Difference between revisions of "2012 AIME I Problems/Problem 7"
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| + | ==Problem 7== | ||
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| + | == Solution == | ||
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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''. | 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''. | ||
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''.) | 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''.) | ||
| − | 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>\ | + | 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. |
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| + | == See also == | ||
| + | {{AIME box|year=2012|n=I|num-b=6|num-a=8}} | ||
Revision as of 00:20, 17 March 2012
Problem 7
Solution
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.
We see that the center student gives away 5a coins (having 5 connections to other students) and receives 5b coins in total from his neighbors. We know from the problem statement that these two numbers are equal, so 5a = 5b and a = b. Establishing similar ratios for the other groups of students reveals that a = b = c = d. (a + 2c = 3b substitutes to b + 2c = 2b, so c = b; 2c + 2d = 4d, so c = d.)
Having established this equality, we see that the center student has 5a coins, each of the 5 students in the next ring has 3a coins, and each of the 10 students in the two outermost rings has 4a coins. This sums to a total of 5a + 15a + 40a = 60a 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 5a coins, so he has 
coins.
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 | ||
