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Difference between revisions of "2013 AMC 10B Problems/Problem 22"

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==Solution 2==
 
==Solution 2==
  
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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*4*6*8*3=\boxed{\textbf{(C) }1152}</math>.
+
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*4*6*8*3=\boxed{\textbf{(C) }1152}</math>. In reality tho, Nobody GAF
  
 
== See also ==
 
== See also ==

Revision as of 17:07, 2 January 2019

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

[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 $J$ 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 $A+E = B+F = C+G = D+H$.

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 $2*4*6*8*3=\boxed{\textbf{(C) }1152}$. In reality tho, Nobody GAF

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

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