NOT A REAL PLANE ON FIRE. Source.

Appropriately for May Day, this question will cause you to send distress signals. Prize this week: you get to strut around all weekend telling people how much smarter you are than they, and they have to humbly agree. You may force up to three strangers to “kiss the ring.”

Note: Figure not drawn to scale.

In the figure above, P is the center of a circle, and Q and R lie on the circle. If the area of the sector PQR is 25π3 and the length of arc QR is 5π2, what is the measure of ∠RPQ?

UPDATE: props and ring kisses to Elias, who answered on Facebook, and to JD, who did his thing right here on the site (and spelled out the solution quite elegantly). Side note: I’d love to be able to integrate those two commenting systems — is that possible?

Solution below the cut:

##### Solution

This difficult circle question its, at its core, a ratio question. Remember that, with circles, part/whole = part/whole, or:

(For a more rigorous explanation and practice with this, see this post.)

We can set set up ratios using this framework for both the area of the sector and the length of the arc:

Sector Area:
Arc Length:

At this point, you could just use Wolfram Alpha, but surely you’d rather do it the long way. I know your pulse quickened a bit when you noticed that both equations share the same left side, and can therefore be combined:

And neatly solved for r:

$\dpi{150}&space;\fn_cm&space;\frac{25\pi&space;^2}{r^2}=\frac{5\pi}{2&space;r}$
$\dpi{150}&space;\fn_cm&space;25\pi&space;^2(2r)=5\pi&space;(r^2)$
$\dpi{150}&space;\fn_cm&space;50\pi&space;^2r=5\pi&space;r^2$
$\dpi{150}&space;\fn_cm&space;10\pi=r$

Sweet. If r = 10π, we can now solve for x using one of our original ratios:

$\dpi{150}&space;\fn_cm&space;\frac{x}{360}=\frac{5\pi^2}{2\pi(10\pi)}$
$\dpi{150}&space;\fn_cm&space;\frac{x}{360}=\frac{5\pi^2}{20\pi^2}$
$\dpi{150}&space;\fn_cm&space;\frac{x}{360}=\frac{1}{4}$

$\dpi{150}&space;\fn_cm&space;x=90$

Oh heck yes. Laser show.

JD says:

I get 90 degrees.

Let X = the angle we’re looking for so that X/360 represents the portion of the 360 degree circle.

Circumference:

X/360 * 2πr = 5π^2

X/360 = 5π/2r

Area:

X/360 * πr^2 = 25π^3

X/360 = 25π^2 / r^2

Set both formulas equal to onn another (because they both equal X/360) and solve for the radius:

r= 10π

plug that back into the original formulas and you get X=90 in both of them.

I agree, not drawn to scale!