China Southeastern Mathematical Olympiad

We use $A,B,C,D$ to represent these four colors, respectively, and the points in the same color by lower letters as $a_1,a_2,a_3;b_1,b_2,b_3;c_1,c_2,c_3$ and $d_1,d_2,d_3$ respectively.

Now consider color $A$. If, in $n$ quadrilaterals, the number of points $a_1,a_2,a_3$ in color $A$ are $n_1,n_2,n_3$ respectively, then $n_1+n_2+n_3=n$. Suppose that $n_1\ge n_2\ge n_3$. If $n\ge 10$, then $n_1+n_2\ge 7$. Consider these seven quadrilaterals (its $A$ color vertex is either $a_1$ or $a_2$), if the numbers of points $b_1,b_2,b_3$ in color $B$ are $m_1,m_2,m_3$, respectively, then $m_1+m_2+m_3=7$. By symmetricity, we may suppose that $m_1\ge m_2\ge m_3$, then $m_3\le 2$, that is, $m_1+m_2\ge 5$.

Consider these five quadrilaterals (its $A$ color point is either $a_1$ or $a_2$, and its $B$ color point is either $b_1$ or $b_2$), if the numbers of points $c_1,c_2,c_3$ in color $C$ are $k_1,k_2,k_3$, respectively, then $k_1+k_2+k_3=5$. By symmetricity, we may suppose that $k_1\ge k_2\ge k_3$, then $k_3\le 1$, that is, $k_1+k_2\ge 4$.

Consider these four quadrilaterals, denoted as $T_1,T_2,T_3,T_4$ (its color $A$ point is either $a_1$ or $a_2$, its color $B$ point is either $b_1$ or $b_2$, and its color $C$ point is either $c_1$ or $c_2$). Since there are only three points in color $D$, there are two quadrilaterals that have the same color $D$ point. Suppose that the same color $D$ point of $T_1,T_2$ is $d_1$.

Then, in three quadrilaterals $T_1,T_2,T_3$, whatever be the color of the vertex, there are repeated points, which contradicts condition (2). Hence, $n\le 9$.

We show the maximal number $n=9$ by construction of these nine quadrilaterals.



We draw three “concentric annulus” with four points on each radius representing four vertices and the color. So nine radii represent nine quadrilaterals, which satisfy condition (1).

Next, we show that they also satisfy condition (2). Take any three radii (or three quadrilaterals). If these three radii come from a concentric annulus, for each color except $A$, there are three points. If these three radii come from three concentric annuli, then for color $A$, there are three points. If these three radii come from two concentric annuli, call these three figures Fig. 1, Fig. 2 and Fig. 3. The radius directions are called “up radius”, “left radius” and “right radius”, and denoted, respectively, by $S$, $Z$ and $Y$. If three radii have three directions, then there are three points of color $B$ in three quadrilaterals. If the three radii have only two directions, then there are all cases as shown in the tables below, where 1, 2 and 3 stand for Fig. 1, Fig. 2 and Fig. 3, respectively.

Here, the color in figure means that the three quadrilaterals have color with different figures.

C D

	S	1, 3	1, 3	1		3
Z	3		1, 3	1, 3	1
Y		1		3	1, 3

C

	S	1, 3	1, 3	3		1
Z	1		1, 3	1, 3	3
Y		3		1	1, 3

D

	S	2, 3	2, 3	3		2
Z	2		2, 3	2, 3	3
Y		3		2	2, 3

C

	S	2, 3	2, 3	2		3
Z	3		2, 3	2, 3	2
Y		2		3	2, 3

D

Thus, the maximal number of $n$ is 9. $\square$