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2005 AMC 12B Problems/Problem 23

Revision as of 20:09, 24 December 2020 by Arthy00009 (talk | contribs) (→‎Solution 1)

Problem

Let $S$ be the set of ordered triples $(x,y,z)$ of real numbers for which

\[\log_{10}(x+y) = z \text{ and } \log_{10}(x^{2}+y^{2}) = z+1.\] There are real numbers $a$ and $b$ such that for all ordered triples $(x,y.z)$ in $S$ we have $x^{3}+y^{3}=a \cdot 10^{3z} + b \cdot 10^{2z}.$ What is the value of $a+b?$

$\textbf{(A)}\ \frac {15}{2} \qquad \textbf{(B)}\ \frac {29}{2} \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ \frac {39}{2} \qquad \textbf{(E)}\ 24$

Solution 1

Let $x + y = s$ and $x^2 + y^2 = t$. Then, $\log(s)=z$ implies $\log(10s) = z+1= \log(t)$,so $t=10s$. Therefore, $x^3 + y^3 = s\cdot\dfrac{3t-s^2}{2} = s(15s-\dfrac{s^2}{2})$. Since $s = 10^z$, we find that $x^3 + y^3 = 15\times10^{2z} - (1/2)\times10^{3z}$. Thus, $a+b = \frac{29}{2}$ $\Rightarrow$ $\boxed{B}$

Solution 2

First, remember that $x^3 + y^3$ factors to $(x + y) (x^2 - xy + y^2)$. By the givens, $x + y = 10^z$ and $x^2 + y^2 = 10^{z + 1}$. These can be used to find $xy$: \[(x + y)^2 = 10^{2z}\] \[x^2 + 2xy + y^2 = 10^{2z}\] \[2xy = 10^{2z} - 10^{z + 1}\] \[xy = \frac{10^{2z} - 10^{z + 1}}{2}\]

Therefore, \[x^3 + y^3 = a \cdot 10^{3z} + b \cdot 10^{2z} = 10^z\left(10^{z + 1} - \frac{10^{2z} - 10^{z + 1}}{2}\right)\] \[= 10^z\left(10^{z + 1} - \frac{10^{2z} - 10^{z + 1}}{2}\right)\] \[= 10^{2z + 1} - \frac{10^{3z} - 10^{2z + 1}}{2}\] \[= -\frac{1}{2} \cdot 10^{3z} + \frac{3}{2} \cdot 10^{2z + 1}\] \[= -\frac{1}{2} \cdot 10^{3z} + 15 \cdot 10^{2z}.\]

It follows that $a = -\frac{1}{2}$ and $b = 15$, thus $a + b = \frac{29}{2}.$

Solution 3

We can rearrange $\log_{10}(x+y)=z$ into $x+y=10^{z}$ and $\log_{10}(x^2+y^2)=z+1$ into $x^2+y^2=10^{z+1}$

We can then put $x+y$ to the third power or $(x+y)^{3}=10^{3z}$. Basic polynomial multiplication shows us that $(x+y)^{3}=x^3+3x^{2}y+3xy^{2}+y^3=10^{3z}$ Thus, $x^3+y^3=10^{3z}-3x^{2}y-3xy^{2}$ or $x^3+y^3=10^{3z}-3xy(x+y)$. We know that $x+y=10^{z}$ so we have $x^3+y^3=10^{3z}-3xy(10^{z})$.

Now we need to find out what $xy$ is equal to in terms of $z$. We will find $xy$ by squaring $x+y$. It is basic polynomial multiplication to figure out that $(x+y)^{2}=x^2+2xy+y^2$. We are also given that $x^2+y^2=10^{z+1}$ and $x+y=10^{z}$. Thus $(10^{z})^{2}=10^{z+1}+2xy$ or $10^{2z}=10^{z+1}+2xy$. Rearranging the terms of this equation we obtain that $xy=\frac{10^{2z}-10^{z+1}}{2}$ or $xy=\frac{10^z(10^z-10)}{2}$. Now plugging this equation into our original equation $x^3+y^3=10^{3z}-3xy(10^{z})$, we obtain the equation $x^3+y^3=10^{3z}-3(\frac{10^z(10^z-10)}{2})(10^{z})$ Simple rearranging of this equation yields the result that $x^3+y^3=10^{3z}-3(\frac{10^{3z}}{2})+30(\frac{10^{2z}}{2})$. Combining like terms we obtain the equation $x^3+y^3= -(\frac{10^{3z}}{2})+30(\frac{10^{2z}}{2})$.

Now we know the coefficients of $10^{3z}$ and $10^{2z}$ which are $\frac{-1}{2}$ and $\frac{30}{2}$ respectively. Adding the two coefficients we obtain an answer of $\frac{29}{2}.$ $\Rightarrow$ $\boxed{B}$ (Author: David Camacho)

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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