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Difference between revisions of "2012 AIME I Problems/Problem 7"
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==Problem 7==
 
==Problem 7==
 +
At each of the sixteen circles in the network below stands a student. A total of <math>3360</math> coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally.
 +
 +
<center><asy>
 +
import cse5;
 +
unitsize(6mm);
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defaultpen(linewidth(.8pt));
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dotfactor = 8;
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pathpen=black;
 +
 +
pair A = (0,0);
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pair B = 2*dir(54), C = 2*dir(126), D = 2*dir(198), E = 2*dir(270), F = 2*dir(342);
 +
pair G = 3.6*dir(18), H = 3.6*dir(90), I = 3.6*dir(162), J = 3.6*dir(234), K = 3.6*dir(306);
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pair M = 6.4*dir(54), N = 6.4*dir(126), O = 6.4*dir(198), P = 6.4*dir(270), L = 6.4*dir(342);
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pair[] dotted = {A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P};
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D(A--B--H--M);
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D(A--C--H--N);
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D(A--F--G--L);
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D(A--E--K--P);
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D(A--D--J--O);
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D(B--G--M);
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D(F--K--L);
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D(E--J--P);
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D(O--I--D);
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D(C--I--N);
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D(L--M--N--O--P--L);
 +
 +
dot(dotted);
 +
 +
</asy></center>
  
 
== Solution ==
 
== Solution ==
 +
Say the student in the center starts out with <math>a</math> coins, the students neighboring the center students each start with <math>b</math> coins, and all other students start out with <math>c</math> coins. Then all the <math>a</math> students have five neighbors, all the <math>b</math> students have three neighbors, and all the <math>c</math> students have four neighbors.
  
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Define the number of coins the center student gives out to each neighbor as ''a'', the number of coins each student in the innermost pentagon (each having three connections) gives as ''b'', the number of coins each student in the next ring of five gives as ''c'', and the number that each in the outermost ring gives as ''d''.
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Now in order for each student's number of coins to remain equal after the trade, the number of coins given by each student must equal the number received, and thus
  
−
We see that the center student gives away 5''a'' coins (having 5 connections to other students) and receives 5''b'' coins in total from his neighbors. We know from the problem statement that these two numbers are equal, so 5''a'' = 5''b'' and ''a'' = ''b''. Establishing similar ratios for the other groups of students reveals that ''a'' = ''b'' = ''c'' = ''d''. (''a'' + 2''c'' = 3''b'' substitutes to ''b'' + 2''c'' = 2''b'', so ''c'' = ''b''; 2''c'' + 2''d'' = 4''d'', so ''c'' = ''d''.)
+
<cmath>
 +
\begin{align*}
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a &= 5 \cdot \frac{b}{3}\\
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b &= \frac{a}{5} + 2 \cdot \frac{c}{4}\\
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c &= 2 \cdot \frac{c}{4} + 2 \cdot \frac{b}{3}.
 +
\end{align*}
 +
</cmath>
  
−
Having established this equality, we see that the center student has 5''a'' coins, each of the 5 students in the next ring has 3''a'' coins, and each of the 10 students in the two outermost rings has 4''a'' coins. This sums to a total of 5''a'' + 15''a'' + 40''a'' = 60''a'' coins, and we know that there are 3360 coins in total, so solving for ''a'' is simply dividing 3360 by 60, which yields 56. The student in the center has 5''a'' coins, so he has <math>\boxed{280}</math> coins.
+
Solving these equations, we see that <math>\frac{a}{5} = \frac{b}{3} = \frac{c}{4}.</math> Also, the total number of coins is <math>a + 5b + 10c = 3360,</math> so <math>a + 5 \cdot \frac{3a}{5} + 10 \cdot \frac{4a}{5} = 3360 \rightarrow a = \frac{3360}{12} = \boxed{280.}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2012|n=I|num-b=6|num-a=8}}

Revision as of 17:19, 17 March 2012

Problem 7

At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally.

[asy] import cse5; unitsize(6mm); defaultpen(linewidth(.8pt)); dotfactor = 8; pathpen=black; pair A = (0,0); pair B = 2*dir(54), C = 2*dir(126), D = 2*dir(198), E = 2*dir(270), F = 2*dir(342); pair G = 3.6*dir(18), H = 3.6*dir(90), I = 3.6*dir(162), J = 3.6*dir(234), K = 3.6*dir(306); pair M = 6.4*dir(54), N = 6.4*dir(126), O = 6.4*dir(198), P = 6.4*dir(270), L = 6.4*dir(342); pair[] dotted = {A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P}; D(A--B--H--M); D(A--C--H--N); D(A--F--G--L); D(A--E--K--P); D(A--D--J--O); D(B--G--M); D(F--K--L); D(E--J--P); D(O--I--D); D(C--I--N); D(L--M--N--O--P--L); dot(dotted); [/asy]

Solution

Say the student in the center starts out with $a$ coins, the students neighboring the center students each start with $b$ coins, and all other students start out with $c$ coins. Then all the $a$ students have five neighbors, all the $b$ students have three neighbors, and all the $c$ students have four neighbors.

Now in order for each student's number of coins to remain equal after the trade, the number of coins given by each student must equal the number received, and thus

\begin{align*} a &= 5 \cdot \frac{b}{3}\\ b &= \frac{a}{5} + 2 \cdot \frac{c}{4}\\ c &= 2 \cdot \frac{c}{4} + 2 \cdot \frac{b}{3}. \end{align*}

Solving these equations, we see that $\frac{a}{5} = \frac{b}{3} = \frac{c}{4}.$ Also, the total number of coins is $a + 5b + 10c = 3360,$ so $a + 5 \cdot \frac{3a}{5} + 10 \cdot \frac{4a}{5} = 3360 \rightarrow a = \frac{3360}{12} = \boxed{280.}$

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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