Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 9"

m (Problem)
(Standardized answer choices; eqnarray -> align; minor edits)
 
Line 2: Line 2:
 
If <math>x=\sqrt[3]{4}</math> and <math>y=\sqrt[3]{6}-\sqrt[3]{3}</math>, then which of the following is correct?
 
If <math>x=\sqrt[3]{4}</math> and <math>y=\sqrt[3]{6}-\sqrt[3]{3}</math>, then which of the following is correct?
  
A. <math>x=y</math>
+
<math>\mathrm{(A)}\ x=y\qquad\mathrm{(B)}\ x<y\qquad\mathrm{(C)}\ x=2y\qquad\mathrm{(D)}\ x>2y\qquad\mathrm{(E)}\ \text{None of these}</math>
 
 
B. <math>x<y</math>
 
 
 
C. <math>x=2y</math>
 
 
 
D. <math>x>2y</math>
 
 
 
E. None of these
 
  
 
==Solution==
 
==Solution==
The question is asking us for an approximation of the [[ratio]] between <math>x : y</math>. Thus we are allowed to multiply both sides by a constant. So (by [[difference of cube]]s)
+
The question is asking us for an approximation of the [[ratio]] between <math>x : y</math>. Thus, we are allowed to [[multiply]] both sides by a [[constant]].
 
 
<cmath>\begin{eqnarray*}\sqrt[3]{4}(\sqrt[3]{36}+\sqrt[3]{18}+\sqrt[3]{9}) &:& (\sqrt[3]{6}-\sqrt[3]{3})(\sqrt[3]{36}+\sqrt[3]{18}+\sqrt[3]{9})\\
 
2\sqrt[3]{18} + 2\sqrt[3]{9} + \sqrt[3]{36} &:& 3</cmath>
 
  
 +
By [[difference of cube]]s,
 +
<cmath>
 +
\begin{align*}\sqrt[3]{4}(\sqrt[3]{36}+\sqrt[3]{18}+\sqrt[3]{9})&:(\sqrt[3]{6}-\sqrt[3]{3})(\sqrt[3]{36}+\sqrt[3]{18}+\sqrt[3]{9})\\
 +
2\sqrt[3]{18}+2\sqrt[3]{9}+\sqrt[3]{36}&:3\end{align*}
 +
</cmath>
 
We can approximate the terms on the LHS; <math>2\sqrt[3]{18} > 4</math>, <math>2\sqrt[3]{9} > 4</math>, <math>\sqrt[3]{36} > 3</math>, so the sum on the left side <math>> 11</math>. Hence <math>x > 2y</math>, and the answer is <math>\mathrm{(D)}</math>.  
 
We can approximate the terms on the LHS; <math>2\sqrt[3]{18} > 4</math>, <math>2\sqrt[3]{9} > 4</math>, <math>\sqrt[3]{36} > 3</math>, so the sum on the left side <math>> 11</math>. Hence <math>x > 2y</math>, and the answer is <math>\mathrm{(D)}</math>.  
  

Latest revision as of 10:40, 27 April 2008

Problem

If $x=\sqrt[3]{4}$ and $y=\sqrt[3]{6}-\sqrt[3]{3}$, then which of the following is correct?

$\mathrm{(A)}\ x=y\qquad\mathrm{(B)}\ x<y\qquad\mathrm{(C)}\ x=2y\qquad\mathrm{(D)}\ x>2y\qquad\mathrm{(E)}\ \text{None of these}$

Solution

The question is asking us for an approximation of the ratio between $x : y$. Thus, we are allowed to multiply both sides by a constant.

By difference of cubes, \begin{align*}\sqrt[3]{4}(\sqrt[3]{36}+\sqrt[3]{18}+\sqrt[3]{9})&:(\sqrt[3]{6}-\sqrt[3]{3})(\sqrt[3]{36}+\sqrt[3]{18}+\sqrt[3]{9})\\ 2\sqrt[3]{18}+2\sqrt[3]{9}+\sqrt[3]{36}&:3\end{align*} We can approximate the terms on the LHS; $2\sqrt[3]{18} > 4$, $2\sqrt[3]{9} > 4$, $\sqrt[3]{36} > 3$, so the sum on the left side $> 11$. Hence $x > 2y$, and the answer is $\mathrm{(D)}$.

Remark: There doesn't seem to be any direct way to calculate a simple ratio between the two terms, but various variations can involve approximating terms by multiplying by certain quantities.

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 8
Followed by
Problem 10
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 26 27 28 29 30