Difference between revisions of "2015 AMC 12B Problems/Problem 23"
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==Solution 3== | ==Solution 3== | ||
− | Find that 2ab+2bc+2ac = abc | + | Find that <math>2ab+2bc+2ac = abc</math> |
I'm not going to go through the bashing process of this solution, I will just give the outline | I'm not going to go through the bashing process of this solution, I will just give the outline | ||
Line 50: | Line 50: | ||
Split it up into cases: | Split it up into cases: | ||
− | Case 1: a=b=c | + | Case 1: <math>a=b=c</math> |
− | You should get a=6 | + | You should get <math>a=6</math> |
− | Case 2: a=b | + | Case 2: <math>a=b</math> |
− | You should get a^2(2-c)+4ac = 0. The easiest way to bash this is to set c equal to an integer, starting from 3 (so that a^2 is negative). You should get a=12, 8, 6, 5, but cut off 6 as we have already counted that in Case 1. Thus, three cases here. However, we can also do b=c and a=c on top of this - thus, we get 3*3 = 9 cases. | + | You should get <math>a^2(2-c)+4ac = 0</math>. The easiest way to bash this is to set <math>c</math> equal to an integer, starting from 3 (so that a^2 is negative). You should get <math>a=12, 8, 6, 5</math>, but cut off <math>6</math> as we have already counted that in Case 1. Thus, three cases here. However, we can also do <math>b=c</math> and <math>a=c</math> on top of this - thus, we get <math>3*3 = 9</math> cases. |
Case 3: Can we have a/b/c independent of each other? | Case 3: Can we have a/b/c independent of each other? | ||
Simply put, no. It's just not possible for there to be no overlap. You can verify this by setting persay a equal to some integer, plugging it in to the equation, and then using SFFT, but you will be to no avail. | Simply put, no. It's just not possible for there to be no overlap. You can verify this by setting persay a equal to some integer, plugging it in to the equation, and then using SFFT, but you will be to no avail. | ||
− | A practical approach here is to just look at the answer choices. If a/b/c were independent of each other, and a/b/c have nothing differentiating them, then for each independent ordered triple {a,b,c}, we should have 6 solutions. That would mean that the solution has to be 10+6n, where n is some integer. Obviously the only answer choice here that works is B, as the the other answer choices we cannot plug in an integral n to get them. Thus, either by POE or complicated induction, we find the answer to be 10. | + | A practical approach here is to just look at the answer choices. If <math>a/b/c</math> were independent of each other, and <math>a/b/c</math> have nothing differentiating them, then for each independent ordered triple {a,b,c}, we should have 6 solutions. That would mean that the solution has to be <math>10+6n</math>, where n is some integer. Obviously the only answer choice here that works is B, as the the other answer choices we cannot plug in an integral n to get them. Thus, either by POE or complicated induction, we find the answer to be <math>10.</math> |
iron | iron |
Revision as of 11:17, 2 January 2020
Contents
[hide]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
, 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.
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.
Solution 3
Find that
I'm not going to go through the bashing process of this solution, I will just give the outline
Split it up into cases:
Case 1:
You should get
Case 2:
You should get
. The easiest way to bash this is to set
equal to an integer, starting from 3 (so that a^2 is negative). You should get
, but cut off
as we have already counted that in Case 1. Thus, three cases here. However, we can also do
and
on top of this - thus, we get
cases.
Case 3: Can we have a/b/c independent of each other? Simply put, no. It's just not possible for there to be no overlap. You can verify this by setting persay a equal to some integer, plugging it in to the equation, and then using SFFT, but you will be to no avail.
A practical approach here is to just look at the answer choices. If were independent of each other, and
have nothing differentiating them, then for each independent ordered triple {a,b,c}, we should have 6 solutions. That would mean that the solution has to be
, where n is some integer. Obviously the only answer choice here that works is B, as the the other answer choices we cannot plug in an integral n to get them. Thus, either by POE or complicated induction, we find the answer to be
iron
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.