Difference between revisions of "2024 AMC 10B Problems/Problem 24"
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==Problem== | ==Problem== | ||
| − | ==Solution== | + | ==Solution #1== |
| + | Let x,y,z be 3 sides of triangle, r = <math>\frac{s}{p} </math>, 2p = x+y+z , 2s = ax=by=cz =2pr | ||
| + | <cmath>s= rs( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) </cmath> | ||
| + | <cmath> r( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = 1 </cmath> | ||
| + | <cmath>r =1,2,3 </cmath> | ||
| + | case r=1: | ||
| + | <cmath>( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = 1 </cmath> | ||
| + | given that 1<=a<=b<=c<=9 | ||
| + | <cmath> 1 = ( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) <= \frac{3}{a} </cmath> | ||
| + | <cmath> a <= 3 </cmath> | ||
| + | case 1.1 no solution for b,c <cmath> a= 2 , ( \frac{1}{2} + \frac{1}{b} +\frac{1}{c} ) = 1 </cmath> | ||
| + | case 1.2 <cmath> a= 3 , ( \frac{1}{3} + \frac{1}{b} +\frac{1}{c} ) = 1 , (b,c) =(3,3) </cmath> | ||
| + | case r=2: | ||
| + | <cmath>( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = \frac{1}{2} </cmath> | ||
| + | <cmath> (a,b,c) =(6,6,6) </cmath> | ||
| + | case r=3: | ||
| + | <cmath>( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = \frac{1}{3} </cmath> | ||
| + | <cmath> (a,b,c) =(9,9,9) </cmath> | ||
| + | answer <math>\boxed{\textbf{(B) } 3}</math> | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
==See also== | ==See also== | ||
Revision as of 01:41, 14 November 2024
- The following problem is from both the 2024 AMC 10B #24 and 2024 AMC 12B #18, so both problems redirect to this page.
Problem
Solution #1
Let x,y,z be 3 sides of triangle, r = 
, 2p = x+y+z , 2s = ax=by=cz =2pr
![\[s= rs( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} )\]](http://latex.artofproblemsolving.com/4/f/f/4fff2317e613a83cf1f5caf0c1900857ccc52878.png)
![\[r( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = 1\]](http://latex.artofproblemsolving.com/2/5/e/25e14c1e9ccda2dd290852cef4e8d15bc9ea4d75.png)
![\[r =1,2,3\]](http://latex.artofproblemsolving.com/e/c/6/ec67d7a580b089eb76ecb41d198a157beb207000.png)
case r=1:
![\[( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = 1\]](http://latex.artofproblemsolving.com/4/6/8/46836cb28e77a6fa884a3f894146d9eb2898e7f0.png)
given that 1<=a<=b<=c<=9
![\[1 = ( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) <= \frac{3}{a}\]](http://latex.artofproblemsolving.com/d/3/e/d3e296df17cd03b468132c3bf6f851b4f1953ed8.png)
![\[a <= 3\]](http://latex.artofproblemsolving.com/4/d/a/4da116a78a85bdcba6af31602284ed41cc73e887.png)
case 1.1 no solution for b,c ![\[a= 2 , ( \frac{1}{2} + \frac{1}{b} +\frac{1}{c} ) = 1\]](http://latex.artofproblemsolving.com/8/4/e/84e4adea8105777cb2365e9be4e98bbcd293788f.png)
case 1.2 ![\[a= 3 , ( \frac{1}{3} + \frac{1}{b} +\frac{1}{c} ) = 1 , (b,c) =(3,3)\]](http://latex.artofproblemsolving.com/6/b/4/6b4424c99a5dd3df42f3d61575f87b92b89c6483.png)
case r=2:
![\[( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = \frac{1}{2}\]](http://latex.artofproblemsolving.com/0/e/f/0efe6407f848755476b0d031aff9471e877cc39c.png)
![\[(a,b,c) =(6,6,6)\]](http://latex.artofproblemsolving.com/7/a/4/7a4ae6e80ceeaf259c38c06488d9d81c06460f03.png)
case r=3:
![\[( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = \frac{1}{3}\]](http://latex.artofproblemsolving.com/e/3/4/e34e4ba17b23148ef532751d08dceacc0a174042.png)
![\[(a,b,c) =(9,9,9)\]](http://latex.artofproblemsolving.com/8/4/0/8406b848bf34f26405a7f248fafcc14a9f21a39d.png)
answer 
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
| 2024 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 17 |
Followed by Problem 19 |
| 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. 
