Difference between revisions of "2015 AMC 12B Problems/Problem 23"
(→Solution) |
Indefintense (talk | contribs) (→Note) |
||
(12 intermediate revisions by 8 users not shown) | |||
Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
− | |||
− | Divide both sides by <math>2abc</math> | + | We need <cmath>abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).</cmath> |
+ | Since <math>ab, ac \le bc</math>, 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>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>, 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>. | ||
+ | Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | The surface area is <math>2(ab+bc+ca)</math>, and the volume is <math>abc</math>, so equating the two yields | ||
+ | |||
+ | <cmath>2(ab+bc+ca)=abc.</cmath> | ||
+ | |||
+ | Divide both sides by <math>2abc</math> to obtain <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>. | 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\ | + | Also note that <math>c \geq b \geq a > 0</math>, hence <math>\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}</math>. |
− | Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ | + | Thus, <math>\frac{1}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{3}{a}</math>, so <math>a \leq 6</math>. |
So we have <math>a=3, 4, 5</math> or <math>6</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>. | + | 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} \geq \frac{1}{b}+\frac{1}{c}=k</math>, we have <math>b \leq \frac{2}{k}</math>. Thus <math>\frac{1}{k}<b \leq \frac{2}{k}</math>. |
− | + | When <math>a=3</math>, we get <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{6}</math>, so <math>b=7, 8, 9, 10, 11, 12</math>. We find the solutions <math>(a, b, c)=(3, 7, 42)</math>, <math>(3, 8, 24)</math>, <math>(3, 9, 18)</math>, <math>(3, 10, 15)</math>, <math>(3, 12, 12)</math>, for a total of <math>5</math> solutions. | |
− | When <math>a= | + | When <math>a=4</math>, we get <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{4}</math>, so <math>b=5, 6, 7, 8</math>. We find the solutions <math>(a, b, c)=(4, 5, 20)</math>, <math>(4, 6, 12)</math>, <math>(4, 8, 8)</math>, for a total of <math>3</math> solutions. |
− | When <math>a= | + | When <math>a=5</math>, we get <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= | + | 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 <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> 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. -Indefintense | ||
==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 21:03, 5 July 2020
Contents
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
We need Since , 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 , which is the same as . We get only one root (corresponding to ) .
- If we need . Then . 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.
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. -Indefintense
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.