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Difference between revisions of "2000 AIME I Problems/Problem 7"

 
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== Problem ==
 
== Problem ==
 +
Suppose that <math>x,</math> <math>y,</math> and <math>z</math> are three positive numbers that satisfy the equations <math>xyz = 1,</math> <math>x + \frac {1}{z} = 5,</math> and <math>y + \frac {1}{x} = 29.</math> Then <math>z + \frac {1}{y} = \frac {m}{n},</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>.
  
−
== Solution ==
+
 
 +
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques
 +
 
 +
==Solution 1==
 +
We can rewrite <math>xyz=1</math> as <math>\frac{1}{z}=xy</math>.
 +
 
 +
Substituting into one of the given equations, we have
 +
<cmath>x+xy=5</cmath>
 +
<cmath>x(1+y)=5</cmath>
 +
<cmath>\frac{1}{x}=\frac{1+y}{5}.</cmath>
 +
 
 +
We can substitute back into <math>y+\frac{1}{x}=29</math> to obtain
 +
<cmath>y+\frac{1+y}{5}=29</cmath>
 +
<cmath>5y+1+y=145</cmath>
 +
<cmath>y=24.</cmath>
 +
 
 +
We can then substitute once again to get
 +
<cmath>x=\frac15</cmath>
 +
<cmath>z=\frac{5}{24}.</cmath>
 +
Thus, <math>z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}</math>, so <math>m+n=\boxed{005}</math>.
 +
 
 +
==Solution 2==
 +
Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>.
 +
 
 +
<cmath>
 +
\begin{align*}
 +
(5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\
 +
&=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\
 +
&=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\
 +
&=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\
 +
&=2 + 5 + 29 + r\\
 +
&=36 + r
 +
\end{align*}
 +
</cmath>
 +
 
 +
Thus <math>145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}</math>. So <math>m + n = 1 + 4 = \boxed{5}</math>.
 +
 
 +
== Solution 3 ==
 +
 
 +
Since <math>x+(1/z)=5, 1=z(5-x)=xyz</math>, so <math>5-x=xy</math>. Also, <math>y=29-(1/x)</math> by the second equation. Substitution gives <math>x=1/5</math>, <math>y=24</math>, and <math>z=5/24</math>, so the answer is 4+1 which is equal to <math>5</math>.
 +
 
 +
== Solution 4 ==
 +
(Hybrid between 1/2)
 +
 
 +
Because <math>xyz = 1, \hspace{0.15cm} \frac{1}{x} = yz, \hspace{0.15cm} \frac{1}{y} = xz, </math> and <math>\hspace{0.05cm}\frac{1}{z} = xy</math>. Substituting and factoring, we get <math>x(y+1) = 5</math>, <math>\hspace{0.15cm}y(z+1) = 29</math>, and <math>\hspace{0.05cm}z(x+1) = k</math>. Multiplying them all together, we get, <math>xyz(x+1)(y+1)(z+1) = 145k</math>, but <math>xyz</math> is <math>1</math>, and by the Identity property of multiplication, we can take it out. So, in the end, we get <math>(x+1)(y+1)(z+1) = 145k</math>. And, we can expand this to get <math>xyz+xy+yz+xz+x+y+z+1 = 145k</math>, and if we make a substitution for <math>xyz</math>, and rearrange the terms, we get <math>xy+yz+xz+x+y+z = 145k-2</math> This will be important.
 +
 
 +
 
 +
Now, lets add the 3 equations <math>x(y+1) = 5, \hspace{0.15cm}y(z+1) = 29 </math>, and <math>\hspace{0.05cm}z(x+1) = k</math>. We use the expand the Left hand sides, then, we add the equations to get <math>xy+yz+xz+x+y+z = k+34</math> Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus <math>145k-2 = k+34</math> We move all constant terms to the right, and all linear terms to the left, to get <math>144k = 36</math>, so <math>k = \frac{1}{4}</math> which gives an answer of <math>1+4 = \boxed{005}</math>
 +
 
 +
-AlexLikeMath
 +
 
 +
==Solution 5==
 +
Get rid of the denominators in the second and third equations to get <math>xz-5z=-1</math> and <math>xy-29x=-1</math>. Then, since <math>xyz=1</math>, we have <math>\tfrac 1y-5z=-1</math> and <math>\tfrac 1z-29x=-1</math>. Then, since we know that <math>\tfrac 1z+x=5</math>, we can subtract these two equations to get that <math>30x=6\implies x=5</math>. The result follows that <math>z=\tfrac 5{24}</math> and <math>y=24</math>, so <math>z+\tfrac 1y=\tfrac 1{24}+\tfrac 5{24}=\tfrac 14</math>, and the requested answer is <math>1+4=\boxed{005}.</math>
 +
 
 +
==Solution 6==
 +
Rewrite the equations in terms of x.
 +
 
 +
<math>x+\frac{1}{z}=5</math> becomes <math>z=\frac{1}{x+5}</math>.
 +
 
 +
<math>y+\frac{1}{x}=29</math> becomes <math>y=29-\frac{1}{x}</math>
 +
 
 +
Now express <math>xyz=1</math> in terms of x.
 +
 
 +
<math>\frac{1}{5-x}\cdot(29-\frac{1}{x})\cdot x=1</math>.
 +
 
 +
This evaluates to <math>29x-1=5-x</math>, giving us <math>x=\frac{1}{5}</math>. We can now plug x into the other equations to get <math>y=24</math> and <math>z=\frac{5}{24}</math>.
 +
 
 +
Therefore, <math>z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{6}{24}=\frac{1}{4}</math>.
 +
 
 +
<math>1+4=\boxed{5}</math>, and we are done.
 +
~MC413551
  
 
== See also ==
 
== See also ==
−
* [[2000 AIME I Problems]]
+
{{AIME box|year=2000|n=I|num-b=6|num-a=8}}
 +
 
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 14:02, 20 April 2024

Problem

Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.


note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques

Solution 1

We can rewrite $xyz=1$ as $\frac{1}{z}=xy$.

Substituting into one of the given equations, we have \[x+xy=5\] \[x(1+y)=5\] \[\frac{1}{x}=\frac{1+y}{5}.\]

We can substitute back into $y+\frac{1}{x}=29$ to obtain \[y+\frac{1+y}{5}=29\] \[5y+1+y=145\] \[y=24.\]

We can then substitute once again to get \[x=\frac15\] \[z=\frac{5}{24}.\] Thus, $z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$, so $m+n=\boxed{005}$.

Solution 2

Let $r = \frac{m}{n} = z + \frac {1}{y}$.

\begin{align*} (5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ &=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\ &=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\ &=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\ &=2 + 5 + 29 + r\\ &=36 + r \end{align*}

Thus $145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}$. So $m + n = 1 + 4 = \boxed{5}$.

Solution 3

Since $x+(1/z)=5, 1=z(5-x)=xyz$, so $5-x=xy$. Also, $y=29-(1/x)$ by the second equation. Substitution gives $x=1/5$, $y=24$, and $z=5/24$, so the answer is 4+1 which is equal to $5$.

Solution 4

(Hybrid between 1/2)

Because $xyz = 1, \hspace{0.15cm} \frac{1}{x} = yz, \hspace{0.15cm} \frac{1}{y} = xz,$ and $\hspace{0.05cm}\frac{1}{z} = xy$. Substituting and factoring, we get $x(y+1) = 5$, $\hspace{0.15cm}y(z+1) = 29$, and $\hspace{0.05cm}z(x+1) = k$. Multiplying them all together, we get, $xyz(x+1)(y+1)(z+1) = 145k$, but $xyz$ is $1$, and by the Identity property of multiplication, we can take it out. So, in the end, we get $(x+1)(y+1)(z+1) = 145k$. And, we can expand this to get $xyz+xy+yz+xz+x+y+z+1 = 145k$, and if we make a substitution for $xyz$, and rearrange the terms, we get $xy+yz+xz+x+y+z = 145k-2$ This will be important.


Now, lets add the 3 equations $x(y+1) = 5, \hspace{0.15cm}y(z+1) = 29$, and $\hspace{0.05cm}z(x+1) = k$. We use the expand the Left hand sides, then, we add the equations to get $xy+yz+xz+x+y+z = k+34$ Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus $145k-2 = k+34$ We move all constant terms to the right, and all linear terms to the left, to get $144k = 36$, so $k = \frac{1}{4}$ which gives an answer of $1+4 = \boxed{005}$

-AlexLikeMath

Solution 5

Get rid of the denominators in the second and third equations to get $xz-5z=-1$ and $xy-29x=-1$. Then, since $xyz=1$, we have $\tfrac 1y-5z=-1$ and $\tfrac 1z-29x=-1$. Then, since we know that $\tfrac 1z+x=5$, we can subtract these two equations to get that $30x=6\implies x=5$. The result follows that $z=\tfrac 5{24}$ and $y=24$, so $z+\tfrac 1y=\tfrac 1{24}+\tfrac 5{24}=\tfrac 14$, and the requested answer is $1+4=\boxed{005}.$

Solution 6

Rewrite the equations in terms of x.

$x+\frac{1}{z}=5$ becomes $z=\frac{1}{x+5}$.

$y+\frac{1}{x}=29$ becomes $y=29-\frac{1}{x}$

Now express $xyz=1$ in terms of x.

$\frac{1}{5-x}\cdot(29-\frac{1}{x})\cdot x=1$.

This evaluates to $29x-1=5-x$, giving us $x=\frac{1}{5}$. We can now plug x into the other equations to get $y=24$ and $z=\frac{5}{24}$.

Therefore, $z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{6}{24}=\frac{1}{4}$.

$1+4=\boxed{5}$, and we are done. ~MC413551

See also

2000 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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