HMMT Spring Team Round, 2021

Solution:

In all solutions, let $O$ be the center of $\omega$. Then $\angle DOE = 90^{\circ}$, so $\angle DFE = 45^{\circ}$, so $\angle DXE = 135^{\circ}$. Let $\Gamma$ be the circle centered at $A$ with radius $AD = AE$, and let $X' = \overrightarrow{AX} \cap \Gamma$. Then $\angle DXE = \angle DX'E = 135^{\circ}$, so $X = X'$.

Solution:

Let $N$ be the midpoint of arc $FDEG$. Note that $DOEA$ is a square. Also, $DXEN$ is a parallelogram; one way to see this is that by considering inscribed angles, $\angle NEX = \angle NEF = 45^{\circ}$, $\angle NDX = \angle NDG = 45^{\circ}$, and $\angle END = 135^{\circ}$. This means that $\triangle AXD \cong \triangle ONE$, because $AD = OE$, $XD = NE$, and $\angle ADX = \angle OEN$ by considering parallel lines. So $AX = ON = OE = AD$.

Solution:

Let $Y = DF \cap EG$. Since $GX \perp FY$ and $FX \perp GY$, $X$ is the orthocenter of $\triangle FGY$. Since $(AODE)$ passes through the midpoint of $FG$, and the feet of altitudes from $F$ and $G$, $(AODE)$ is the nine-point circle of $\triangle FGY$. Since $AO$ is the diameter of $(AODE)$, it follows that $A$ is the midpoint of $XY$, so $A$ is the center of $(DXEY)$, so $AX = AD$.

Solution:

Let $Z = DE \cap FG$, possibly at infinity. Then $A$ and $X$ are on the polar of $Z$ with respect to $\omega$, so $AX \perp BC$. Let $J = AX \cap BC$. Then $(XJFD)$ is cyclic, so $\angle ADX = \angle DFJ = \angle DXA$, so $AX = AD$.

Solution:

We use complex numbers, with $(FDEG)$ being the unit circle, and $d = -1$, $e = -i$. As $FG$ is a diameter of the unit circle, we have $g = -f$. We have $a = -1 - i$, from either intersecting the tangents to the unit circle at $D$ and $E$ or noting that $ADOE$ is a square.
Now, intersecting the chords $DG$ and $EF$, we obtain
$$x = \frac{dg(e+f) - ef(d+g)}{dg - ef} = \frac{f(i+f) + if(1-f)}{f + if} = \frac{f - i - i - if}{1 + i} = -1 - i + f \cdot \frac{1 - i}{1 + i}.$$
So,
$$|x - a| = \left|f \cdot \frac{1 - i}{1 + i}\right| = 1 \cdot \frac{\sqrt{2}}{\sqrt{2}} = 1$$
So $AX$ and $AD$ have the same length (1 unit), as desired.