Difference between revisions of "2001 AMC 10 Problems/Problem 10"
(→Solution 2) |
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| Line 31: | Line 31: | ||
We square root: | We square root: | ||
<math>xyz = 288</math>. | <math>xyz = 288</math>. | ||
| − | Aha! We divide each of the given equations into this, yielding <math>x = 4</math>, <math>y = 6</math>, and <math>z = 12</math>. The desired sum is <math>4+6+12 = \boxed{22}</math>, so the answer is <math>\boxed{\ | + | Aha! We divide each of the given equations into this, yielding <math>x = 4</math>, <math>y = 6</math>, and <math>z = 12</math>. The desired sum is <math>4+6+12 = \boxed{22}</math>, so the answer is <math>\boxed{\textbf{(D)}}</math>. |
== See Also == | == See Also == | ||
{{AMC10 box|year=2001|num-b=9|num-a=11}} | {{AMC10 box|year=2001|num-b=9|num-a=11}} | ||
Revision as of 20:01, 2 June 2011
Contents
Problem
If 
, 
, and 
are positive with 
, 
, and 
, then 
is
Solution 1
Look at the first two equations in the problem.
and 
.
We can say that 
.
Given , we can substitute for 
and find
.
We can replace y into the first equation.
.
Since we know every variable's value, we can substitute it in for 
.
Solution 2
These equations are symmetric, and furthermore, they use multiplication. This makes us think to multiply them all. This gives 
.
We square root:
.
Aha! We divide each of the given equations into this, yielding 
, 
, and 
. The desired sum is 
, so the answer is 
.
See Also
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 10 Problems and Solutions | ||
