How to solve?

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

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 7 years, 5 months ago

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 - 7 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$