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Monday, 30 April 2007
Re: China Challenge
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All right, already.  Never let it be said I copped out of a math problem.

(1) Sketch the quadrilateral ABCD with the information given.  It's just two 30-60-90 triangles joined symmetrically along the hypoteneuse.  All the angles are 30-60-90-120, all the lengths combinations of 1, 2, and sqr(3).

(2)  Question (i) is now trivial, from symmetry.  (Though of course you'd have to find a politer way to say that in an examination paper.)

(3)  For question (ii), again from symmetry, it is sufficient to calculate the angle A1-E-C1, which can easily be done from the lengths involved, since this is yet another 30-60-90 triangle.  Answer: 90 degrees.

(4)  For question (iii), first parallel-transport the triangle B-C-C1 until B coincides with A.  You now have a solid angle at A formed by three intersecting planes, two of them—both with angle 30 degrees at the apex—at right angles to each other (i.e. the planes A-B-C-D and B-C-C1).

(One of those 30 angles is C-B-C1.  The other you get from parallel-transporting BC until B coincides with A, since angle DAB is 120 and angle ABC is 90.)

Chopping off these two planes by a vertical one to make 30-60-90 triangles with sides length 1-2-sqr(3), one other face of the tetrahedron you just made—the one with an angle at the vertex A—has sides 2, 2, sqr(2).  Its vertex angle is therefore 90 degrees.
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Posted on 04/30/2007 9:49 AM by John Derbyshire
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