1st Mexico 1987 Problem B4 ABCD is a tetrahedron. The plane ABC is perpendicular to the line BD. ∠ADB = ∠CDB = 45 o and ∠ABC = 90 o . Find ∠ADC. A plane through A perpendicular to DA meets the line BD at Q and the line CD at R. If AD = 1, find AQ, AR, and QR.
60 o , 1, √3, √2 Solution The lines BA, BC, BD are mutually perpendicular. The triangles BAD and BDC are isosceles, so BA = BD = BC. Hence ACD is equilateral and ∠ADC = 60 o . It is convenient to take Cartesian axes with origin B. Take A as (1/√2, 0, 0), C as (0, 1/√2, 0, 0), D as (0, 0, 1/√2). Then Q is (0, 0, -1/√2) and R is (0, √2, -1/√2). So we find (by Pythagoras) AQ = 1, AR = √3, QR = √2. 1st Mexican 1987 © John Scholes jscholes@kalva.demon.co.uk 21 February 2004 Last corrected/updated 21 Feb 04