The South African Mathematical Olympiad Third Round

Consider first the case that one of the areas is zero, e.g. $\text{area}(O{A}_1{B}_1)=0$. Then either $a_1=0$ or $b_1=0$. If $a_1=0$, then $\text{area}(O{A}_1{B}_k)=a_1{b}_k=0$ for all $k$, which means that all $B_k$ lie on a straight line through $O$ and $A_1$. Likewise, if $b_1=0$, then all $A_j$ lie on a straight line through $O$ and $B_1$. So we can assume from now on that none of the areas is zero.

Suppose there are two points among $A_1,A_2,\dots ,A_n$ that are not on the same line through $O$ (since the numbering does not matter, we can assume that $A_1$ and $A_2$ have this property), and at the same time there are two points (again, we can assume them to be $B_1$ and $B_2$) that are not on the same line through $O$. We have $$\frac{\text{area}(O{A}_1{B}_2)}{\text{area}(O{A}_1{B}_1)}=\frac{a_1{b}_2}{a_1{b}_1}=\frac{b_2}{b_1}=\frac{a_2{b}_2}{a_2{b}_1}=\frac{\text{area}(O{A}_2{B}_2)}{\text{area}(O{A}_2{B}_1)}.$$ Since triangles $O{A}_1{B}_1$ and $O{A}_1{B}_2$ have the same base ($O{A}_1$), their heights must be in a $b_1/b_2$-ratio. The same is true for the heights of triangles $O{A}_2{B}_1$ and $O{A}_2{B}_2$. If $B_1{B}_2$ is parallel to $O{A}_1$, then $b_1=b_2$, so $B_1{B}_2$ is parallel to $O{A}_2$. In this case, $O,A_1$ and $A_2$ lie on one line, contradicting the assumption. The same argument applies if $B_1{B}_2$ is parallel to $O{A}_2$.

If neither $O{A}_1$ nor $O{A}_2$ is parallel to $B_1{B}_2$, let $X$ be the intersection of $O{A}_1$ and $B_1{B}_2$, and let $Y$ be the intersection of $O{A}_2$ and $B_1{B}_2$.



Since $A_1$ and $A_2$ are in the first quadrant, $X$ and $Y$ are either in the first or third quadrant. In either case, they are not between $B_1$ and $B_2$. Using similar triangles, we see that the ratio of the heights of $O{A}_1{B}_1$ and $O{A}_1{B}_2$ is $|X{B}_1|/|X{B}_2|$, which must be $b_1/b_2$. The same is true (analogously) for the ratio $|Y{B}_1|/|Y{B}_2|$. Hence $$\frac{|X{B}_1|}{|X{B}_2|}=\frac{b_1}{b_2}=\frac{|Y{B}_1|}{|Y{B}_2|}$$ If $X$ and $Y$ are on different sides of $B_1{B}_2$, one of these ratios is greater than 1, the other less than 1, a contradiction. Thus we assume that $B_1$ is closer to both $X$ and $Y$ (otherwise, we just interchange the roles of $B_1$ and $B_2$), as in the figure. We get $$1-\frac{|B_1{B}_2|}{|X{B}_2|}=\frac{|X{B}_2|-|B_1{B}_2|}{|X{B}_2|}=\frac{|Y{B}_2|-|B_1{B}_2|}{|Y{B}_2|}=1-\frac{|B_1{B}_2|}{|Y{B}_2|}$$ and thus $|X{B}_2|=|Y{B}_2|$. This means that $X$ and $Y$ coincide, so $X,Y,O,A_1,A_2$ lie on one line, and we get a contradiction to our assumption again. This completes the proof.

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Alternative solution.

We argue as in the first proof and assume again that $A_1$ and $A_2$ do not lie on a common line through $O$, and that $B_1$ and $B_2$ do not lie on a common line through $O$. Let ${\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2$ be the angles enclosed by $O{A}_1,O{A}_2,O{B}_1$ and $O{B}_2$ with the x-axis. Then $$\text{area}(O{A}_1{B}_1)\cdot \text{area}(O{A}_2{B}_2)=a_1{b}_1{a}_2{b}_2=\text{area}(O{A}_1{B}_2)\cdot \text{area}(O{A}_2{B}_1)$$ and thus $$\frac{|O{A}_1||O{B}_1|\sin ({\beta }_1-{\alpha }_1)}{2}\cdot \frac{|O{A}_2||O{B}_2|\sin ({\beta }_2-{\alpha }_2)}{2}=\frac{|O{A}_1||O{B}_2|\sin ({\beta }_2-{\alpha }_1)}{2}\cdot \frac{|O{A}_2||O{B}_1|\sin ({\beta }_1-{\alpha }_2)}{2}$$ It follows that $$\sin ({\beta }_1-{\alpha }_1)\sin ({\beta }_2-{\alpha }_2)=\sin ({\beta }_2-{\alpha }_1)\sin ({\beta }_1-{\alpha }_2).$$ Now we use the trigonometric identity $\sin x\sin y=\frac{1}{2}(\cos (x-y)-\cos (x+y))$ to get $$\frac{1}{2}(\cos ({\beta }_1-{\alpha }_1+{\alpha }_2-{\beta }_2)-\cos ({\beta }_1-{\alpha }_1+{\beta }_2-{\alpha }_2))=\frac{1}{2}(\cos ({\beta }_2-{\alpha }_1+{\alpha }_2-{\beta }_1)-\cos ({\beta }_2-{\alpha }_1+{\beta }_1-{\alpha }_2)).$$ This implies $$\cos ({\beta }_1-{\alpha }_1+{\alpha }_2-{\beta }_2)=\cos ({\beta }_2-{\alpha }_1+{\alpha }_2-{\beta }_1),$$ and by the addition theorem for the cosine $$\cos ({\beta }_1-{\beta }_2)\cos ({\alpha }_1-{\alpha }_2)+\sin ({\beta }_1-{\beta }_2)\sin ({\alpha }_1-{\alpha }_2)=\cos ({\beta }_1-{\beta }_2)\cos ({\alpha }_1-{\alpha }_2)-\sin ({\beta }_1-{\beta }_2)\sin ({\alpha }_1-{\alpha }_2),$$ so finally $$\sin ({\beta }_1-{\beta }_2)\sin ({\alpha }_1-{\alpha }_2)=0,$$ which means that either ${\beta }_1={\beta }_2$ or ${\alpha }_1={\alpha }_2$, contradicting our assumption and thus completing the proof.