2011 AMC 12B Problems/Problem 21
Contents
Problem
The arithmetic mean of two distinct positive integers 
and 
is a two-digit integer. The geometric mean of and is obtained by reversing the digits of the arithmetic mean. What is 
?
Solution
Answer: (D)
\frac{x + y}{2} = 10 a+b
1\le a\le 9 
0\le b\le 9
\sqrt{xy} = 10 b+a$$ (Error compiling LaTeX. Unknown error_msg)100 a^2 + 20 ab + b^2 = \frac{x^2 + 2xy + y^2}{4}$$ (Error compiling LaTeX. Unknown error_msg)xy = 100b^2 + 20ab + a^2$$ (Error compiling LaTeX. Unknown error_msg)\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)
|x-y| = 2\sqrt{99(a^2 - b^2)}
(a^2 - b^2)
11n
n
a
9n = 1
4
n = 1|x-y| = 66
n = 4|x-y| = 132
a^2 -b^2 = 44
(a-b)(a+b) = 44
a-b = 1
24
a+b = 442211
a+b \le 18a+b = 11a-b = 4
ab$.
In addition:
Note that$ (Error compiling LaTeX. Unknown error_msg)11n
n = 1
a = 6b = 5
a^2 - b^2 = 36 - 25 = 11$.
Sidenote
It is easy to see that 
is the only solution. This yields 
. Their arithmetic mean is 
and their geometric mean is 
.
See also
| 2011 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 20 |
Followed by Problem 22 |
| 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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