Difference between revisions of "2005 AMC 12B Problems/Problem 23"
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It follows that <math>a = -\frac{1}{2}</math> and <math>b = 15</math>, thus <math>a + b = \frac{29}{2}.</math> | It follows that <math>a = -\frac{1}{2}</math> and <math>b = 15</math>, thus <math>a + b = \frac{29}{2}.</math> | ||
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+ | == Solution 3 == | ||
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+ | We can rearrange <math>\log_{10}(x+y)=z</math> into <math>x+y=10^{z}</math> and <math>\log_{10}(x^2+y^2)=z+1</math> into <math>x^2+y^2=10^{z+1}</math> | ||
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+ | We can then put <math>x+y</math> to the third power or <math>(x+y)^{3}=10^{3z}</math>. Basic polynomial multiplication shows us that <math>(x+y)^{3}=x^3+3x^{2}y+3xy^{2}+y^3=10^{3z}</math> Thus, <math>x^3+y^3=10^{3z}-3x^{2}y-3xy^{2}</math> or <math>x^3+y^3=10^{3z}-3xy(x+y)</math>. We know that <math>x+y=10^{z}</math> so we have <math>x^3+y^3=10^{3z}-3xy(10^{z})</math>. | ||
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+ | Now we need to find out what <math>xy</math> is equal to in terms of z. We will find <math>xy</math> by squaring <math>x+y</math>. It is basic polynomial multiplication to figure out that <math>(x+y)^{2}=x^2+2xy+y^2</math>. We also given that<math>x^2+y^2=10^{z+1}</math> and <math>x+y=10^{z}</math>. Thus <math>(10^{z})^{2}=10^{z+1}+2xy</math> or <math>10^{2z}=10^{z+1}+2xy</math>. Rearranging the terms of this equation we obtain that <math>xy=\frac{10^{2z}-10^{z+1}}{2}</math> or <math>xy=\frac{10^z(10^z-10)}{2}</math>. Now plugging this equation into our original equation <math>x^3+y^3=10^{3z}-3xy(10^{z})</math>, we obtain the equation <math>x^3+y^3=10^{3z}-3(\frac{10^z(10^z-10)}{2})(10^{z})</math> | ||
+ | Simple rearranging of this equation yields the result that <math>x^3+y^3=10^{3z}-3(\frac{10^{3z}}{2})+30(\frac{10^{2z}}{2})</math>. | ||
+ | Combining like terms we obtain the equation <math>x^3+y^3= -(\frac{10^{3z}}{2})+30(\frac{10^{2z}}{2})</math>. | ||
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+ | Now we know the coefficients of <math>10^{3z}</math> and <math>10^{2z}</math> which are <math>\frac{-1}{2}</math> and <math>\frac{30}{2}</math> respectively. Adding the two coefficients we obtain an answer of <math>\frac{29}{2}.</math> <math>\Rightarrow</math> <math>\boxed{B}</math> | ||
== See Also == | == See Also == |
Revision as of 15:08, 18 June 2018
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 z. We will find by squaring . It is basic polynomial multiplication to figure out that . We 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.