19. ( 3 marks) There are 5 points on the plane. The following steps are used to construct lines. In step 1 , connect all possible pairs of the points; it is found that no two lines are parallel, nor any two lines perpendicular to each other, also no three lines are concurrent. In step 2, perpendicular lines are drawn from each of the five given points to straight lines connecting any two of the other four points. What is the maximum number of points of intersection formed by the lines drawn in step 2 , including the 5 given points?
(3 分) 在平面上有 5 點。現用以下步驟畫上直線。
步驟 1 : 將任何 2 點連接起來。
直線畫好之後, 發覺沒有兩條線是平行的, 沒有兩條線是互相垂直的, 也沒有三條線交於一點。
步驟 2 : 由原先的 5 點任何一點出發，畫直線使垂直於其餘四點間任二點的連線。
問在步䝒 2 的過程中所形成的線之聞最多共有多少個交點(包括原先的 5 點 $)$ ?
19. Let the five given points be $A, B, C, D$ and $E$. The number of straight lines constructed in step 1 is $C_{2}^{5}=10$. From each of the five pints $A, B, C, D$ and $E$, a total of $C_{2}^{4}=6$ perpendicular lines can be drawn. It seems that the number of intersection points should be $C_{2}^{30}=435$. However for any straight line, say $A B$, the three perpendicular lines drawn from $C, D$ and $E$ are parallel, thus will not meet. Hence the number of intersection points should be cut down by $10 \times C_{2}^{3}=30$. Next for any of the $C_{3}^{5}=10$ triangle, formed, say $\triangle A B C$, there are three perpendicular lines which are concurrent. Hence the number of intersection points should be further cut down by $10 \times\left(C_{2}^{3}-1\right)=20$. Finally the points $A, B, C, D$ and $E$, each represents the intersection of six lines (lines drawn from each of them), hence the number of intersection points should be cut down by $5 \times\left(C_{2}^{6}-1\right)=70$. The required number of points of intersection is therefore $435-30-20-70=315$.