Difference between revisions of "1976 AHSME Problems/Problem 30"
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== Solution == | == Solution == | ||
− | The first equation suggests the substitution <math>a = x</math>, <math>b = 2y</math>, and <math>c = 4z</math>. Then <math>x = a</math>, <math>y = b/2</math>, and <math>z = c/4</math>. Substituting into the given equations, we get | + | The first equation suggests the substitution <math>a = x</math>, <math>b = 2y</math>, and <math>c = 4z</math>. Then <math>x = a</math>, <math>y = b/2</math>, and <math>z = c/4</math>. Substituting into the given equations, we get |
− | + | ||
− | a + b + c | + | |
− | + | a + b + c = 12 | |
− | + | ||
− | + | ab + ac + bc = 44 | |
− | + | ||
− | + | abc = 48. | |
− | |||
− | ab + ac + bc | ||
− | abc | ||
− | |||
Then by Vieta's formulas, <math>a</math>, <math>b</math>, and <math>c</math> are the roots of the equation | Then by Vieta's formulas, <math>a</math>, <math>b</math>, and <math>c</math> are the roots of the equation | ||
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Since our substitution was not symmetric, each possible solution <math>(a,b,c)</math> leads to a different solution <math>(x,y,z)</math>, as follows: | Since our substitution was not symmetric, each possible solution <math>(a,b,c)</math> leads to a different solution <math>(x,y,z)</math>, as follows: | ||
− | + | ||
− | + | a | b | c | x | y | z | |
− | + | ----------------------- | |
− | 2 | + | 2 | 4 | 6 | 2 | 2 | 3/2 |
− | 2 | + | |
− | 4 | + | 2 | 6 | 4 | 2 | 3 | 1 |
− | 4 | + | |
− | 6 | + | 4 | 2 | 6 | 4 | 1 | 3/2 |
− | 6 | + | |
− | + | 4 | 6 | 2 | 4 | 3 | 1/2 | |
− | + | ||
+ | 6 | 2 | 4 | 6 | 1 | 1 | ||
+ | |||
+ | 6 | 4 | 2 | 6 | 2 | 1/2 | ||
+ | |||
Hence, there are <math>\boxed{6}</math> solutions in <math>(x,y,z)</math>. The answer is (E). | Hence, there are <math>\boxed{6}</math> solutions in <math>(x,y,z)</math>. The answer is (E). |
Revision as of 03:48, 15 June 2020
Problem 30
How many distinct ordered triples satisfy the equations
Solution
The first equation suggests the substitution , , and . Then , , and . Substituting into the given equations, we get
a + b + c = 12
ab + ac + bc = 44
abc = 48.
Then by Vieta's formulas, , , and are the roots of the equation which factors as Hence, , , and are equal to 2, 4, and 6 in some order.
Since our substitution was not symmetric, each possible solution leads to a different solution , as follows:
a | b | c | x | y | z
2 | 4 | 6 | 2 | 2 | 3/2
2 | 6 | 4 | 2 | 3 | 1
4 | 2 | 6 | 4 | 1 | 3/2
4 | 6 | 2 | 4 | 3 | 1/2
6 | 2 | 4 | 6 | 1 | 1
6 | 4 | 2 | 6 | 2 | 1/2
Hence, there are solutions in . The answer is (E).