Difference between revisions of "2002 AMC 12P Problems/Problem 10"
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Divide by 2 on both sides to get <cmath>3f_{4}(x)-2f_{6}(x)=f_{2}(x)</cmath> | Divide by 2 on both sides to get <cmath>3f_{4}(x)-2f_{6}(x)=f_{2}(x)</cmath> | ||
Substituting the definitions of <math>f_{2}(x)</math>, <math>f_{4}(x)</math>, and <math>f_{6}(x)</math>, we may rewrite the expression as | Substituting the definitions of <math>f_{2}(x)</math>, <math>f_{4}(x)</math>, and <math>f_{6}(x)</math>, we may rewrite the expression as |
Revision as of 14:18, 10 March 2024
Problem
Let For how many in is it true that
Solution 1
Divide by 2 on both sides to get Substituting the definitions of , , and , we may rewrite the expression as We now simplify each term separately using some algebraic manipulation and the Pythagorean identity.
We can rewrite as , which is equivalent to .
As for , we may factor it as which can be rewritten as , and then as , which is equivalent to .
Putting everything together, we have or . Therefore, the given equation is true for all real , meaning that there are more than 8 values of that satisfy the given equation and so the answer is .
See also
2002 AMC 12P (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.