Geometry - Area Chasing 4

Geometry Level 3

The triangle $ABC$ shown above is equilateral. If $BP=3,AP=5$ and $CP=4$ , find the area of $\triangle ABC$ . If your answer can be expressed as $\dfrac{a}{b}\sqrt{c}+d$ where $a,b,c$ and $d$ are positive coprime integers and $d$ is square free, find $a+b+c+d$ .

1 solution

A Former Brilliant Member
Oct 24, 2017

Rotate $\triangle ABC$ $60^\circ$ in the counter-clockwise direction at point $A$ . Since $\phi=\phi$ , $\angle PAP’=60$ . So $\triangle PAP’$ is equilateral.

$\angle BCC’=2(60)=120$

$\angle PCB + \angle P’CC’=120-90=30$

But $\angle P’CC=\angle PBC$ so $\angle PBC+\angle PCB=30$ . Hence $\angle BPC=150$ .

$A_{ABC}=A_{AP’CP}+A_{BPC}$

$=\dfrac{\sqrt{3}}{4}(5^2)+\dfrac{1}{2}(4)(3)+\dfrac{1}{2}(3)(4)(\sin 150)=\dfrac{25}{4}\sqrt{3}+9$

Finally, the desired answer is

$a+b+c+d=25+4+3+9=41$

I think you meant for c to be square free (and not d)?

Kurt Kleinberg - 2 years, 2 months ago

Geometry - Area Chasing 4

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