, 2020

Solution:

Using Fact 5, we know that $II_{A}$ intersects the circle $(ABC)$ at $M_{A}$, which is the center of $\left(II_{A}BCXY\right)$. Let $R$ be the radius of the latter circle. We have $R=\frac{1}{\sqrt{3}}$.

We have $\angle AIM=\angle YII_{A}=\angle YIX=\frac{\pi}{3}$. Also, $\angle II_{A}M=\angle IMI_{A}$ by calculating the angles from the equilateral triangle. Using 90-60-30 triangles, we have:
$$\begin{gathered} AI=\frac{1}{2} MI=\frac{1}{2} II_{A}=R \\ AM=\frac{\sqrt{3}}{2} MI=\sqrt{3} R \\ MM_{A}^{2}=AM^{2}+AM_{A}^{2}=7R^{2} \end{gathered}$$
Now, let $J$ and $N$ be the feet of the altitudes from $A$ and $B$ respectively on $MM_{A}$. Note that as $M$ is an arc midpoint of $BC$, $N$ is actually the midpoint of $BC$.
$$\begin{aligned} & M_{A}J=\frac{AM_{A}^{2}}{MM_{A}}=\frac{4}{\sqrt{7}} R \\ & M_{A}N=\frac{BM_{A}^{2}}{MM_{A}}=\frac{1}{\sqrt{7}} R \end{aligned}$$
Thus $JN=\frac{3}{\sqrt{7}} R$. Also, we have,
$$BN^{2}=M_{A}N \cdot MN=\frac{6}{7} R^{2}$$
Now, $[ABC]=\frac{1}{2} JN \cdot BC=JN \cdot BN=\frac{3 \sqrt{6}}{7} R^{2}=\frac{\sqrt{6}}{7}$.

Solution:

By Fact 5, we construct the diagram first with $\triangle XYI_{A}$ as the reference triangle.

Let $M_{A}$ be the circumcenter of $\triangle XYI_{A}$ and let $\Omega$ be the circumcircle, which has circumradius $R=\frac{1}{\sqrt{3}}$. Then by Fact 5, $M_{A}$ is the midpoint of minor arc $\widehat{BC}$, and $B, C \in \Omega$. Now we show the following result:

Claim. $b+c=2a$.

Proof. Letting $D$ be the intersection of $II_{A}$ with $BC$, we have $M_{A}D=\frac{1}{2}R$. Then by the Shooting Lemma,
$$M_{A}A \cdot M_{A}D=R^{2} \Longrightarrow M_{A}A=2R$$
On the other hand, by Ptolemy,
$$M_{A}B \cdot AC + M_{A}C \cdot AB = M_{A}A \cdot BC \Longrightarrow R \cdot (b+c) = 2R \cdot a$$
whence the result follows.

By the triangle area formula, we have
$$[ABC]=sr=\frac{1}{2}(a+b+c)r=\frac{3}{2}ar.$$
Therefore, we are left to compute $a$ and $r$.

Since $M_{A}$ is the antipode of $M$ on $(ABC)$ we get $MB, MC$ are tangent to $\Omega$; in particular, $MBM_{A}C$ is a kite with $MB=MC$ and $M_{A}B=M_{A}C$. Then by Power of a Point, we get
$$MB^{2}=MC^{2}=MI \cdot MY=2R \cdot 3R=2$$
where $MI=II_{A}=2R$ and $IY=R$.

Since $ID=DM_{A}$, we know that $r$ is the length of the projection from $M_{A}$ to $BC$. Finally, given our above information, we can compute
$$\begin{aligned} & r=\frac{BM_{A}^{2}}{M_{A}M}=\frac{BM_{A}^{2}}{\sqrt{BM_{A}^{2}+MB^{2}}}=\frac{1}{\sqrt{21}} \\ & a=2r \frac{BM_{A}}{MB}=\frac{2 \sqrt{2}}{7} \end{aligned}$$
The area of $ABC$ is then
$$$$
$\frac{3}{2}\left(\frac{2 \sqrt{2}}{\sqrt{7}}\right)\left(\frac{1}{\sqrt{21}}\right)=\frac{\sqrt{6}}{7}.$