, 2020

Bus A takes 12 minutes to complete one round trip that begins and ends at $P$. Since $PX=XS$, it takes Bus A $12\div4=3$ minutes to travel from $P$ to $X$, 6 minutes to travel from $X$ to $S$ to $X$ (3 minutes from $X$ to $S$ and 3 minutes from $S$ to $X$), and 6 minutes to travel from $X$ to $P$ to $X$. That is, Bus A first arrives at $X$ at 1:03 and then continues to return to $X$ every 6 minutes. We write times that Bus A arrives at $X$ in the table below. Bus A 1:03 1:09 1:15 1:21 1:27 1:33 1:39 1:45 1:51 1:57 2:03 Notice that Bus A arrives at $X$ at 1:03 and exactly one hour later at 2:03. This makes sense since Bus A returns to $X$ every 6 minutes and 60 minutes (one hour) is divisible by 6. This tells us that Bus A will continue to arrive at the same number of minutes past each hour, or 2:03, 2:09, 2:15, $\dots$, 3:03, 3:09, $\dots$, 5:03, 5:09, $\dots$, 9:03, 9:09, $\dots$, 9:51, 9:57. Bus B takes 20 minutes to complete one round trip that begins and ends at $Q$. Since $QX=XT$, it takes Bus B $\frac{20}{4}=5$ minutes to travel from $Q$ to $X$, 10 minutes to travel from $X$ to $T$ to $X$ (5 minutes from $X$ to $T$ and 5 minutes from $T$ to $X$), and 10 minutes to travel from $X$ to $Q$ to $X$. That is, Bus B first arrives at $X$ at 1:05 and then continues to return to $X$ every 10 minutes. We write times that Bus B arrives at $X$ in the table below. Bus B 1:05 1:15 1:25 1:35 1:45 1:55 2:05 Notice that Bus B arrives at $X$ at 1:05 and exactly one hour later at 2:05. This makes sense since Bus B returns to $X$ every 10 minutes and 60 minutes (one hour) is divisible by 10. This tells us that Bus B will continue to arrive at the same number of minutes past each hour, or 2:05, 2:15, 2:25, $\dots$, 3:05, 3:15, $\dots$, 5:05, 5:15, $\dots$, 9:05, 9:15, $\dots$, 9:45, 9:55. From the two tables above, we see that Bus A and Bus B both arrive at $X$ at 15 minutes and 45 minutes past each hour. Thus between 5:00 p.m. and 10:00 p.m., these two buses will meet $2\times 5=10$ times at $X$. These times are: 5:15, 5:45, 6:15, 6:45, 7:15, 7:45, 8:15, 8:45, 9:15, and 9:45. Bus C takes 28 minutes to complete one round trip that begins and ends at $R$. Since $RX=XU$, it takes Bus C $\frac{28}{4}=7$ minutes to travel from $R$ to $X$, $2\times7=14$ minutes to travel from $X$ to $U$ to $X$, and 14 minutes to travel from $X$ to $R$ to $X$. That is, Bus C first arrives at $X$ at 1:07 and then continues to return to $X$ every 14 minutes. Unlike Bus A and Bus B, Bus C will not arrive at $X$ at consistent times past each hour since 60 is not divisible by 14. What is the first time after 5:00 p.m. that Bus C arrives at $X$? Since 238 is a multiple of 14 ($14\times17=238$), Bus C will arrive at $X$ 238 minutes after first arriving at $X$ at 1:07 p.m. Since 238 minutes is 2 minutes less than 4 hours ($4\times60=240$), Bus C will arrive at $X$ at 5:05 p.m. (This is the first time after 5:00 p.m. that Bus C arrives at $X$.) Bus B also arrives at $X$ at 5:05 p.m. Are there other times after 5:05 p.m. (and before 10:00 p.m.) that Bus B and Bus C arrive at $X$ at the same time? Bus B arrives at $X$ every 10 minutes and Bus C arrives at $X$ every 14 minutes. Since the lowest common multiple of 10 and 14 is 70, then Bus B and Bus C will each arrive at $X$ every 70 minutes after 5:05 p.m., or at 6:15 p.m., 7:25 p.m., 8:35 p.m., and at 9:45 p.m. Next we determine if there are times when Bus A and Bus C arrive at $X$ at the same time. Bus C arrives at $X$ every 14 minutes after 5:05 p.m., or 5:19 p.m., 5:33 p.m., and so on. Bus A also arrives at $X$ at 5:33 p.m. Are there other times after 5:33 p.m. (and before 10:00 p.m.) that Bus A and Bus C arrive at $X$ at the same time? Bus A arrives at $X$ every 6 minutes and Bus C arrives at $X$ every 14 minutes. Since the lowest common multiple of 6 and 14 is 42, then Bus A and Bus C will each arrive at $X$ every 42 minutes after 5:33 p.m., or at 6:15 p.m., 6:57 p.m., 7:39 p.m., 8:21 p.m., 9:03 p.m., and at 9:45 p.m. The times when each pair of buses meet at $X$ at the same time between 5:00 p.m. and 10:00 p.m. are listed below. Bus A and Bus B: 15 and 45 minutes past each hour Bus B and Bus C: 5:05 p.m., 6:15 p.m., 7:25 p.m., 8:35 p.m., 9:45 p.m. Bus A and Bus C: 5:33 p.m., 6:15 p.m., 6:57 p.m., 7:39 p.m., 8:21 p.m., 9:03 p.m., 9:45 p.m. Finally, we determine the number of different times that two or more buses arrive at $X$ at the same time. Bus A and Bus B arrive at $X$ at 10 different times. Bus B and Bus C arrive at $X$ at 5 different times; however 2 of these times (6:15 p.m. and 9:45 p.m.) have already been counted, so there are 3 new times. Bus A and Bus C arrive at $X$ at 7 different times; however 2 of these times (6:15 p.m. and 9:45 p.m.) have already been counted, so there are 5 new times. The number of times that two or more buses arrive at $X$ between 5:00 p.m. and 10:00 p.m. is $10+3+5=18$.

We begin by determining the area of $\triangle FGH$. The base $GH$ has length 10, the perpendicular height of the triangle from $GH$ to $F$ is also 10, and so the area of $\triangle FGH$ is $\frac12\times10\times10=50$. We wish to determine which of the 41 points is a possible location for $P$ so that $\triangle FPG$ or $\triangle GPH$ or $\triangle HPF$ has area $\frac12\times50=25$. We begin by considering the possible locations for $P$ so that $\triangle GPH$ has area 25. The base $GH$ has length 10, and so the perpendicular height from $GH$ to $P$ must be 5 (since $\frac12\times10\times5=25$). Since the distance between two parallel lines remains constant, any point $P$ lying on a line that is parallel to $GH$ and that is 5 units from $GH$ will give a $\triangle GPH$ whose area is 25. The line labelled $\ell$ is parallel to $GH$ and lies 5 units above $GH$, and so any point that lies on $\ell$ is a distance of 5 units from base $GH$. There are 5 points that lie on $\ell$ (that are at the intersection of gridlines) and that are inside $\triangle FGH$. These 5 points and one of the 5 possibilities for $\triangle GPH$ are shown. [[IMAGE0]] Next, we consider the possible locations for $P$ so that $\triangle FPG$ has area 25. Consider the point $X$ on $GH$ so that $FX$ is perpendicular to $GH$. [[IMAGE1]] Since $\triangle FGH$ is isosceles, then $FX$ divides the area of $\triangle FGH$ in half and so $\triangle FXG$ has area 25. However, if $X$ lies on $GH$, then $X$ does not lie inside $\triangle FGH$. For this reason, $X$ is not a possible location for $P$, however it does provide some valuable information and insight. If we consider the base of $\triangle FXG$ to be $FG$, then the perpendicular distance from $X$ to $FG$ is equal to the height required from base $FG$ so that $\triangle FXG$ has area 25. Any point $P$ that lies on a line that is parallel to $FG$ and that is the same distance from $FG$ as $X$ will give a $\triangle FPG$ whose area is also 25. (This is the same property that we saw previously for $\triangle GPH$, with base $GH$.) How do we create a line that passes through $X$ and that is parallel to $FG$? Beginning at $G$, if we move 5 units right and 10 units up we arrive at $F$. Begin at $X$, move 5 units right and 10 units up, and call this point $Y$. Can you explain why the line segment $XY$ is parallel to $FG$? Any point that lies on $XY$ is a distance from base $FG$ equal to the required height of $\triangle FPG$. There are 2 points that lie on $XY$ (that are at the intersection of gridlines) and that are inside $\triangle FGH$. (We note that we may move right 5 and up 10 by moving in ‘steps’ of right 1 and up 2 to arrive at each of these 2 points.) These 2 points and one possibility for $\triangle FPG$ are shown. [[IMAGE2]] Finally, we consider the possible locations for $P$ so that $\triangle HPF$ has area 25. As a result of symmetry, this case is identical to the previous case. Thus, there are 2 possible locations for $P$ so that $\triangle HPF$ has area 25. In total, there are $5+2+2=9$ triangles that have an area that is exactly half of the area of $\triangle FGH$.