Difference between revisions of "2024 AMC 10B Problems/Problem 24"

Revision as of 01:44, 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.

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

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 ==Problem==
-==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==