South African Mathematical Olympiad Third Round

Consider first the case that one of the areas is zero, e.g. $\text{area}(OA_1B_1) = 0$. Then either $a_1 = 0$ or $b_1 = 0$. If $a_1 = 0$, then $\text{area}(OA_1B_k) = a_1b_k = 0$ 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
$$\frac{\text{area}(OA_1B_2)}{\text{area}(OA_1B_1)} = \frac{a_1b_2}{a_1b_1} = \frac{b_2}{b_1} = \frac{a_2b_2}{a_2b_1} = \frac{\text{area}(OA_2B_2)}{\text{area}(OA_2B_1)}.$$
Since triangles $OA_1B_1$ and $OA_1B_2$ have the same base ($OA_1$), their heights must be in a $b_1/b_2$-ratio. The same is true for the heights of triangles $OA_2B_1$ and $OA_2B_2$. If $B_1B_2$ is parallel to $OA_1$, then $b_1 = b_2$, so $B_1B_2$ is parallel to $OA_2$. In this case, $O, A_1$ and $A_2$ lie on one line, contradicting the assumption. The same argument applies if $B_1B_2$ is parallel to $OA_2$.

If neither $OA_1$ nor $OA_2$ is parallel to $B_1B_2$, let $X$ be the intersection of $OA_1$ and $B_1B_2$, and let $Y$ be the intersection of $OA_2$ and $B_1B_2$.

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 $OA_1B_1$ and $OA_1B_2$ is $|XB_1|/|XB_2|$, which must be $b_1/b_2$. The same is true (analogously) for the ratio $|YB_1|/|YB_2|$. Hence
$$\frac{|XB_1|}{|XB_2|} = \frac{b_1}{b_2} = \frac{|YB_1|}{|YB_2|}$$
If $X$ and $Y$ are on different sides of $B_1B_2$, 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
$$1 - \frac{|B_1B_2|}{|XB_2|} = \frac{|XB_2| - |B_1B_2|}{|XB_2|} = \frac{|YB_2| - |B_1B_2|}{|YB_2|} = 1 - \frac{|B_1B_2|}{|YB_2|}$$
and thus $|XB_2| = |YB_2|$. 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.

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 $OA_1, OA_2, OB_1$ and $OB_2$ with the x-axis. Then
$$\text{area}(OA_1B_1) \cdot \text{area}(OA_2B_2) = a_1b_1a_2b_2 = \text{area}(OA_1B_2) \cdot \text{area}(OA_2B_1)$$
and thus
$$\frac{|OA_1||OB_1|\sin(\beta_1 - \alpha_1)}{2} \cdot \frac{|OA_2||OB_2|\sin(\beta_2 - \alpha_2)}{2} = \frac{|OA_1||OB_2|\sin(\beta_2 - \alpha_1)}{2} \cdot \frac{|OA_2||OB_1|\sin(\beta_1 - \alpha_2)}{2}$$
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.