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

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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.
 +
 +
==Solution 2==
 +
 +
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==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}}
 
{{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:58, 10 February 2019

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

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 2

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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