Difference between revisions of "2005 AMC 12B Problems/Problem 23"
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== Solution == | == Solution == | ||
Call <math>x + y = s</math> and <math>x^2 + y^2 = t</math>. Then, we note that <math>\log(s)=z</math> which implies that <math>\log(10s) = z+1= \log(t)</math>. Therefore, <math>t=10s</math>. Let us note that <math>x^3 + y^3 = \frac{3st}{2}-\frac{s^3}{2} = s(15s-\frac{s^2}{2})</math>. Since <math>s = 10^z</math>, we find that <math>x^3 + y^3 = 15\times10^2 - (1/2)\times10^3</math>. Thus, <math>a+b = \frac{29}{2}</math>. <math>\boxed{\text{B}}</math> is the answer. | Call <math>x + y = s</math> and <math>x^2 + y^2 = t</math>. Then, we note that <math>\log(s)=z</math> which implies that <math>\log(10s) = z+1= \log(t)</math>. Therefore, <math>t=10s</math>. Let us note that <math>x^3 + y^3 = \frac{3st}{2}-\frac{s^3}{2} = s(15s-\frac{s^2}{2})</math>. Since <math>s = 10^z</math>, we find that <math>x^3 + y^3 = 15\times10^2 - (1/2)\times10^3</math>. Thus, <math>a+b = \frac{29}{2}</math>. <math>\boxed{\text{B}}</math> is the answer. | ||
+ | |||
+ | ==Alternate Solution== | ||
+ | First, remember that <math>x^3 + y^3</math> factors to <math>(x + y) (x^2 - xy + y^2)</math>. By the givens, <math>x + y = 10^z</math> and <math>x^2 + y^2 = 10^{z + 1}</math>. These can be used to find <math>xy</math>: | ||
+ | <cmath>(x + y)^2 = 10^{2z}</cmath> | ||
+ | <cmath>x^2 + 2xy + y^2 = 10^{2z}</cmath> | ||
+ | <cmath>2xy = 10^{2z} - 10^{z + 1}</cmath> | ||
+ | <cmath>xy = \frac{10^{2z} - 10^{z + 1}}{2}</cmath> | ||
+ | |||
+ | Therefore, | ||
+ | <cmath>x^3 + y^3 = a \cdot 10^{3z} + b \cdot 10^{2z} = 10^z\left(10^{z + 1} - \frac{10^{2z} - 10^{z + 1}}{2}\right)</cmath> | ||
+ | <cmath>= 10^z\left(10^{z + 1} - \frac{10^{2z} - 10^{z + 1}}{2}\right)</cmath> | ||
+ | <cmath>= 10^{2z + 1} - \frac{10^{3z} - 10^{2z + 1}}{2}</cmath> | ||
+ | <cmath>= -\frac{1}{2} \cdot 10^{3z} + \frac{3}{2} \cdot 10^{2z}</cmath> | ||
+ | <cmath>= -\frac{1}{2} \cdot 10^{3z} + 15 \cdot 10^{2z}.</cmath> | ||
+ | |||
+ | It follows that <math>a = -\frac{1}{2}</math> and <math>b = 15</math>, thus <math>a + b = \frac{29}{2}.</math> | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2005|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2005|ab=B|num-b=22|num-a=24}} |
Revision as of 20:41, 29 June 2013
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
Call and
. Then, we note that
which implies that
. Therefore,
. Let us note that
. Since
, we find that
. Thus,
.
is the answer.
Alternate Solution
First, remember that factors to
. By the givens,
and
. These can be used to find
:
Therefore,
It follows that and
, thus
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 |