Difference between revisions of "2015 AMC 12B Problems/Problem 23"
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We need <cmath>abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).</cmath> | We need <cmath>abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).</cmath> | ||
− | Since <math> | + | Since <math>a\le b, ac \le bc</math>, from the first equation we get <math>abc \le 6bc</math>. Thus <math>a\le 6</math>. From the second equation we see that <math>a > 2</math>. Thus <math>a\in \{3, 4, 5, 6\}</math>. |
*If <math>a=3</math> we need <math>bc = 6(b+c) \Rightarrow (b-6)(c-6)=36</math>. We get '''five''' roots <math>\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}</math>. | *If <math>a=3</math> we need <math>bc = 6(b+c) \Rightarrow (b-6)(c-6)=36</math>. We get '''five''' roots <math>\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}</math>. |
Revision as of 17:50, 18 October 2022
- The following problem is from both the 2015 AMC 10B #25 and 2015 AMC 12B #23, so both problems redirect to this page.
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 1
We need
Since
, from the first equation we get
. Thus
. From the second equation we see that
. Thus
.
- If
we need
. We get five roots
.
- If
we need
. We get three roots
.
- If
we need
. We get one root
.
- If
we need
. We get one root
.
Thus, there are solutions.
Solution 2
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.
Simplification of Solution 2
The surface area is , the volume is
, so
.
Divide both sides by , we have:
First consider the bound of the variable
. Since
we have
, or
.
Also note that , we have
. Thus,
, so
.
So we have or
.
We can say , where
.
Notice that
. This is our key step.
Then we can say
,
. If we clear the fraction about b and c (do the math), our immediate result is that
. Realize also that
.
Now go through cases for and you end up with the same result. However, now you don't have to guess solutions. For example, when
, then
and
.
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 |
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.