Difference between revisions of "2015 AMC 10B Problems/Problem 25"
Indefintense (talk | contribs) |
m (→Simplification of Solution 1) |
||
(29 intermediate revisions by 11 users not shown) | |||
Line 4: | Line 4: | ||
<math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math> | <math>\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21 \qquad\textbf{(E)}\; 26</math> | ||
− | ==Solution== | + | ==Solution 1== |
The surface area is <math>2(ab+bc+ca)</math>, the volume is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>. | The surface area is <math>2(ab+bc+ca)</math>, the volume is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>. | ||
Line 30: | Line 30: | ||
Thus, our answer is <math>\boxed{\textbf{(B)}\;10}</math> | Thus, our answer is <math>\boxed{\textbf{(B)}\;10}</math> | ||
− | ==Simplification of Solution== | + | ==Simplification of Solution 1== |
The surface area is <math>2(ab+bc+ca)</math>, the volume is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>. | The surface area is <math>2(ab+bc+ca)</math>, the volume is <math>abc</math>, so <math>2(ab+bc+ca)=abc</math>. | ||
Line 43: | Line 43: | ||
We can say <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{q}</math>, where <math>\frac{1}{q} = \frac{1}{2}-\frac{1}{a}</math>. | We can say <math>\frac{1}{b}+\frac{1}{c}=\frac{1}{q}</math>, where <math>\frac{1}{q} = \frac{1}{2}-\frac{1}{a}</math>. | ||
− | Notice <math>immediately</math> that <math>b, c > q</math> This is our key step. | + | Notice <math>\emph{\text{immediately}}</math> that <math>b, c > q</math>. This is our key step. |
Then we can say <math>b=q+d</math>, <math>c=q+e</math>. If we clear the fraction about b and c (do the math), our immediate result is that <math>de = q^2</math>. Realize also that <math>d \leq e</math>. | Then we can say <math>b=q+d</math>, <math>c=q+e</math>. If we clear the fraction about b and c (do the math), our immediate result is that <math>de = q^2</math>. Realize also that <math>d \leq e</math>. | ||
Now go through cases for <math>a</math> and you end up with the same result. However, now you don't have to guess solutions. For example, when <math>a=3</math>, then <math>de = 36</math> and <math>d=1, 2, 3, 4, 6</math>. | Now go through cases for <math>a</math> and you end up with the same result. However, now you don't have to guess solutions. For example, when <math>a=3</math>, then <math>de = 36</math> and <math>d=1, 2, 3, 4, 6</math>. | ||
− | - | + | ==Solution 2== |
+ | 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. | Thus, there are <math>5+3+1+1 = \boxed{\textbf{(B)}\; 10}</math> solutions. | ||
− | Solution | + | -minor edit by Bobbob |
− | The surface area is <math>2(ab+bc+ca)</math>, and the volume is <math>abc</math>, so equating the two yields | + | |
+ | ==Solution 3 (Basically the exact same as Solution 1)== | ||
+ | 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> | <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> | + | 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>. | ||
Line 81: | Line 85: | ||
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. | ||
+ | |||
+ | -minor edit by Snow52 | ||
+ | |||
+ | -minor edit by Bobbob | ||
+ | |||
+ | == Note == | ||
+ | This is also 2015 AMC 12B Problem 23, but the pages are separate. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}} | {{AMC10 box|year=2015|ab=B|after=Last Question|num-b=24}} | ||
+ | {{AMC12 box|year=2015|ab=B|num-a=24|num-b=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
− | [[Category: Introductory | + | [[Category: Introductory Geometry Problems]] |
Revision as of 21:53, 4 August 2021
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 1
The surface area is , the volume is , so .
Divide both sides by , we have:
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, our answer is
Simplification of Solution 1
The surface area is , the volume is , so .
Divide both sides by , we have: First consider the bound of the variable . Since we have , or .
Also note that , we have . Thus, , so .
So we have or .
We can say , where .
Notice that . This is our key step. Then we can say , . If we clear the fraction about b and c (do the math), our immediate result is that . Realize also that .
Now go through cases for and you end up with the same result. However, now you don't have to guess solutions. For example, when , then and .
Solution 2
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.
-minor edit by Bobbob
Solution 3 (Basically the exact same as Solution 1)
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.
-minor edit by Snow52
-minor edit by Bobbob
Note
This is also 2015 AMC 12B Problem 23, but the pages are separate.
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.