# A locus

Something else I stumbled across during my ramblings on GeoGebra:

Extend each side of the triangle ABC in both directions. The point P lies on BC; the circle ABP cuts AC again at S; the circle ACP cuts AB  again at T. Circles BPT and CPS intersect at P and at X.

A) As P moves along the line BC, X moves around a fixed circle.
B) Also, this fixed circle has the same radius as the circle ABC.

This holds whether ABC is acute, right angled or obtuse; P is also not constrained to be between B and C, but can lie anywhere on the line – similarly, the points T and S do not have to lie on the interiors of AB and AC.
Here is a diagram of when P is not located on the interior of BC:

The green circle, along which X moves, has not changed.

Proof of A)
Case 1
First we will show that X is constrained to lie on a fixed circle when P is located between B and C.
Let ∠ABC = b and ∠ACB = c

By opposite angles in the cyclic quad BPSA, ∠PSA = 180 – b
∠PSC = b (straight line)
∠PXC = b (same segment in cyclic quad PCSX)
Similarly, by opposite angles in the cyclic quad PCAT, ∠PTA = 180 – c
∠PTB = c (straight line)
∠PXB = c (same segment in cyclic quad BPXT)
So ∠BXC = b + c
This is independent of the position of P; therefore, for all P on the interior of BC, the angles ∠BXC are all equal, and hence all the X’s lie on a common circle that passes through B and C (by the converse of the same segment theorem).
As a side-note, we can deduce that BXS and CXT are straight lines; ∠PXB = c, and ∠PXS = 180 – c (opposite angles in cyclic quadrilateral PCSX); these add to 180 so P, X and S are collinear.
Similarly, C, X and T are collinear.

Case 2
Now we will show that when P is not between B and C, X still moves along a fixed circle, and also that this fixed circle is the same circle as in Case 1.

Using same segment theorem in cyclic quadrilateral ABSP, ∠ABP = ∠ASP = b
Using same segment theorem in cyclic quadrilateral CXSP, ∠CSP = ∠CXP = b
Also, ∠ACP = 180 – c (straight line)
∠ATP = c (opposite angle in cyclic quadrilateral ACPT)
∠BXP = 180 – c (opposite angle in cyclic quadrilateral  BXPT)
So ∠BXC = ∠BXP – ∠CXP = 180 – b – c
Again, this is independent of the location of P, so X moves along a fixed circle. Also, this angle is 180 minus the angle BXC in Case 1. Therefore, by the converse of opposite angles in a cyclic quadrilateral are supplementary, this circle is the same circle as the circle in Case 1.
Note that X will never completely trace round the circle; as P travels along the line BC to infinity, X will approach, but never meet, the intersection of the circle and the perpendicular bisector of BC.

Proof of B)
This is very quick. Refer to the diagram from B) case 1:

We employ the full sine rule:
$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R$
where R is the radius of the circumcircle.
In triangle ABC, the diameter of the circumcircle is:
$\frac{BC}{sin(BAC)}$
In triangle XBC, the diameter of the circumcircle is:
$\frac{BC}{sin(BXC)}$
But ∠BAC = ∠BXC = 180 – b – c
Hence the lengths of the diameters and radii of circles ABC and XBC are equal.