Difference between revisions of "2015 AMC 12B Problems/Problem 23"
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+ | {{duplicate|[[2015 AMC 10B Problems#Problem 25|2015 AMC 10B #25]] and [[2015 AMC 12B Problems#Problem 23|2015 AMC 12B #23]]}} | ||
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==Problem== | ==Problem== | ||
− | + | ||
A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c</math>. The volume and the surface area of the box are numerically equal. How many ordered triples <math>(a,b,c)</math> are possible? | A rectangular box measures <math>a \times b \times c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers and <math>1\leq a \leq b \leq c</math>. The volume and the surface area of the box are numerically equal. How many ordered triples <math>(a,b,c)</math> are possible? | ||
<math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math> | <math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math> | ||
− | ==Solution== | + | ==Solution 1== |
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</math> and <math>a,b,c</math> are all positive<math>,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=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. | ||
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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. | ||
− | == | + | ==Simplification of Solution 2== |
− | This is | + | The surface area is <math>2(ab+bc+ca)</math>, the volume is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>. |
− | + | ||
− | + | Divide both sides by <math>2abc</math>, we have: <cmath>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.</cmath> | |
+ | First consider the bound of the variable <math>a</math>. Since <math>\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},</math> we have <math>a>2</math>, or <math>a\ge 3</math>. | ||
+ | |||
+ | Also note that <math>c\ge b\ge a>0</math>, we have <math>\frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}</math>. Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le \frac{3}{a}</math>, so <math>a\le 6</math>. | ||
+ | |||
+ | So we have <math>a=3, 4, 5</math> or <math>6</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. | ||
+ | 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>. | ||
+ | |||
+ | Now go through cases for <math>a</math> and you end up with the same result. However, now you don't have to guess solutions. For example, when <math>a=3</math>, then <math>de = 36</math> and <math>d=1, 2, 3, 4, 6</math>. | ||
==See Also== | ==See Also== |
Latest revision as of 03:41, 12 June 2024
- 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
and
are all positive
. 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.