IGO Advanced P1s used to be better than this in my opinion. See the image below.

Attachments:

Let $D, E, F$ the midpoints of $AB, BC, CA$ respectively. As we have that $AB=BC=CA$, $BX=CY=AZ$, $XA=YB=ZA$. Then: $AD=BE, \angle BAX= \angle DAZ = \angle EBX, AZ=BX \Rightarrow ADZ \cong EBX \Rightarrow DZ=XE$. Then, it’s well known that $(BDE)=\frac{1}{4} * (ABC)= (XYZ)$, and $BDE$ and $XYZ$ are equilateral, so $BDE \cong XYZ \Rightarrow ED =XZ$ All of that tell us that $DXEZ$ is an isosceles trapezoid, so $DXEZ$ must be cyclic. $XD$ and $YE$ are homologous cevians on $AXB$ and $BYC$ (medians to $AB$ and $AC$) so $XD=YE$. $XZ=XY$, so $DE=XY$ All of that tell us that $DXEY$ is cyclic because is an isosceles trapezoid. We conclude that $Z, D, X, E, Y$ are on the same circumference. Analogous, because equilateral triangle $ABC$ symmetry, $X, E, Y, F, Z$ are on the same circumference. That implies that $Z, D, X, E, Y, F$ are on the same circumference. But note that, in an equilateral triangle, the midpoints are the tangency points of the incircle, so $X, Y, Z$ are on the incircle. I’m still wondering if it was possible to supose that $AZ=CY=BX$ and $ZC=BY=AX$ by the imagen on the test or the order as the congruent triangles are named in the statement.

Let $a$ be a side length of $\triangle{ABC}$ and let $S$ be four equal areas. Then $4S=\frac{(a^2)\sqrt{3}}{4}$. Since $\triangle{XYZ}$ is also equilateral with area $S$, we can find $XY=\frac{a}{2}$. It's easy to compute $AI=BI$ and inradius. By noting $AZ=BX$, apply Ptolemy's theorem on $(AIXB)$ to get $XI=inradius$. $\blacksquare$

Let's Bary! Let $[A_1A_2A_3] = \text{Area of } \triangle A_1A_2A_3$ Consider point $Y$ $\frac{[YBC]}{[ABC]} = \frac{1}{4}$ Let $\frac{[YCA]}{[BCA]} = p$ and $\frac{[YAB]}{[CAB]} = q$ $\implies Y \equiv (\frac{1}{4},p,q)$ with $p+q=\frac{3}{4}$ By symmetry, $X \equiv (p,q,\frac{1}{4})$ and $Z \equiv (q,\frac{1}{4},p)$ We know $\frac{[XYZ]}{[ABC]} = \frac{1}{4} = \begin{vmatrix} p & q & \frac{1}{4}\\ \frac{1}{4} & p & q \\ q & \frac{1}{4} & p \\ \end{vmatrix}$ Solving we get, $\frac{15}{16} = 4p^3 + 4q^3 - 3pq$ Factorizing and spamming $p+q = \frac{3}{4} \implies pq = \frac{1}{16} $ If $a$ is the side of equilateral $\triangle ABC$, then the incircle has equation: $-a^2(ay+yz+zx) + \frac{a^2}{4}(x+y+z)^2$ $\forall$ $X$, $Y$, $Z$: $x+y+z = 1$ and $xy+yz+zx = \frac{1}{4}$ Easy to check that they lie on the incircle!

Simple trig exercise. WLOG let the side length of triangle $ABC$ be 2. By the area condition we must have: \[ [XYZ] = \frac{1}{4}[ABC] \implies XY = \frac{1}{2}a = 1\]since the triangles are similar. Clearly the centers of the triangles coincide, since they $XYZ$ are constructed symetrically. Call the center $O$. We now calculate the incircle radius r, as the distance from $O$ to $M$, the midpoint of $BC$ as: \[ \frac{1}{\sqrt{3}} = \tan(30) = \tan(\angle CAI) = \frac{OM}{MB} = r \]Similarly compute $ R = OX$ with the cosine rule in $OXY$: \[ 1 = XY^2 = XO^2 + YO^2 - 2\cdot XO \cdot YO \cdot \cos(120) = 2R^2 - 2R^2 \cdot \frac{-1}{2} = 3R^2 \implies R = \frac{1}{\sqrt{3}}\]Thus we have $r = R$ and are done

Probably a new approach. Note that due to congruences we have that $AZ\cdot AX=BX\cdot BY=CY\cdot CZ \iff$ powers of $A$, $B$, and $C$ wrt to $(XYZ)$ are equal. This means that $(XYZ)$ and $(ABC)$ are concentric, but since the triangles are equilateral we can say that triangles $ABC$ and $XYZ$ have the same centers (note that incenter and circumecenters are the same here). Now denote by $r$ the radius of $(XYZ)$, $R$ the inradius of triangle $ABC$, and by $M$ the midpoint of side $AB$, then by POP, we have that $AI^2-r^2=AZ\cdot AX=BX\cdot AX=2\frac{[BXA]}{\sin 120^{\circ}}=\frac{[ABC]}{\sqrt{3}}=\frac{AB\cdot AC\cdot\sin 60^{\circ}}{2\sqrt{3}}=\frac{AB \cdot AB\cdot\sqrt{3}}{2\cdot 2\cdot \sqrt{3}}=AM^2=AI^2-R^2\Rightarrow r=R$ and we are done.