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The question is given in the following picture:enter image description here

The question and its answer is given in the following picture:enter image description here

I wonder why in the question the author used the word minimal and not minimum, could anyone explain this for me?

Also, the question said the minimal distance between any point on sphere 1 and any point on sphere 2, but the answer speaks about 2 point exactly that are shown in the pink color in the following picture:enter image description here

Could anyone clarify this for me?

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  • $\begingroup$ Can you check whether you asked this the wrong way around? I see minimum in the question. Generally, do not get hung up on the difference in situations in which the distinction is not important. $\endgroup$
    – Carsten S
    Jun 20, 2017 at 8:51
  • $\begingroup$ In the unanswered version of the exam it is written minimal @CarstenS $\endgroup$
    – user426277
    Jun 20, 2017 at 9:01
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    $\begingroup$ Please type out your question, don't just post a photo. Photos can take time to load, are not searchable on the site, and are not visible to some members on the site. (NB: The diagram of the circles is accceptable!) Here is the MathJax tutorial for writing Maths on the site, should you need it $\endgroup$
    – lioness99a
    Jun 20, 2017 at 9:04
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    $\begingroup$ Imagine that two years from now I want to find the minimal (or minimum!) distance between two circles and search for previous questions on that topic in StackExchange. The search engine won't find your question because the words are not typed but just part of the photo... $\endgroup$
    – Evargalo
    Jun 20, 2017 at 9:17
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    $\begingroup$ @Idonotknow Evargalo's answer is correct for the first half, the second half of the sentence just means that on some devices, photos just don't load and therefore people can't see the content $\endgroup$
    – lioness99a
    Jun 20, 2017 at 9:26

2 Answers 2

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I think the question meant some thing like this: take any point $(x_1,y_1)$ on first sphere and $(x_2,y_2)$ on second sphere.calculate the distance between these two points. Now by varying $(x_1,y_1)$ on sphere 1 and $(x_2,y_2)$ on sphere 2 we get different distances. The question is to find the pair of points such that this distance is minimum.

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  • $\begingroup$ but the choices contain distances and not points. $\endgroup$
    – user426277
    Jun 20, 2017 at 9:29
  • $\begingroup$ yes. to find that distance between that pair of points. $\endgroup$
    – praveen kr
    Jun 20, 2017 at 9:32
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Minimum and minimal are I'm sure intended to mean the same thing in this situation. As for your question 'why are they only considering the distance between $2$ specific points?':

Imagine taking $2$ points $x,y$, one on each circle such that the $2$ points are not the exact ones in your picture. Then draw the line connecting them and add in the radii from each point the respective centres of the circles that the points lie on. You should now have a piecewise linear curve connecting the $2$ radii. Since in Euclidean geometry, the shortest curve between $2$ points is a straight line, the length of this curve (which is the sum of the lengths of the radii plus the straight line distance between $x$ and $y$) is strictly greater than the straight line distance between the centres of the circles. Since this distance is the sum of the $2$ radii plus the distance between the $2$ points in your picture we can conclude that the shortest distance between $2$ point on the circles, is attained by the $2$ points in your picture (i.e the points the answer considers).

More mathematically: Let $r_1$, $r_2$ be the lengths of the radii of the circles. Let $x,y$ be points - one on each circle. Let $u,v$ be the points in your picture (i.e the points of intersection of the circles and the straight line between centres). Then by what I described above for all $(x,y) \neq (u,v)$,

$$r_1 + r_2 + \text{distance}(x,y) > r_1 + r_2 + \text{distance}(u,v)$$ so $$\text{distance}(x,y) > \text{distance}(u,v).$$

Hence the minimal distance is attained by the straight line distance between $u$ and $v$.

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  • $\begingroup$ Great answer, but I think your inequalities should use $\geqslant$ and not $>$. $\endgroup$
    – Evargalo
    Jun 20, 2017 at 9:25
  • $\begingroup$ I said at the beginning that $(x,y) \neq (u,v)$, but I should definitely make that more clear - thanks! $\endgroup$ Jun 20, 2017 at 9:26
  • $\begingroup$ Oh yeah sure. I just read too fast, it is clear actually ! $\endgroup$
    – Evargalo
    Jun 20, 2017 at 9:26

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