Difference between revisions of "2015 AMC 12B Problems/Problem 23"
Pi over two (talk | contribs) (→Problem) |
(→Solution) |
||
Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
+ | The surface area is <math>2(ab+bc+ca)</math>, the volumn is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>. | ||
+ | Divide both sides by <math>2abc</math>, we get that <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\geqslant3</math>. | ||
+ | |||
+ | Also note that <math>c\geqslant b\geqslant a>0</math>, we have <math>\frac{1}{a}>\frac{1}{b}>\frac{1}{c}</math>. | ||
+ | Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geqslant \frac{3}{a}</math>, so <math>a\leqslant6</math>. | ||
+ | |||
+ | So we have <math>a=3, 4, 5</math> or <math>6</math>. | ||
+ | |||
+ | Before the casework, let's consider the possible range for <math>b</math> if <math>\frac{1}{b}+\frac{1}{c}=k>0</math>. | ||
+ | |||
+ | From <math>\frac{1}{b}<k</math>, we have <math>b>\frac{1}{k}</math>. From <math>\frac{2}{b}\geqslant\frac{1}{b}+\frac{1}{c}=k</math>, we have <math>b\leqslant\frac{2}{k}</math>. Thus <math>\frac{1}{k}<b\leqslant\frac{2}{k}</math> | ||
+ | |||
+ | When <math>a=3</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{6}</math>, so <math>b=7, 8, \cdots, 12</math>. The solutions we find are <math>(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)</math>, for a total of <math>5</math> solutions. | ||
+ | |||
+ | When <math>a=4</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{4}</math>, so <math>b=5, 6, 7, 8</math>. The solutions we find are <math>(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)</math>, for a total of <math>3</math> solutions. | ||
+ | |||
+ | When <math>a=5</math>, <math>\frac{1}{b}+\frac{1}{c}=\frac{3}{10}</math>, so <math>b=5, 6</math>. The only solution in this case is <math>(a, b, c)=(5, 5, 10)</math>. | ||
+ | |||
+ | When <math>a=6</math>, <math>b</math> is forced to be <math>6</math>, and thus <math>(a, b, c)=(6, 6, 6)</math>. | ||
+ | |||
+ | Thus, there are 11 solutions | ||
==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 00:33, 4 March 2015
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
The surface area is , the volumn is
, so
.
Divide both sides by , we get that
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, there are 11 solutions
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.