2024 AMC 10B Problems/Problem 24

Revision as of 01:41, 14 November 2024 by Cyantist (talk | contribs) (Solution)
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 = $\frac{s}{p}$, 2p = x+y+z , 2s = ax=by=cz =2pr \[s= rs( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} )\] \[r( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = 1\] \[r =1,2,3\] case r=1: \[( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = 1\] given that 1<=a<=b<=c<=9 \[1 = ( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} )  <= \frac{3}{a}\] \[a <= 3\] case 1.1 no solution for b,c \[a= 2 ,  ( \frac{1}{2} + \frac{1}{b} +\frac{1}{c} ) = 1\] case 1.2 \[a= 3 ,  ( \frac{1}{3} + \frac{1}{b} +\frac{1}{c} ) = 1 , (b,c) =(3,3)\] case r=2: \[( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = \frac{1}{2}\] \[(a,b,c) =(6,6,6)\] case r=3: \[( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) = \frac{1}{3}\] \[(a,b,c) =(9,9,9)\] answer $\boxed{\textbf{(B) } 3}$

~luckuso

See also

2024 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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