2005 AMC 12B Problems/Problem 23
Problem
Let be the set of ordered triples of real numbers for which
There are real numbers and such that for all ordered triples in we have What is the value of
Solution 1
Let and . Then, implies ,so . Therefore, . Since , we find that . Thus,
Solution 2
First, remember that factors to . By the givens, and . These can be used to find :
Therefore,
It follows that and , thus
Solution 3
We can rearrange into and into
We can then put to the third power or . Basic polynomial multiplication shows us that Thus, or . We know that so we have .
Now we need to find out what is equal to in terms of . We will find by squaring . It is basic polynomial multiplication to figure out that . We are also given that and . Thus or . Rearranging the terms of this equation we obtain that or . Now plugging this equation into our original equation , we obtain the equation Simple rearranging of this equation yields the result that . Combining like terms we obtain the equation .
Now we know the coefficients of and which are and respectively. Adding the two coefficients we obtain an answer of (Author: David Camacho)
Solution 4 (Alcumus)
From the given conditions it follows that\[ x+y=10^z, \quad x^2+y^2=10\cdot 10^z\quad\text{and}\quad 10^{2z}=(x+y)^2=x^2+2xy+y^2.
\[\]\] (Error compiling LaTeX. Unknown error_msg)
ThusAlsowhich yieldsand .
No other value of is possible for all members of , because the triple is in , and for this ordered triple, the equation reduces to .
somethings wrong with the code, pls help :(
See Also
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 |
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