Difference between revisions of "2015 AMC 12B Problems/Problem 23"
Pi over two (talk | contribs) m (→Solution) |
(→Solution) |
||
Line 14: | Line 14: | ||
Also note that <math>c \geq b \geq a > 0</math>, hence <math>\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}</math>. | Also note that <math>c \geq b \geq a > 0</math>, hence <math>\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}</math>. | ||
− | Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ | + | Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}</math>, so <math>a \leq 6</math>. |
So we have <math>a=3, 4, 5</math> or <math>6</math>. | So we have <math>a=3, 4, 5</math> or <math>6</math>. |
Revision as of 16:29, 7 March 2015
Problem
A rectangular box measures , where
,
, and
are integers and
. The volume and the surface area of the box are numerically equal. How many ordered triples
are possible?
Solution
The surface area is , and the volume is
, so equating the two yields
Divide both sides by to obtain
First consider the bound of the variable . Since
we have
, or
.
Also note that , hence
.
Thus,
, so
.
So we have or
.
Before the casework, let's consider the possible range for if
. From
, we have
. From
, we have
. Thus
.
When , we get
, so
. We find the solutions
,
,
,
,
, for a total of
solutions.
When , we get
, so
. We find the solutions
,
,
, for a total of
solutions.
When , we get
, so
. The only solution in this case is
.
When ,
is forced to be
, and thus
.
Thus, there are solutions.
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.