China Mathematical Olympiad

Lemma The area of a triangle inscribed in a rectangle is no more than half of the area of the rectangle.

Denote $E$, $F$, $H$ and $G$ the midpoints of $AB$, $CD$, $BC$ and $AD$, respectively, and $O$ the intersection of line segments $EF$ and $GH$. $EF$ and $GH$ divide the rectangle $ABCD$ into 4 small rectangles, it follows that there exists certain small rectangle, say $AEOG$, in which there are at least two points (say, $M$ and $N$) out of those 5 points, see Figure 5.

Figure 5

(1) If there is no more than one given point in the rectangle $OHCF$, consider any given point $X$ which is different from $M$ and $N$ and not in the rectangle $OHCF$. It is easy to verify that triple $(M,N,X)$ is either in rectangle $ABHG$, or in rectangle $AEFD$. By the above lemma $(M,N,X)$ is good. Since there are at least two such points $X$, we have at least two such good triples.

(2) If there exist at least two given points in rectangle $OHCF$, we suppose that $P$ and $Q$ are the given points in rectangle $OHCF$. Consider the final given point $R$. If $R$ is in the rectangle $OFDG$, then $(M,N,R)$ is in the rectangle $AEFD$, and $(P,Q,R)$ is in the rectangle $GHCD$. So they are all good. It follows that there are at least two good triples. Similarly, when point $R$ is in the rectangle $EBHO$ there are at least two good triples too. If point $R$ is in the rectangle $OHCF$ or the rectangle $AEOG$, suppose that $R$ is in the rectangle $OHCF$. Consider the smallest convex polygon containing the 5 points $M$, $N$, $P$, $Q$ and $R$. The polygon must be contained in the convex hexagon $AEHCFG$ (see Figure 6). But the area $$S_{AEHCFG}=1-\frac{1}{8}-\frac{1}{8}=\frac{3}{4}.$$

Figure 6

i) Suppose the convex polygon generated by $M,N,P,Q$ and $R$ is a convex pentagon, (no loss of generality) say $MNPQR$ (as seen in Figure 7). In this case $$S_{△MQR}+S_{△MNQ}+S_{△NPQ}\le \frac{3}{4},$$ it follows that there is at least one good triple among $(M,Q,R)$, $(M,N,Q)$ and $(N,P,Q)$. Moreover, $(P,Q,R)$ is clearly good for it is in the rectangle $OHCF$. Thus there are at least two good triples.

Figure 7

ii) If the convex polygon generated by $M,N,P,Q$ and $R$ is a convex quadrilateral, say $A_1{A}_2{A}_3{A}_4$, and the fifth point is $A_5$ (as seen in Figure 8) where $A_i\in \{M,N,P,Q,R\}(i=1,2,3,4,5)$. Draw line segments $A_5{A}_i$ ($i=1,2,3,4$), then $$S_{△A_1{A}_2{A}_5}+S_{△A_2{A}_3{A}_5}+S_{△A_3{A}_4{A}_5}+S_{△A_4{A}_1{A}_5}=S_{△A_1{A}_2{A}_3{A}_4}\le \frac{3}{4}.$$ Therefore, there are at least two good triples among $(A_1,A_2,A_5)$, $(A_2,A_3,A_5)$, $(A_3,A_4,A_5)$ and $(A_1,A_4,A_5)$.

Figure 8

iii) If the convex polygon generated by $M,N,P,Q$ and $R$ is a triangle, say $△A_1{A}_2{A}_3$, and the remaining two points are $A_4$ and $A_5$ (as seen in Figure 9), where $A_i\in \{M,N,P,Q,R\}(i=1,2,3,4,5)$. Draw line segments $A_4{A}_i(i=1,2,3)$, then $$S_{△A_1{A}_2{A}_4}+S_{△A_2{A}_3{A}_4}+S_{△A_3{A}_1{A}_4}=S_{△A_1{A}_2{A}_3}\le \frac{3}{4}.$$ Therefore, there is at least one good triple among $(A_1,A_2,A_4)$, $(A_2,A_3,A_4)$ and $(A_1,A_3,A_4)$. Similarly, $A_5$ with two of $A_1,A_2$ and $A_3$ constitutes a good triple. Consequently, in this case there are at least two good triples.

Figure 9

Thus, in any case there are at least two good triples among these 5 points.

In the following we will give some examples to show that the number of good triples may be just two. Pick a point $M$ on the side $AD$ of the rectangle $ABCD$ and a point $N$ on the side $AB$ such that $AN:NB=AM:MD=2:3$ (as shown in Figure 10). Then among 5 points $M,N,B,C$ and $D$ there are just two good triples. In fact, $(B,C,D)$ is clearly not good. Suppose that a triple containing exactly one of two points $M$ and $N$, say $M$. Let $E$ be the midpoint of $AD$, then $$S_{△MBD}>S_{△EBD}=\frac{1}{4}.$$ It follows that $(M,B,D)$ is not good. Thus $$S_{△MBC}=\frac{1}{2},S_{△MCD}>S_{△ECD}=\frac{1}{4}.$$ Hence $(M,B,C)$ and $(M,C,D)$ are not good. If a triangle does contain two points $M$ and $N$, then $$\begin{array}{rl}{S}_{△MNC}& =1-S_{△NBC}-S_{△MCD}-S_{△AMN}\\ & =1-\frac{3}{5}{S}_{△ABC}-\frac{3}{5}{S}_{△ACD}-\frac{4}{25}{S}_{△ABD}\end{array}$$ $$=1-\frac{3}{10}-\frac{3}{10}-\frac{2}{25}=\frac{8}{25}>\frac{1}{4}.$$ So $(M,N,C)$ is not good. But $S_{△MNB}=S_{△MND}=\frac{1}{5}<\frac{1}{4}$, thus among the triples there are only two good triples $(M,N,B)$ and $(M,N,D)$.

Consequently, the minimum number of triangles with area not greater than $\frac{1}{4}$ is $2$.