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This is probably an easy question, but I am not capable of solving it. I have two particles which distance of each other is $r$. One particle is inside a box with diameter $d$. Now I want to determine the smallest distance from particle 1 to the box.

Is there an easy way I don't see to do that?

Here is what I mean:

enter image description here

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    Diameter of a box? – 2017-01-28
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    Are you asking only about the 2D case depicted in your figure? – 2017-01-28
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    Can you clarify more? Is particle 2 always in the middle of the box as in the picture? What are the input variables? Just r and d are not enough to solve the problem. Also, if d is defined as in the picture, I would not call it the diameter, maybe radius. Diameter would be the distance between the corners or the width. – 2017-01-28

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You simply have a square enclosed by a circle. So draw a circle of radius $r$ centered at particle 2 and then you will see that the closest distance is happening when the circle meets the edges of the square.

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    This approach is "circular" in that you need to know the radius $r$ in order to draw such a circle. However the Question asks about the distance of point 1 (outside the box) to the box, not about the distance of point 2 to the box. This makes the possible nearest points slightly more interesting. – 2017-01-30
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    That would still be the case since the distance $r$ is given. – 2017-01-30
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    It is of course easy to find the distance from the center of the box (2?) to any point (1) outside the box of (half)"diameter" $d$. But the circle you thus describe of known radius $r$ need not intersect "the edges of the square". – 2017-01-30
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Going by your drawing, moving particle 1 clockwise around particle 2 at the "center" of the box, its perpendicular distance from the box is least when the perpendicular meets the corner of the box. But continuing the rotation thru the right angle which is vertical to the corner of the box, the distance from the box, now measured radially from the corner, continues to diminish and is least when $r$ passes through the corner. Looks like the closest approach to the box = $r - d\sqrt2$

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    I don't think the Question is about rotating point 1 around the center of the box to minimize the distance to the nearest point on the box. – 2017-01-30
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    @hardmath- How do you understand the question? – 2017-01-30
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    I'm going out on a limb since the OP did not respond to my clarification request, but my guess is that we are given a square centered at point 2, with edges parallel to the coordinate axes of length $2d$, and a point 1 outside the square, find the distance from point 1 to the perimeter of the square. I suspect $r$ simply denotes the distance between the two points and does not suggest a rotation to be performed – 2017-01-30
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    Your interpretation does provide a proof that the distance from point 1 to the box (labelled with a ? in the diagram) cannot be determined solely from $r$ and $d$, since rotating the point 1 around point 2 leaves $r$ (and $d$ of course) unchanged, but you point out that it varies the ? distance. – 2017-01-30
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    Yes, since the problem is characterized as "optimization", something has to be variable. The distance between particles 1 and 2 is given as r, but the position of particle 1 is not given. If particle 1 were a tired swimmer left r meters offshore from the center of a square island, where in that circle of waters would he most like to be tossed, so as to reach land soonest? – 2017-01-30
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    The tag "optimization" by itself is not sufficient to convince me that rotation is intended. Rather the picture illustrates that the shortest distance is determined by choosing the point on an edge closest to point 1. – 2017-01-30
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    @hardmath- On that interpretation, would you agree that the solution is geometrically obvious but impossible as a calculation? I mean, we can drop the perpendicular to the box, but we can't express its length in terms of r and/or d without introducing unknown quantities? – 2017-01-30
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    Yes, I agree with you on that and your Answer points that out. – 2017-01-30