The 16th Japanese Mathematical Olympiad - The First Round

Draw a semicircle whose radius is $1$, and let $A$ and $B$ be the two ends of the arc. Take two points $C$ and $D$ on arc $AB$ so that $A$, $C$, $D$ and $B$ are on the arc in this order. Let $2a=AC$, $2b=AD$, $\angle CAB=\alpha$, $\angle DAB=\beta$. Then $a=\cos \alpha$, $b=\cos \beta$ and area of the triangle $ACD$ is $$\begin{array}{rl}\frac{1}{2}(2a)(2b)\sin (\alpha -\beta )& =2ab(\sin \alpha \cos \beta -\cos \alpha \sin \beta )\\ & =2ab(b\sqrt{1-a^2}-a\sqrt{1-b^2})=2f(a,b).\end{array}$$ Therefore, by taking points $A_1,A_2,\dots ,A_n$ on the arc $AB$ with $A{A}_i=2a_i$ and considering the sum of the area of triangles $A{A}_i{A}_{i+1}$, we conclude that the left side of the given inequality does not exceed $\pi /4$, the half of the area of the semicircle.

Next, we shall prove that $c$ cannot be smaller than $\pi /4$. Take any positive integer $n$. Take points $A=A_0,A_1,A_2,\dots ,A_{2^n-1},A_{2^n}=B$ on the arc $AB$ in this order so that these points divide the arc equally. Let $A{A}_i=2a_i$ ($i=1,2,\dots ,2^n$). Denote by $S_n$ the part of the semicircle not covered by polygon $A_0{A}_1\cdots A_{2^n}$. $S_n$ consists of $2^n$ congruent figures (a fan minus a triangle). Denote each of them by $P_n$. We can easily show that $2|P_{n+1}|<\frac{1}{2}|P_n|$, so $|S_{n+1}|<\frac{1}{2}|S_n|$, hence $|S_n|<\frac{1}{2^{n-1}}|S_1|$. This inequality shows that, if $c$ is less than $\pi /4$, we can take $n$ large enough so that the area of the part which is not covered by polygon $A_0{A}_1\cdots A_{2^n}$ be less than $\pi /4-c$, which yields $c$ inappropriate. Therefore the minimum value of $c$ is $\pi /4$.