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

(Created page with "==Problem== Six spheres of radius <math>1</math> are positioned so that their centers are at the vertices of a regular hexagon of side length <math>2</math>. The six spheres are...")
 
 
(34 intermediate revisions by 21 users not shown)
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==Problem==
 
==Problem==
  
−
Six spheres of radius <math>1</math> are positioned so that their centers are at the vertices of a regular hexagon of side length <math>2</math>. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon.  An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere.  What is the radius of this eighth sphere?
+
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)}\ 384 \qquad\textbf{(B)}\ 576  \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456 </math>
  
−
<math> \textbf{(A)}\ \sqrt2\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ \sqrt3\qquad\textbf{(E)}\ 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",J,SE);
 +
</asy>
  
−
==Solution==
+
==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:
 +
 
 +
<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>
 +
 
 +
==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

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