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>. | + | 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>. | Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>. | ||
− | < | + | <cmath> |
\begin{align*} | \begin{align*} | ||
(5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ | (5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ | ||
Line 14: | Line 35: | ||
&=36 + r | &=36 + r | ||
\end{align*} | \end{align*} | ||
− | </math> | + | </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 == | ||
{{AIME box|year=2000|n=I|num-b=6|num-a=8}} | {{AIME box|year=2000|n=I|num-b=6|num-a=8}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:02, 20 April 2024
Contents
Problem
Suppose that
and
are three positive numbers that satisfy the equations
and
Then
where
and
are relatively prime positive integers. Find
.
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 as
.
Substituting into one of the given equations, we have
We can substitute back into to obtain
We can then substitute once again to get
Thus,
, so
.
Solution 2
Let .
Thus . So
.
Solution 3
Since , so
. Also,
by the second equation. Substitution gives
,
, and
, so the answer is 4+1 which is equal to
.
Solution 4
(Hybrid between 1/2)
Because and
. Substituting and factoring, we get
,
, and
. Multiplying them all together, we get,
, but
is
, and by the Identity property of multiplication, we can take it out. So, in the end, we get
. And, we can expand this to get
, and if we make a substitution for
, and rearrange the terms, we get
This will be important.
Now, lets add the 3 equations , and
. We use the expand the Left hand sides, then, we add the equations to get
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
We move all constant terms to the right, and all linear terms to the left, to get
, so
which gives an answer of
-AlexLikeMath
Solution 5
Get rid of the denominators in the second and third equations to get and
. Then, since
, we have
and
. Then, since we know that
, we can subtract these two equations to get that
. The result follows that
and
, so
, and the requested answer is
Solution 6
Rewrite the equations in terms of x.
becomes
.
becomes
Now express in terms of x.
.
This evaluates to , giving us
. We can now plug x into the other equations to get
and
.
Therefore, .
, and we are done.
~MC413551
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.