Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AMC 12B Problems/Problem 23"
(9 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2015 AMC 12B Problems|2015 AMC 12B #23]] and [[2015 AMC 10B Problems|2015 AMC 10B #25]]}}
 
==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?
  
Line 11: Line 13:
 
*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=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>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=4</math> we need <math>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)</math>. Then <math>b\;|\; c</math>, say <math>c=bk</math>. Now we have <math>3bk=10(1+k)</math> and so <math>k\;|\; 10</math>. We get only '''one''' root (corresponding to <math>k=2</math>) <math>(5,5,10)</math>.  
+
*If <math>a=5</math> we need <math>3bc = 10(b+c)</math>, which is the same as <math>9bc=30(b+c)\Rightarrow (3b-10)(3c-10)=100</math>. We get only '''one''' root (corresponding to <math>100=5\cdot 20</math>) <math>(5,5,10)</math>.  
 
*If <math>a=6</math> we need <math>4bc = 12(b+c)</math>. Then <math>(b-3)(c-3)=9</math>. We get ''' one''' root <math>(6,6,6)</math>.  
 
*If <math>a=6</math> we need <math>4bc = 12(b+c)</math>. Then <math>(b-3)(c-3)=9</math>. We get ''' one''' root <math>(6,6,6)</math>.  
 
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.
Line 42: Line 44:
 
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 3==
+
==Note==
−
 
+
This is also AMC 10B Problem 25, but the pages are separate. Someone should fix this.
−
Find that 2ab+2bc+2ac = abc
+
EDIT: fixed it, but someone help with the link
−
 
 
−
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
+
EDIT #2: fixed all
  
 
==See Also==
 
==See Also==
 +
{{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}}
 
{{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 00:17, 18 January 2021

The following problem is from both the 2015 AMC 12B #23 and 2015 AMC 10B #25, so both problems redirect to this page.

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.

Note

This is also AMC 10B Problem 25, but the pages are separate. Someone should fix this. EDIT: fixed it, but someone help with the link

EDIT #2: fixed all

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_23&oldid=142585"