# Let a,b,c,d real numbers such that $|a-b|=2,|b-c|=3,|c-a|=4.$Then the sum of all possible values of $|a-d|$is...

Let $a,b,c,d$ real numbers such that $\lvert\,a-b\,\rvert=2,\lvert\,b-c\,\rvert=3,\lvert\,c-d\,\rvert=4$.Then the sum of all possible values of $\lvert\,a-d\,\rvert$ is...

Now obviously one method is to take off the modulus and and take cases... which would obviously be a lengthy method.

Is there any method which would take this sum down in $6$ or $7$ lines...

• $d$ isn't represented in your absolute differences; it seems completely unrestrained.
– Tom
Oct 24, 2017 at 14:37
• questio has been editted Oct 25, 2017 at 7:46

I'm reading your third condition as $|c-d|=4$.
I'm lining up the points on a figure showing the number line with ticks. WLOG we may assume $a=0$, $b=2$ and then obtain for $c$ the two possibilities $c_1=-1$, $c_2=5$. This leads to the four possible $d$-values $d_{11}=-5$, $d_{12}=3$, $d_{21}=1$, $d_{22}=9$. The sum of the absolute values of these four numbers is $18$.