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Re: As good as it gets
Posted:
Dec 10, 2010 1:24 PM


> Only condition to feel the magic of following problem > is to stick to pure geometry > > Q.In a triangle ABC, point P is taken so that angle > PAB=30 degree, angle PAC=80 degree, angle PCA=20 > degree and angle PCB=20 degree.Find angle PBC.
I am giving its solution Q.In a triangle ABC, point P is taken so that angle PAB=30 degree, angle PAC=80 degree, angle PCA=20 degree and angle PCB=20 degree.Find angle PBC. Sol.Contruct an equilateral triangle BMC, so that M is on the same side of BC as of A.Join AM. Considering triangles BAM and BAC. BA is common side. Angle ABM=angle ABC=30 degree. BC=BM So, triangle BAM is congruent to BAC. So, angle BMA=40 degree. Angle AMC=6040 =20 degree=angle ACM AM=AC Angle CAP=angle CPA=80 degree. Hence, AC=PC Considering triangles, PCB and ACM PC=AC Angle PCB=angle ACM=20 degree. BC=MC. Hence, triangle PCB is congruent to triangle ACM. So, angle PBC=20 degree.
I gave this problem different title because I was most thrilled after solving this problem, not because I use different concept for solving this problem but because final diagram that came out looked really beautiful because of symmetry of the triangles.And I was reminded of the following problem, which again is very popular one.This is one of the 5 Hardy's perennial along with Langley problem as given in the book ?Geometry Revisited? by H.S.M. Coxeter and S.L. Greitzer.
Q.In a square ABCD point M is taken so that MA=MB and angle MAB=15 degree.Find angle CMD.



