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Let \(\triangle{ABC}\) be equilateral. \(P\) is a point in the triangle such that \(AP, BP, CP = 2, 2\sqrt{3},4\) respectively. Find \(\angle{BPC}\)

Note by Victor Loh
2 years, 9 months ago

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Let \(X,Y,Z\) be the reflections of \(P\) w.r.t. \(AB,BC,CA\). Then \(XYZ\) is a triangle with side lengths \(a, \sqrt{3}a, 2a\) where \(a=2\sqrt{3}\). It's obvious that \(XYZ\) is a right triangle with \(\angle{XYZ}=30^{\circ}\). Hence \(\angle{BPC}=\angle{BYC} = \angle{BYX} + \angle{XYZ} + \angle{ZYC} = 30^{\circ}+30^{\circ}+30^{\circ}=90^{\circ}\). George G · 2 years, 9 months ago

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One way can be finding the area of the \(ΔABC\) considering each side of the equilateral triangle be \(x\) using Heron's formula and then finding the area of the three small triangles inside \(ΔABC\) in a similar way and then comparing them we get value of \(x\).Now using the formula \(1/2 \times BP \times PC \times \sin \angle BPC=ar(ΔBPC)\) and solving for \(\sin \angle BPC\) we get our answer. Bhargav Das · 2 years, 9 months ago

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