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 $$ \left\lfloor \frac{A_1B_1}{2A_2B_2} + \frac{A_1B_1}{2A_3B_3} + ... + \frac{A_1B_1}{2A_{2008}B_{2008}} \right\rfloor$$
Problem
Source: Indonesia INAMO Shortlist 2009 G9 https://artofproblemsolving.com/community/c1101409_
Tags: floor function, equal areas, geometry
x3yukari
11.12.2021 17:17
which line is $A_1B_1?$ I assume it's either the shortest or longest, but which
parmenides51
11.12.2021 17:34
there was nothing more mentioned in the official wording
GravitoelectromagnetismTM
12.12.2021 22:11
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 $A_1 B_1$?
x3yukari
12.12.2021 22:22
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 $2007 \cdot \frac{1}{2}$
GravitoelectromagnetismTM
12.12.2021 22:47
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.)