Difference between revisions of "2001 AMC 8 Problems/Problem 12"
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When you expand the general form of <math>(a\otimes b)\otimes c</math>, you get <cmath>(a\otimes b)\otimes c = \dfrac{a\otimes b + c}{a\otimes b - c}</cmath> | When you expand the general form of <math>(a\otimes b)\otimes c</math>, you get <cmath>(a\otimes b)\otimes c = \dfrac{a\otimes b + c}{a\otimes b - c}</cmath> | ||
<cmath> (a\otimes b)\otimes c = \dfrac{\dfrac{a + b}{a - b} + c}{\dfrac{a + b}{a - b} - c} </cmath> | <cmath> (a\otimes b)\otimes c = \dfrac{\dfrac{a + b}{a - b} + c}{\dfrac{a + b}{a - b} - c} </cmath> | ||
| − | <cmath> (a\otimes b)\otimes c = \dfrac{\dfrac{a + b + ac - | + | <cmath> (a\otimes b)\otimes c = \dfrac{\dfrac{a + b + ac - bc}{a - b}}{\dfrac{a + b - ac + bc}{a - b}} </cmath> |
| − | <cmath> (a\otimes b)\otimes c = \dfrac{a + b + ac - | + | <cmath> (a\otimes b)\otimes c = \dfrac{a + b + ac -bc}{a + b - ac + bc} </cmath> |
Now, substituting <math>a=6</math>, <math>b=4</math>, and <math>c=3</math>: | Now, substituting <math>a=6</math>, <math>b=4</math>, and <math>c=3</math>: | ||
| Line 21: | Line 21: | ||
<cmath> (6\otimes 4)\otimes 3 = \dfrac{16}{4} </cmath> | <cmath> (6\otimes 4)\otimes 3 = \dfrac{16}{4} </cmath> | ||
<cmath> (6\otimes 4)\otimes 3 = 4 </cmath> | <cmath> (6\otimes 4)\otimes 3 = 4 </cmath> | ||
| − | <math>\boxed{\text {(A)}</math> | + | <math>\boxed{\text {(A)}}</math> |
~megaboy6679 | ~megaboy6679 | ||
Latest revision as of 13:10, 22 December 2024
Problem
If 
, then 
Solution 1
.
Solution 2 (Overkill)
When you expand the general form of 
, you get ![\[(a\otimes b)\otimes c = \dfrac{a\otimes b + c}{a\otimes b - c}\]](http://latex.artofproblemsolving.com/5/3/c/53c5b44dacef376f74c8e3db4f212d3411c79905.png)
![\[(a\otimes b)\otimes c = \dfrac{\dfrac{a + b}{a - b} + c}{\dfrac{a + b}{a - b} - c}\]](http://latex.artofproblemsolving.com/f/f/b/ffb5bc846f22c8345fad734ace42feb5da4cd680.png)
![\[(a\otimes b)\otimes c = \dfrac{\dfrac{a + b + ac - bc}{a - b}}{\dfrac{a + b - ac + bc}{a - b}}\]](http://latex.artofproblemsolving.com/e/b/a/ebaa482e2514980dad0d82621161149a593aa515.png)
![\[(a\otimes b)\otimes c = \dfrac{a + b + ac -bc}{a + b - ac + bc}\]](http://latex.artofproblemsolving.com/a/e/b/aeb7b12f0d4a9b39f3c64d83b315703d00746ab0.png)
Now, substituting 
, 
, and 
:
![\[(6\otimes 4)\otimes 3 = \dfrac{6 + 4 + 18 - 12}{6 + 4 - 18 + 12}\]](http://latex.artofproblemsolving.com/8/5/a/85a8469752049315230646d1833a993d68921547.png)
![\[(6\otimes 4)\otimes 3 = \dfrac{16}{4}\]](http://latex.artofproblemsolving.com/6/c/7/6c725614dc79b5d8a1d752d7738c930b1bac4a32.png)
![\[(6\otimes 4)\otimes 3 = 4\]](http://latex.artofproblemsolving.com/e/8/f/e8f54e528d3ba0f43368e148e9a11627ca39cdde.png)
~megaboy6679
Video Solution-Cooler Method
https://www.youtube.com/watch?v=ZfwtAiH_6PI&t=36s
See Also
| 2001 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
