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April 4

Curious geometry problem

I saw this puzzle in a journal my institution subscribes to (it is a recreational maths journal) and don't know if I did it right--it's been quite some time since maths in university for me! It goes: hexagon ABCDEF is inscribed in a circle with the following side lengths: AB=BC=DE=EF=22 units, CD=AF=20 units. What is the radius of the circle? This is how I solved it.

(1) Consider the rectangle ACDF. AD=CF=2r (a diameter), and call AC=DF 'x'. Pythagoras tells us $20^{2}+x^{2}=(2r)^{2}$ .

(2) Next consider the isosceles trapezoid BEFA. BF=AE so call this y. BE is a diameter. Ptolemy's theorem tells us that $20(2r)+22^{2}=y^{2}$ .

(3) Finally consider the trapezoid ABCF. Its diagonals are AC (x) and BF (y). Ptolemy's once more gives us $22(20)+22(2r)=xy$ .

Square this last equation and substitute in the known expressions y^2 and x^2 for the rhs. I get r=√252. Can anybody check my solution? 72.128.82.131 (talk) 23:16, 4 April 2013 (UTC)[reply]

I've not done maths on Wiki before, so bear with me. I agree with your points (1), (2), and (3). But that's not the answer I get when I square the last equation and equate it with the other two. So as not to spoil your fun, I won't give my answer unless you ask, but I note that

x^{2}

is a difference of two squares with a factor of

r+10

, and you can cancel that with one of the

r+10

on the (3)-squared side, to reduce to a quadratic with none of the terms going to zero. So your answer must be a surd, not a simple square root. TrohannyEoin (talk) 01:53, 5 April 2013 (UTC)[reply]

Here's a different solution. Draw the isosceles trapezoid BCDE. BE is a diameter of the circle. Call the centre of the circle O. If H is the projection of C onto the diameter BE, then OH = 10. Thus OHC is a right triangle with leg OH = 10 and hypotenuse OC=r. BHC is also a right triangle, with leg HB = r - 10 and hypotenuse BC = 22. Apply Pythagoras to each triangle and equate the expressions obtained for CH. Then solve for r. I get an answer that's approximately 21.34. 64.140.121.87 (talk) 03:06, 6 April 2013 (UTC)[reply]

BTW, you can check these answers geometrically, with a compass and a ruler:

1) Set the compass to the radius you came up with and make a full circle.

2) Mark and label a starting point on the circle.

3) Set the radius to either 20 or 22, then measure off that distance from that point to the next points on the circle. Label those.

4) Continue until all points have been found. If the last two are the correct distance apart, then your answer is correct. StuRat (talk) 05:34, 6 April 2013 (UTC)[reply]