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 \(\angle{BPC}=\angle{BYC} = \angle{BYX} + \angle{XYZ} + \angle{ZYC} = 30^{\circ}+30^{\circ}+30^{\circ}=90^{\circ}\).

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.

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestLet \(X,Y,Z\) 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 \(\angle{BPC}=\angle{BYC} = \angle{BYX} + \angle{XYZ} + \angle{ZYC} = 30^{\circ}+30^{\circ}+30^{\circ}=90^{\circ}\).

Log in to reply

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.

Log in to reply