Latest Maths Olympiad Aptitude Question SOLUTION: Let AC be a line segment in the plane and B a point between A and C. Construct isosceles triangles PAB and QBC on one side of the segment AC such that \APB = \BQC = 120 and an isos

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Previous Question: HW:8'VXHM:701N>fBiHI4~70jm:²lp+VY Next Question: Produce AP and CQ to meet at K. Observe that AKCR is a rhombus and BQKP is a parallelogram. Put AP = x,CQ = y. Then PK = BQ = y, KQ = PB = x and AR = RC = CK = KA = x + y. Using cosine rule in triangle PKQ, we get PQ^2 = x^2 + y^2 - 2xy cos120 = x^2 + y^2 + xy. Similarly cosine rule in triangle QCR gives QR^2 = y^2 +(x+y)^2 - 2xy cos60 = x^2 +y^2+xy and cosine rule in triangle PAR gives RP^2 = x^2 + (x + y)^2 - 2xy cos 60 = x^2 + y^2 + xy. It follows that PQ = QR = RP.

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Let AC be a line segment in the plane and B a point between A and C. Construct isosceles triangles PAB and QBC on one side of the segment AC such that APB = BQC = 120 and an isosceles triangle RAC on the otherside of AC such that ARC = 120. Show that PQR is an equilateral triangle.: Asked In Maths Olympiad MAN (11 years ago); Unsolved; Is this Puzzle helpful? (1) (1) Submit Your Solution

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Previous Question: HW:8'VXHM:701N>fBiHI4~70jm:²lp+VY Next Question: Produce AP and CQ to meet at K. Observe that AKCR is a rhombus and BQKP is a parallelogram. Put AP = x,CQ = y. Then PK = BQ = y, KQ = PB = x and AR = RC = CK = KA = x + y. Using cosine rule in triangle PKQ, we get PQ^2 = x^2 + y^2 - 2xy cos120 = x^2 + y^2 + xy. Similarly cosine rule in triangle QCR gives QR^2 = y^2 +(x+y)^2 - 2xy cos60 = x^2 +y^2+xy and cosine rule in triangle PAR gives RP^2 = x^2 + (x + y)^2 - 2xy cos 60 = x^2 + y^2 + xy. It follows that PQ = QR = RP.

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