[sum A_kB_k/2A_kB_k] where 2008 // lines divide ABC in 2008 equal areas

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[sum A_kB_k/2A_kB_k] where 2008 // lines divide ABC in 2008 equal areas

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#1 h Dec 11, 2021, 8:03 AM

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Given triangle $ABC$ . Let $A_1B_1$ , $A_2B_2$ , $...$ , $A_{2008}B_{2008}$ be $2008$ lines parallel to $AB$ which divide triangle $ABC$ into $2009$ equal areas. Calculate the value of $\left\lfloor \frac{A_1B_1}{2A_2B_2} + \frac{A_1B_1}{2A_3B_3} + ... + \frac{A_1B_1}{2A_{2008}B_{2008}} \right\rfloor$

This post has been edited 3 times. Last edited by parmenides51, Dec 16, 2022, 1:01 AM

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#2 h Dec 11, 2021, 2:17 PM • 1 Y

Y by parmenides51

which line is $A_1B_1?$ I assume it's either the shortest or longest, but which

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#3 h Dec 11, 2021, 2:34 PM • 1 Y

Y by x3yukari

there was nothing more mentioned in the official wording

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GravitoelectromagnetismTM

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GravitoelectromagnetismTM

#4 h Dec 12, 2021, 7:11 PM

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x3yukari wrote:

which line is $A_1B_1?$ I assume it's either the shortest or longest, but which

I'm assuming then that the value is the same no matter the choice of $A_1 B_1$ ?

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#5 h Dec 12, 2021, 7:22 PM

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GravitoelectromagnetismTM wrote:

x3yukari wrote:

which line is $A_1B_1?$ I assume it's either the shortest or longest, but which

I'm assuming then that the value is the same no matter the choice of $A_1 B_1$ ?

i'm almost certain that is not possible; if $A_1B_1$ is the shortest line, then the expression is significantly less than $2007 \cdot \frac{1}{2}$ whereas if $A_1B_1$ is the longest line, then the expression is significantly greater than $2007 \cdot \frac{1}{2}$

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GravitoelectromagnetismTM

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#6 h Dec 12, 2021, 7:47 PM

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There should be some sort of diagram...

I'm pretty sure this is just area ratios. Assuming $A_1 B_1$ is the shortest length, which is probably more logical, $\frac{A_1 B_1}{2A_n B_n} = \frac{1}{2} \frac{1}{\sqrt{n}}$ so the sum will be $\lfloor \frac{1}{2} \sum\limits_{n = 2}^{2008} \frac{1}{\sqrt{n}} \rfloor = 43$ .

(you could also do this assuming $A_1 B_1$ is the longest length, the sum will be instead $\lfloor \frac{1}{2} \sum\limits_{n = 2}^{2008} \sqrt{n} \rfloor$ , which evaluates to something rather large.)

This post has been edited 4 times. Last edited by GravitoelectromagnetismTM, Dec 12, 2021, 7:50 PM

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