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Monday, 30 April 2007
Re: China Challenge
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.
Posted on 04/30/2007 9:49 AM by John Derbyshire
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