math> ways to switch them (i.e. do <math>2</math> <math>9</math> instead of <math>9</math> <math>2</math>).   
  
 
Thus, there are <math>16(24) = 384</math> ways for each possible J value.  There are <math>3</math> possible J values that still preserve symmetry: <math>384(3) = \boxed{\textbf{(C) }1152}</math>
 
Thus, there are <math>16(24) = 384</math> ways for each possible J value.  There are <math>3</math> possible J values that still preserve symmetry: <math>384(3) = \boxed{\textbf{(C) }1152}</math>
 +
 +
==Solution 2==
 +
 +
As in solution 1, <math>J</math> must be <math>1</math>, <math>5</math>, or <math>9</math> giving us 3 choices. Additionally <math>A+E = B+F = C+G = D+H</math>.  This means once we choose <math>J</math> there are <math>8</math> remaining choices. Going clockwise from <math>A</math> we count, <math>8</math> possibilities for <math>A</math>. Choosing <math>A</math> also determines <math>E</math> which leaves <math>6</math> choices for <math>B</math>, once <math>B</math> is chosen it also determines  <math>F</math> leaving <math>4</math> choices for <math>C</math>. Once <math>C</math> is chosen it determines <math>G</math> leaving <math>2</math> choices for <math>D</math>. Choosing <math>D</math> determines <math>H</math>, exhausting the numbers. Additionally, there are three possible values for <math>J</math>. To get the answer we multiply <math>2\cdot4\cdot6\cdot8\cdot3=\boxed{\textbf{(C) }1152}</math>.
 +
 +
==Remark==
 +
 +
Solutions 1 and 2 state that <math>J=1, 5, 9</math> without rigorous analysis. Here it is:
 +
 +
Let <math>S=A+J+E=B+J+F=C+J+G=D+J+H</math>
 +
<math>4S=A+B+C+D+E+F+G+H+4J</math>
 +
<math>4S=45+3J</math>
 +
<math>4S=3(15+J)</math>
 +
 +
Because <math>gcd(4,3)=1</math>, so <math>4|(15+J)</math>, but <math>15 \equiv 3 \pmod 4</math>, so <math>J \equiv 1 \pmod 4</math>, <math>J=1,5,9</math>
 +
 +
When <math>J=1, S=12, A+E=B+F=C+G=D+H=11</math>
 +
When <math>J=5, S=15, A+E=B+F=C+G=D+H=10</math>
 +
When <math>J=9, S=18, A+E=B+F=C+G=D+H=9</math>
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 +
==Video Solution by Pi Academy==
 +
 +
https://youtu.be/0djZt1Fvuvw?si=TnPgQi3DnrI5IsE8
 +
 +
~ Pi Academy
 +
 +
==Video Solution 2==
 +
https://youtu.be/3MDr_t1ZzQk
 +
 +
~IceMatrix
 +
 +
== See also ==
 +
{{AMC10 box|year=2013|ab=B|num-b=21|num-a=23}}
 +
 +
[[Category:Introductory Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 21:24, 10 October 2024

Problem

The regular octagon $ABCDEFGH$ has its center at $J$. Each of the vertices and the center are to be associated with one of the digits $1$ through $9$, with each digit used once, in such a way that the sums of the numbers on the lines $AJE$, $BJF$, $CJG$, and $DJH$ are all equal. In how many ways can this be done?

$\textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456$

[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label("A",A,NNW); label("B",B,NNE); label("C",C,ENE); label("D",D,ESE); label("E",E,SSE); label("F",F,SSW); label("G",G,WSW); label("H",H,WNW); label("J",J,SE); [/asy]

Solution 1

First of all, note that $J$ must be $1$, $5$, or $9$ to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:

[asy] pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label("A",A,NNW); label("B",B,NNE); label("C",C,ENE); label("D",D,ESE); label("E",E,SSE); label("F",F,SSW); label("G",G,WSW); label("H",H,WNW); label("J $(1, 5, 9)$",J,SE); [/asy]

We also notice that $A+E = B+F = C+G = D+H$.

WLOG, assume that $J = 1$. Thus the pairs of vertices must be $9$ and $2$, $8$ and $3$, $7$ and $4$, and $6$ and $5$. There are $4! = 24$ ways to assign these to the vertices. Furthermore, there are $2^{4} = 16$ ways to switch them (i.e. do $2$ $9$ instead of $9$ $2$).

Thus, there are $16(24) = 384$ ways for each possible J value. There are $3$ possible J values that still preserve symmetry: $384(3) = \boxed{\textbf{(C) }1152}$

Solution 2

As in solution 1, $J$ must be $1$, $5$, or $9$ giving us 3 choices. Additionally $A+E = B+F = C+G = D+H$. This means once we choose $J$ there are $8$ remaining choices. Going clockwise from $A$ we count, $8$ possibilities for $A$. Choosing $A$ also determines $E$ which leaves $6$ choices for $B$, once $B$ is chosen it also determines $F$ leaving $4$ choices for $C$. Once $C$ is chosen it determines $G$ leaving $2$ choices for $D$. Choosing $D$ determines $H$, exhausting the numbers. Additionally, there are three possible values for $J$. To get the answer we multiply $2\cdot4\cdot6\cdot8\cdot3=\boxed{\textbf{(C) }1152}$.

Remark

Solutions 1 and 2 state that $J=1, 5, 9$ without rigorous analysis. Here it is: 

Let $S=A+J+E=B+J+F=C+J+G=D+J+H$
$4S=A+B+C+D+E+F+G+H+4J$
$4S=45+3J$
$4S=3(15+J)$

Because $gcd(4,3)=1$, so $4|(15+J)$, but $15 \equiv 3 \pmod 4$, so $J \equiv 1 \pmod 4$, $J=1,5,9$ 

When $J=1, S=12, A+E=B+F=C+G=D+H=11$
When $J=5, S=15, A+E=B+F=C+G=D+H=10$
When $J=9, S=18, A+E=B+F=C+G=D+H=9$

~isabelchen

Video Solution by Pi Academy

https://youtu.be/0djZt1Fvuvw?si=TnPgQi3DnrI5IsE8

~ Pi Academy

Video Solution 2

https://youtu.be/3MDr_t1ZzQk

~IceMatrix

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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=2013_AMC_10B_Problems/Problem_22&oldid=229561"