Problem 1281

Solution 1: (Ankan Bhattacharya ${ }^{1}$ ) Let $\triangle$ denote the triangle, o denote the circle, and $\triangle^{\prime}$ denote the reflection of the triangle in the center of the circle. Letting overline denote complement,
$$\begin{array}{c} {[\Delta \cap \circ] \leq \frac{1}{3}[\Delta]+\frac{1}{2}[0]} \\ \Longleftrightarrow\left[\Delta \cap \Delta^{\prime} \cap \circ\right]+\left[\Delta \cap \overline{\Delta^{\prime}} \cap \circ\right] \leq \frac{1}{3}[\Delta]+\frac{1}{2}\left[\left[\circ \cap \Delta \cap \triangle^{\prime}\right]+2\left[0 \cap \Delta \cap \overline{\Delta^{\prime}}\right]+\left[0 \cap \bar{\Delta} \cap \overline{\Delta^{\prime}}\right]\right] \\ \Longleftrightarrow \frac{1}{2}\left[\Delta \cap \Delta^{\prime} \cap \circ\right] \leq \frac{1}{3}[\Delta]+\frac{1}{2}\left[\circ \cap \bar{\Delta} \cap \overline{\Delta^{\prime}}\right] \\ \Longleftrightarrow\left[\Delta \cap \triangle^{\prime} \cap \circ\right] \leq \frac{2}{3}[\Delta]+\left[\circ \cap \bar{\Delta} \cap \overline{\Delta^{\prime}}\right] . \end{array}$$

Since $\Delta \cap \triangle^{\prime} \cap \circ$ is centrally symmetric, it's enough to show the following well-known lemma.

Lemma. A centrally symmetric region $\mathcal{R}$ of a triangle can have area at most $\frac{2}{3}$ that of the triangle. Let the triangle be $A B C$ with medial triangle $D E F$. We take two cases.
- Case 1. The center of symmetry is in one of the outer triangles, say $\triangle A E F$. Then the maximal possible $\mathcal{R}$ is a parallelogram with one vertex $A$ and two other vertices on $\overline{A B}$ and $\overline{A C}$. By enlargening the parallelogram, we may assume its fourth vertex lies on $\overline{B C}$. If this vertex divides $\overline{B C}$ into an $a: b$ ratio with $a+b=1$, then the fraction of the area taken up by the parallelogram is $1-a^{2}-b^{2} \leq \frac{1}{2}$.
- Case 2. The center of symmetry is in $\triangle D E F$. Then the maximal possible $\mathcal{R}$ is a centrally symmetric hexagon with two vertices on each side. Then there are three little triangles similar to $\triangle A B C$ on the hexagon; by equal lengths, the similarity ratios $a, b, c$ sum to 1 . Then the fraction of the area taken up by the hexagon is $1-a^{2}-b^{2}-c^{2} \leq \frac{2}{3}$.

We are done.
Solution 2: It is also possible to approach this as an optimization problem. Fix a triangle $\triangle=A B C$ on the plane, vary the circle $\odot=\odot(O, r)$, and consider the objective function
$$f(O, r):=[\triangle \cap \odot]-\frac{1}{2}[\odot] .$$

By a compactness argument we can show that $f$ reaches a maximum, and at that maximum $\frac{\partial f}{\partial r}=0$ and $\nabla_{O} f=\overrightarrow{0}$ must simultaneously hold.
- We have
$$\begin{aligned} \frac{\partial}{\partial r} f(O, r) & =\frac{\partial}{\partial r}[\triangle \cap \odot]-\frac{1}{2} \frac{\partial}{\partial r}[\odot] \\ & =r \cdot((\text { total angle subtended by } A B \cap \odot, B C \cap \odot, C A \cap \odot \text { at point } O)-\pi), \end{aligned}$$
therefore
$$\frac{\partial}{\partial r} f(O, r)=0 \Longleftrightarrow \text { the total angle subtended by the sides of } \triangle \text { inside } \odot \text { is } \pi .$$
- Let $\vec{n}_{a}, \vec{n}_{b}, \vec{n}_{c}$ denote the normal vectors of $B C, C A$ and $A B$. Define $k_{a}$ to be the length of $B C \cap \odot$, and define $k_{b}, k_{c}$ similarly.
$$\begin{aligned} \nabla_{O} f(O, r) & =\nabla_{O}[\triangle \cap \odot]-\frac{1}{2} \nabla_{O}[\odot] \\ & =\left(k_{a} \vec{n}_{a}+k_{b} \vec{n}_{b}+k_{c} \vec{n}_{c}\right)-\overrightarrow{0} \end{aligned}$$

It is easy to see that $a \vec{n}_{a}+b \vec{n}_{b}+c \vec{n}_{c}=\overrightarrow{0}$ is the only linear relation between $\vec{n}_{a}, \vec{n}_{b}, \vec{n}_{c}$, so
$$\nabla_{O} f(O, r)=\overrightarrow{0} \Longleftrightarrow k_{a}: k_{b}: k_{c}=a: b: c$$

Now suppose that $\odot(O, r)$ is chosen so that $\frac{\partial f}{\partial r}=0$ and $\nabla_{O} f=\overrightarrow{0}$, so the angle condition in (1) holds, and $\left(k_{a}, k_{b}, k_{c}\right)=k(a, b, c)$ for some $k$. There are two possible cases:
- Case 1. All vertices of $\triangle$ do not lie inside $\odot$. In this case $\odot$ intersects $\triangle$ at six points. The lines joining $O$ and these six points divide $\triangle \cap \odot$ into three sectors and three triangles. By the angle condition in (1), the three sectors have total area $\frac{1}{2}[\odot]$. The three triangles have total area $k[\triangle]$, and it suffices to show that $k \leqslant \frac{1}{3}$.
As the total angle of the triangles at $O$ is $\pi$, we may join two copies of each of the three triangles to form a cyclic hexagon. Therefore,
$$\begin{aligned} 2 k \cdot[\triangle]= & 2 \cdot(\text { total area of three triangles) } \\ = & \text { area of cyclic hexagon with side lengths } k a, k b, k c, k a, k b, k c \\ \geqslant & \text { area of hexagon with side lengths } k a, k b, k c, k a, k b, k c \\ & \quad \text { created by six copies of a triangle with side lengths } k a, k b, k c \\ = & 6 k^{2} \cdot[\triangle] \end{aligned}$$
and our conclusion readily follows.
- Case 2. A vertex of $\triangle$ lies in the interior of $\odot$. WLOG let that vertex be $A$. Let $\odot$ intersect $A B$ and $X Y=D E$, therefore $\square X Y E D$ is a rectangle. From (1), $\angle X O D+\angle Y O E=\pi$, so in fact $\square X Y E D$ is a square. Let $x$ be the side length of $\square X Y D E$, and let $h$ be the height of the altitude from $A$ to $B C$. Clearly $h>x$. As $A$ lies inside $\odot=(X Y D E), \angle B A C>\frac{3 \pi}{4}$, so $a>3 h>3 x$. Now we are done because $k=\frac{x}{a}<\frac{1}{3}$.