Difference between revisions of "2015 AMC 10B Problems/Problem 25"
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We can say <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{q}</math>, where <math>\frac{1}{q} = \frac{1}{2}-\frac{1}{a}</math>. | We can say <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{q}</math>, where <math>\frac{1}{q} = \frac{1}{2}-\frac{1}{a}</math>. | ||
− | Notice <math>\emph{\text{immediately}}</math> that <math>b, c > q</math> This is our key step. | + | Notice <math>\emph{\text{immediately}}</math> that <math>b, c > q</math>. This is our key step. |
Then we can say <math>b=q+d</math>, <math>c=q+e</math>. If we clear the fraction about b and c (do the math), our immediate result is that <math>de = q^2</math>. Realize also that <math>d \leq e</math>. | Then we can say <math>b=q+d</math>, <math>c=q+e</math>. If we clear the fraction about b and c (do the math), our immediate result is that <math>de = q^2</math>. Realize also that <math>d \leq e</math>. | ||
Revision as of 14:55, 29 December 2020
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 AMC 12B Problem 23, but the pages are separate. I don't know how fix this so someone plz help :(. -smileymittens
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.