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Difference between revisions of "2012 AIME I Problems/Problem 7"

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== Solution ==
 
== Solution ==
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Say the student in the center starts out with <math>a</math> coins, the students neighboring the center student each start with <math>b</math> coins, and all other students start out with <math>c</math> coins. Then all the <math>a</math>-coin students have five neighbors, all the <math>b</math>-coin students have three neighbors, and all the <math>c</math>-coin students have four neighbors.
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Say the student in the center starts out with <math>a</math> coins, the students neighboring the center student each start with <math>b</math> coins, and all other students start out with <math>c</math> coins. Then the <math>a</math>-coin student have five neighbors, all the <math>b</math>-coin students have three neighbors, and all the <math>c</math>-coin 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 be equal to the number received, and thus
 
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 be equal to the number received, and thus

Revision as of 13:58, 7 April 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 student each start with $b$ coins, and all other students start out with $c$ coins. Then the $a$-coin student have five neighbors, all the $b$-coin students have three neighbors, and all the $c$-coin 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 be equal to 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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