Square Inscribed in Sector

You are given a circle of radius 5 with center O. A and B lie on the circle. The length of arc AB is 6. A square with side length x is inscribed in sector AOB such that one corner of the square is on OA, one corner on OB, and two corners on arc AB. What is the area of the square (and therefore the value of x)?

In case the solution is not unique, what are the min/max values of x?

Solution:

My answer to every problem is to put it on a coordinate grid, and this puzzle is no exception. I put O on the origin, and considered just the upper half of the given sector. In this way, I "see" only the upper half of the square, so the figure inscribed in the sector is a rectangle that measures k units high by 2k units wide. (I used k instead of x/2 to avoid confusing myself!) At this point, the solution is just a matter of slogging through it. The central angle is the arc length divided by the radius, 3/5 radians. Using this angle, it's easy to identify the coordinates where the rectangle intersects the x axis and the circle. The equation of the circle is

x²+y²=25

So from the point on the circle, I get

(2k+k/tan(3/5))²+k² = 25,

which simplifies to

(cot²(3/5)+4cot(3/5)+5)k² = 25.

So x² = 4k² (the area of the square) is given by

x² = 4k² = 100/(cot²(3/5)+4cot(3/5)+5)

More about this puzzle

I was puzzled by the sentence beginning "In case the answer is not unique..." It got me thinking that perhaps by sliding the square around the arc, it might be possible to let it grow. In this case, the corners on OA and OB will be different distances from O. However, after some thought, I see that's not possible, because if two corners are on the circle, then the other two corners will be equidistant from O.

The only other possibility for an alternate solution that I could think of is this: Looking at the diagram in a mirror, the "left" corners could touch OA and OB, respectively, outside the circle, and the "right" corners could touch the left side of the circle, making a square considerably larger than that of the first solution. However, such a square could not be said to be "inscribed" in the sector, so it's unlikely the author of this puzzle had such a solution in mind. Besides, no such solution is possible given the length of arc AB, which fixes the central angle at 6/5, which is less than pi/2. (On the other hand, it would be fun to rephrase the puzzle to eliminate the word "inscribed", which would leave open the possibility of an alternate solution, making it much more challenging.)

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