Difference between revisions of "2015 AMC 10B Problems/Problem 25"
(→Note) |
m (→Solution 2) |
||
Line 52: | Line 52: | ||
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> | ||
+ | |||
If <math>a=4</math>, we need <math>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=4</math>, we need <math>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)</math>, which is the same as <math>9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100</math>. We get only one root: (corresponding to <math>100=5\cdot 20</math>) <math>(5,5,10)</math>. | If <math>a=5</math>, we need <math>3bc = 10(b+c)</math>, which is the same as <math>9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100</math>. We get only one root: (corresponding to <math>100=5\cdot 20</math>) <math>(5,5,10)</math>. | ||
+ | |||
If <math>a=6</math>, we need <math>4bc = 12(b+c)</math>. Then <math>(b-3)(c-3)=9</math>. We get one root: <math>(6,6,6)</math>. | If <math>a=6</math>, we need <math>4bc = 12(b+c)</math>. Then <math>(b-3)(c-3)=9</math>. We get one root: <math>(6,6,6)</math>. | ||
+ | |||
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 10:21, 21 July 2021
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 1
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
.
Before the casework, let's consider the possible range for if
.
From , we have
. From
, we have
. Thus
When ,
, so
. The solutions we find are
, for a total of
solutions.
When ,
, so
. The solutions we find are
, for a total of
solutions.
When ,
, so
. The only solution in this case is
.
When ,
is forced to be
, and thus
.
Thus, our answer is
Simplification of Solution 1
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
.
Solution 2
We need:Since
, 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
, which is the same as
. We get only one root: (corresponding to
)
.
If , we need
. Then
. We get one root:
.
Thus, there are solutions.
-minor edit by Bobbob
Solution 3 (Basically the exact same as Solution 1)
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.
-minor edit by Snow52
-minor edit by Bobbob
Note
This is also 2015 AMC 12B Problem 23, but the pages are separate.
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.