AB=BC=AC=a, XX1=XX2=x, XP1=XP2=y, XT1=XT2=z, area(ABC)=S

S3/area(T1X1B)=(z/X1B)2 ; area(T1X1B)/S=(X1B/a)2 => S3/S=(z/a)2 =>

=> z/a=sqrt(S3)/sqrt(S)

We have x+y+z=a, and we obtain now

sqrt(S1)/sqrt(S)+sqrt(S2)/sqrt(S)+sqrt(S3)/sqrt(S)=1 =>

sqrt(S)=sqrt(S1)+sqrt(S2)+sqrt(S3) then

S=[sqrt(S1)+sqrt(S2)+sqrt(S3)]2

The result holds for any point X in int(ABC), so the fact that X is on the inscribed circle has not any importance!

You don't even need to consider the similarity.

x = 2 sqrt(S1/sqrt(3)), similar for y and z and then x + y + z = a and S follows.

It works just the same for an arbitrary triangle, you get the same result because you still have similarity and the sum of the corresponding sides in the 3 smaller triangles = the corresponding side of the big one

Gabriel T. Prajitura, Dinu Teodorescu

Thank you for your interest in the task! All of your solutions are correct. I propose attempting to solve this problem with slightly altered conditions:

Given: S1+S2+S3=t.

Find: The area (S) of triangle ABC.

In the case when the only given info is t = S1+S2+S3, and point X belongs to the circle inscribed in the equilateral triangle ABC, a lot of algebraic manipulations give S = 2t ... in 2 steps.

First, from earlier answers and notation

S= S1+S2+S3+2[sqrt(S1S2)+sqrt(S2S3)+sqrt(S1S3)] =

= t + (xy+yz+xz)*sqrt(3)/2. ((1))

In the next step, without loss of generality, assume x < y < z.

Furthermore, x-and-y appear on the "other" side of the inscribed circle because of symmetry in ABC.

Second, for the inscribed circle the tangent-secant theorem implies

https://en.wikipedia.org/wiki/Tangent%E2%80%93secant_theorem

xz=((x+z-y)/2)^2

yz=((y+z-x)/2)^2

xy=((z-x-y)/2)^2

which can be added to obtain:

xy+xz+yz = (x^2 + y^2 + z^2)/2.

However, x^2 + y^2 + z^2 = 4(S1+S2+S3)/sqrt(3) = 4t/sqrt(3).

Combining ((1)) with the last two equalities gives S = 2t.