Difference between revisions of "1976 AHSME Problems/Problem 30"
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− | <cmath> Solution </cmath> | + | == Problem 30 == |
+ | |||
+ | How many distinct ordered triples <math>(x,y,z)</math> satisfy the equations | ||
+ | <cmath>x+2y+4z=12</cmath> | ||
+ | <cmath>xy+4yz+2xz=22</cmath> | ||
+ | <cmath>xyz=6</cmath> | ||
+ | |||
+ | |||
+ | <math>\textbf{(A) }\text{none}\qquad | ||
+ | \textbf{(B) }1\qquad | ||
+ | \textbf{(C) }2\qquad | ||
+ | \textbf{(D) }4\qquad | ||
+ | \textbf{(E) }6 </math> | ||
+ | |||
+ | == 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 | ||
+ | |||
+ | |||
+ | a + b + c = 12 | ||
+ | |||
+ | ab + ac + bc = 44 | ||
+ | |||
+ | abc = 48. | ||
+ | |||
+ | Then by Vieta's formulas, <math>a</math>, <math>b</math>, and <math>c</math> are the roots of the equation | ||
+ | <cmath>x^3 - 12x^2 + 44x - 48 = 0,</cmath> | ||
+ | which factors as | ||
+ | <cmath>(x - 2)(x - 4)(x - 6) = 0.</cmath> | ||
+ | Hence, <math>a</math>, <math>b</math>, and <math>c</math> are equal to 2, 4, and 6 in some order. | ||
+ | |||
+ | 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 | 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 <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).