Abundances Of Absolutes

Algebra Level 2

$\mathcal A=\displaystyle\sum_{i=1}^{2016}\left|x-a_i\right|$

Let $a_1,a_2,a_3,\ldots ,a_{2016}$ form an increasing arithmetic progression (AP) which consists of only positive terms. Let the minimum value of $\mathcal A$ be $2016^2$ for real $x$ . Then find the sum of all possible values of the common difference of AP.

Notation : $| \cdot |$ denotes the absolute value function .

1^2

2^2

3^2

4^2

5^2

2 solutions

Rishabh Jain
Jul 17, 2016

Relevant wiki: Absolute Value Equations - Intermediate

Writing it as:- $|x-a_1|+|x-a_2|+\cdots+|x-a_{1008}|+|a_{1009}-x|+\cdots +|a_{2016}-x|$

Applying general form of triangle inequality,

$\mathcal A\ge |(x-a_1)+(x-a_2)+\cdots +(x-a_{1008})+(a_{1009}-x)+\cdots+(a_{2016}-x)|$

Notice that all $x$ cancel and we are left with:

$(a_{2016}+\cdots+a_{1009})-(a_1+\cdots+a_{1008})$

$=(a_{2016}-a_{1008})+(a_{2015}-a_{1007})+\cdots+(a_{1009}-a_{1})$

$=(1008d)+(1008d)+\cdots\cdots\small{ \text{(1008 times)}}$

$=1008^2 d$

Hence, $\mathcal A\ge 1008^2 d$

We are given that:-

$1008^2 d=2016^2\implies d=\left(\dfrac{2016}{1008}\right)^2=\boxed{2^2}$

$\star~$ NOTE:- Equality in $\mathcal A$ holds when $a_{1008}\le x\le a_{1009}$ .

Also see here for a proof of generalized form of triangle inequality.

Beautiful solution...

Ujjwal Mani Tripathi - 4 years, 10 months ago

I don't understand why equality in A holds when a1008<= x <= a1009? Can you please explain it

Sherzod Khannaev - 4 years, 4 months ago

Why equality of A holds on this interval

pranay pal - 4 years ago

Doesn't the equality hold when x=(a1+a2016)/2 ???

Luke Videckis - 4 years, 11 months ago

$\dfrac{a_1+a_{2016}}2=\dfrac{a_{1008}+a_{1009}}2$ , and since $a_{1008}<\dfrac{a_{1008}+a_{1009}}2<a_{1009}$ , as mentioned in my solution since this value of $x$ lies in interval of equality, equality holds for this value of $x$ also. What you've mentioned is just one value of $x$ at which equality holds. And the most amazing thing is: whatever $x$ you take in the interval in which equality holds, you will always get the same minimum value( $=1008^2 d$ ) each time.

Rishabh Jain - 4 years, 11 months ago

I thought it was just that one value but I understand now. Thanks!

Luke Videckis - 4 years, 11 months ago

@Luke Videckis – No problem... :-)

Rishabh Jain - 4 years, 11 months ago

Harmanjot Singh
Jul 24, 2016

Q statement is wrong. Question should have asked min possible value of 'd'

As someone stated, A is min when x=(a1+a2016)/2 From a geometrical/distance point of view, it is better seen when x=(a1008+a1009)/2, which is basically the same thing Then, if X=(a1007+a1008)/2 A(X)=(X-a1)+...+(X-a1008)+(a1009-X)+...+(a2016-X) if ai=a1+r*(i-1) then both a1 and Xs cancel themselves and it is 1007,5r+1006,5r+...+0,5r+0,r+...+1007,5r

I looked at it geometricaly and it is 2 (r/2)+2 (3r/2)+...+2(2015r/2) which is r+3r+...+2015r=r+3r+...+2017r=r(2016*504) So r=4

Rainer Retzler - 2 years, 3 months ago

Abundances Of Absolutes

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