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>ab, ac \le bc</math>, 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>. | + | Since <math>ab, 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>. |
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+ | *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=4</math> we need <math>2bc = 8(b+c) \Rightarrow bc = 4(b+c) \Rightarrow (b-4)(c-4)=16</math>. We get '''three''' roots <math>\{(4,5,20), (4,6,12), (4,8,8)\}</math>. | ||
+ | |||
+ | *If <math>a=5</math> we need <math>3bc = 10(b+c) \Rightarrow 9bc=30(b+c) \Rightarrow (3b-10)(3c-10)=100</math>. We get '''one''' root <math>\{(5,5,10)\}</math>. | ||
+ | |||
+ | *If <math>a=6</math> we need <math>4bc = 12(b+c) \Rightarrow bc = 3(b+c) \Rightarrow (b-3)(c-3)=9</math>. We get ''' one''' root <math>\{(6,6,6)\}</math>. | ||
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Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions. | Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions. | ||
Revision as of 00:23, 31 October 2021
- The following problem is from both the 2015 AMC 12B #23 and 2015 AMC 10B #25, so both problems redirect to this page.
Contents
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
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.
Note
This is also AMC 10B Problem 25, but the pages are separate. Someone should fix this. EDIT: fixed it, but someone help with the link
EDIT #2: fixed all
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.