Difference between revisions of "2005 AMC 12B Problems/Problem 23"
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<cmath>\log_{10}(x+y) = z \text{ and } \log_{10}(x^{2}+y^{2}) = z+1.</cmath> | <cmath>\log_{10}(x+y) = z \text{ and } \log_{10}(x^{2}+y^{2}) = z+1.</cmath> | ||
− | There are real numbers <math>a</math> and <math>b</math> such that for all ordered triples <math>(x,y | + | There are real numbers <math>a</math> and <math>b</math> such that for all ordered triples <math>(x,y,z)</math> in <math>S</math> we have <math>x^{3}+y^{3}=a \cdot 10^{3z} + b \cdot 10^{2z}.</math> What is the value of <math>a+b?</math> |
<math> | <math> | ||
Line 50: | Line 50: | ||
== Solution 4 (Alcumus)== | == Solution 4 (Alcumus)== | ||
− | From the given conditions it follows that\[ | + | From the given conditions it follows that |
+ | <cmath>\[ | ||
x+y=10^z, \quad x^2+y^2=10\cdot 10^z\quad\text{and}\quad 10^{2z}=(x+y)^2=x^2+2xy+y^2. | x+y=10^z, \quad x^2+y^2=10\cdot 10^z\quad\text{and}\quad 10^{2z}=(x+y)^2=x^2+2xy+y^2. | ||
− | \]Thus\[ | + | \]</cmath> |
+ | Thus<cmath>\[ | ||
xy=\frac{1}{2}(10^{2z}-10\cdot 10^z). | xy=\frac{1}{2}(10^{2z}-10\cdot 10^z). | ||
− | \]Also\[ | + | \]</cmath>Also<cmath>\[ |
(x+y)^3=10^{3z}\quad\text{and}\quad x^3+y^3=(x+y)^3-3xy(x+y), | (x+y)^3=10^{3z}\quad\text{and}\quad x^3+y^3=(x+y)^3-3xy(x+y), | ||
− | \]which yields\begin{align*} | + | \]</cmath>which yields<cmath>\begin{align*} |
x^3+y^3&=10^{3z}-\frac{3}{2}(10^{2z}-10\cdot 10^z)(10^z) \\ | x^3+y^3&=10^{3z}-\frac{3}{2}(10^{2z}-10\cdot 10^z)(10^z) \\ | ||
&=10^{3z}-\frac{3}{2}(10^{3z}-10\cdot 10^{2z}) | &=10^{3z}-\frac{3}{2}(10^{3z}-10\cdot 10^{2z}) | ||
=-\frac{1}{2}10^{3z}+15\cdot 10^{2z}, | =-\frac{1}{2}10^{3z}+15\cdot 10^{2z}, | ||
− | \end{align*}and <math>a+b=-\frac{1}{2}+15=\boxed{29/2}</math>. | + | \end{align*}</cmath>and <math>a+b=-\frac{1}{2}+15=\boxed{29/2}</math>. |
No other value of <math>a + b</math> is possible for all members of <math>S</math>, because the triple <math>\left(\frac{1}{2}(1 + \sqrt{19}), \frac{1}{2}(1 - \sqrt{19}), 0\right)</math> is in <math>S</math>, and for this ordered triple, the equation <math>x^3 + y^3 = a\cdot 10^{3z} + b\cdot 10^{2z}</math> reduces to <math>a + b = 29/2</math>. | No other value of <math>a + b</math> is possible for all members of <math>S</math>, because the triple <math>\left(\frac{1}{2}(1 + \sqrt{19}), \frac{1}{2}(1 - \sqrt{19}), 0\right)</math> is in <math>S</math>, and for this ordered triple, the equation <math>x^3 + y^3 = a\cdot 10^{3z} + b\cdot 10^{2z}</math> reduces to <math>a + b = 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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:36, 19 May 2024
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
Thus
Also
which yields
and
.
No other value of is possible for all members of
, because the triple
is in
, and for this ordered triple, the equation
reduces to
.
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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