Wikipedia:Reference desk/Archives/Mathematics/2013 April 4

= 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)


 * 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)


 * 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)


 * 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)