A circle is centered at O and has an area of 48pi. Let Q and R be points on the circle, and let P be the circumcenter of triangle QRO. If P is contained in triangle QRO and triangle PQR is equilateral, then find the area of triangle PQR. (3sqrt3 is not the answer)

ianw11 Dec 21, 2020

#1**-2 **

OQ = OR = 4√3

Circumradius of a triangle = $\frac{ a b c}{\sqrt{(a + b - c) (a - b + c) (-a + b + c) (a + b + c)}}$

Inputting 4√3, we have $4\sqrt{3} = \frac{48c}{\sqrt{(8\sqrt{3} - c) (c) (c) (8\sqrt{3} + c)}}$

Which simplifies to $4\sqrt{3} = \frac{48c}{\sqrt{(8\sqrt{3} - c)(8\sqrt{3}+c)(c^2)}}$

By difference of squares, $4 \sqrt{3} = \frac{48c}{\sqrt{192-c^2)(c^2}} = \frac{48c}{c \sqrt{192-c^2}} = \frac{48}{\sqrt{192-c^2}} = \frac{48 \sqrt{192-c^2}}{192-c^2}.$

Cross multiplying, $768 \sqrt{3} - 4c^2 \sqrt{3} = 48 \sqrt{192-c^2}$ and $192 \sqrt{3} - c^2 \sqrt{3} = 12 \sqrt{192-c^2}$ and squaring, $(192-c^2)(192-c^2)(3) = 144(192-c^2),$ so $192-c^2 = 48,$ and consequently, $c=12.$ Thus, 12^2 / 4 * √3 = 144/4 * √3 = 36√3.

$\square$

Seems like the person who answered forgot to square the 12 :)

Pangolin14 Dec 21, 2020

#2**+4 **

A circle is centered at O and has an area of 48pi. Let Q and R be points on the circle, and let P be the circumcenter of triangle QRO. If P is contained in triangle QRO and triangle PQR is equilateral, then find the area of triangle PQR.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

OQ = OR = r = 4√3

∠QOR = 30º ∠QOP = 15º ∠QPR = 60º

QR = 2 (OQ * sin∠QOP) ==> QR = 3.586301889

Height of ΔPQR h = sqrt[PQ^{2} - (QR/2)^{2}] = 3.105828541

*[PQR] = h * QR / 2 = 5.569219382 u ^{2}*

jugoslav Dec 22, 2020

#3**+1 **

There seem to be two conflicting answers. If I did make a mistake, can you explain what I did wrong?

Pangolin14
Dec 22, 2020