Complex Euclidean Geometry Question

Question

Let w be the incircle of a fixed equilateral triangle ABC. Let l be a variable line that is tangent to w and meets the interior of segments AC and BC at P and Q, respectively. A point R is chosen such that PR = PA and QR = QB. Find all locations of the point R over all choices of l.

So I drew the diagram, and tried angle chasing, and noted that ARMB is cyclic, where M is the midpoint of BC, but from there, I really didn't know what to do. Please provide some hints, but full proofs would be even better. I really need a full proof, it is kinda urgent. Thank You

Answer 1

This isn't an answer, just a figure to illustrate my comment. It's Community Wiki, so feel free to add stuff. (Preferably not insults, RandomKid!) $\ \ddot\smile$

In the coordinate system from Blue's answer, when the tangent $l$ meets the incircle $\omega$ at $K,$ $$ \begin{array}{ccccc} A = (-\sqrt3, -1), & B = (\sqrt3, -1), & E = (0, -1), & C = (0, 2), & K = (0, 1), \\ P = \left(-\frac1{\sqrt3}, 1\right), & Q = \left(\frac1{\sqrt3}, 1\right), & L = (-\sqrt7, 1), & M = (\sqrt7, 1), & R = (0, 1 + \sqrt5). \end{array} $$ I've taken the radius of the incircle as the unit of measurement. The points $A, B, L, M, R$ all lie on the true locus, as described by Blue's equation (1) with $r = 1$: $$ (x^2 + y^2 - 2)^2 = 4(4y + 5). $$ The equation of the elliptical arc $BMRLA$ that seems to approximate the true locus closely is: $$ \left(\frac{15 + 4\sqrt5}{29}\right)x^2 + \left(y - \frac{15 + 4\sqrt5}{29}\right)^2 = \left(\frac{14 + 25\sqrt5}{29}\right)^2. $$ (I don't know if this can be simplified, nor do I know why the approximation seems so close.)

Answer 2

Here's a trace of my GeoGebra sketch for the case with $|PA|=|PR|$ and $|QB|=|QR|$ (showing two instances of $R$ that trace separate arcs of the complete curve).

For an origin-centered incircle of radius $r$, as shown, the locus is a limaçon-like curve with equation

$$(x^2+y^2-2r^2)^2 = 4 r^3 (4y+5r) \tag{1}$$

---

For the case with $|AR|=|PR|$ and $|BR|=|QR|$ ...

The locus is a hyperbola with equation

$$- x^2 + 3 y^2 + 8 r y + 4 r^2 = 0 \tag{2}$$

---

Neither of these results is particularly "easy" to derive. I got them via coordinate-bashing in Mathematica. It's possible that the hyperbola follows from a nice geometric argument, but OP has stated that the conditions of the limaçon-like curve are in play. As one might expect, the limaçon-like curve is the trickier of the two, requiring a bit of parametric problem-solving, yet OP suggests that the problem shouldn't be that hard. Go figure.