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Difference between revisions of "2015 AMC 12B Problems/Problem 23"

(→‎Solution)
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==Solution 3==
 
==Solution 3==
  
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Find that 2ab+2bc+2ac = abc
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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
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Split it up into cases:
 
Split it up into cases:
  
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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

Problem

A rectangular box measures $a \times b \times c$, where $a$, $b$, and $c$ are integers and $1\leq a \leq b \leq c$. The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?

$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26$

Solution

We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\] Since $ab, ac \le bc$, we get $abc \le 6bc$. Thus $a\le 6$. From the second equation we see that $a > 2$. Thus $a\in \{3, 4, 5, 6\}$.

  • If $a=3$ we need $bc = 6(b+c) \Rightarrow (b-6)(c-6)=36$. We get five roots $\{(3, 7, 42), (3, 8, 24), (3,9,18), (3, 10, 15), (3,12,12)\}.$
  • If $a=4$ we need $bc = 4(b+c) \Rightarrow (b-4)(c-4)=16$. We get three roots $\{(4,5,20), (4,6,12), (4,8,8)\}$.
  • If $a=5$ we need $3bc = 10(b+c)$, which is the same as $9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100$. We get only one root (corresponding to $100=5\cdot 20$) $(5,5,10)$.
  • If $a=6$ we need $4bc = 12(b+c)$. Then $(b-3)(c-3)=9$. We get one root $(6,6,6)$.

Thus, there are $5+3+1+1 = \boxed{\textbf{(B)}\; 10}$ solutions.

Solution 2

The surface area is $2(ab+bc+ca)$, and the volume is $abc$, so equating the two yields

\[2(ab+bc+ca)=abc.\]

Divide both sides by $2abc$ to obtain \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.\]

First consider the bound of the variable $a$. Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we have $a>2$, or $a\geqslant3$.

Also note that $c \geq b \geq a > 0$, hence $\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}$. Thus, $\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}$, so $a \leq 6$.

So we have $a=3, 4, 5$ or $6$.

Before the casework, let's consider the possible range for $b$ if $\frac{1}{b}+\frac{1}{c}=k>0$. From $\frac{1}{b}<k$, we have $b>\frac{1}{k}$. From $\frac{2}{b} \geq \frac{1}{b}+\frac{1}{c}=k$, we have $b \leq \frac{2}{k}$. Thus $\frac{1}{k}<b \leq \frac{2}{k}$.

When $a=3$, we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{6}$, so $b=7, 8, 9, 10, 11, 12$. We find the solutions $(a, b, c)=(3, 7, 42)$, $(3, 8, 24)$, $(3, 9, 18)$, $(3, 10, 15)$, $(3, 12, 12)$, for a total of $5$ solutions.

When $a=4$, we get $\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$, so $b=5, 6, 7, 8$. We find the solutions $(a, b, c)=(4, 5, 20)$, $(4, 6, 12)$, $(4, 8, 8)$, for a total of $3$ solutions.

When $a=5$, we get $\frac{1}{b}+\frac{1}{c}=\frac{3}{10}$, so $b=5, 6$. The only solution in this case is $(a, b, c)=(5, 5, 10)$.

When $a=6$, $b$ is forced to be $6$, and thus $(a, b, c)=(6, 6, 6)$.

Thus, there are $5+3+1+1 = \boxed{\textbf{(B)}\; 10}$ solutions.

Solution 3

Find that $2ab+2bc+2ac = abc$

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: $a=b=c$ You should get $a=6$

Case 2: $a=b$ 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.

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 $a/b/c$ 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 $10.$

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

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