Pancake race

This seems like a logic puzzle at first glance, but it carries a nice geometric intuition. Specifically, imagine time on the $x$-axis and distance from home on the $y$-axis. If Alice, Bob, and Carey (I’ll just call them A, B, C from now on) walk at a constant speed, their paths show up as straight lines.

For the first problem, the optimal thing to do is as follows:

1. A+C ride together, and B rides alone.
2. At some point, A drops off C and returns to meet with B.
3. C continues home alone.
4. A meets up with B, picks them up, and A+B ride home together.

The optimal thing to have happen is for A,B,C to all arrive home at the same time, as illustrated in the diagram below. If C arrived home before A+B, then A should have dropped them off earlier and hence arrived home earlier with B. Likewise, if C arrived home after A+B, then A should have dropped them off later so they could arrive home sooner.

In the diagram, blue lines have “A slope” $\pm 1$, green lines have “B slope” $\pm\frac{1}{2}$, and red lines have “C slope” $\pm\frac{1}{3}$. The points of interest are F, G, H, as shown. I wrote in their coordinates in terms of unknowns $p,q,r$ in units of 10 minutes. We now have three equations:

• FG has slope $1$, therefore $p-\tfrac{1}{2}q = q-p$
• GH has slope $-1$, therefore $1-\tfrac{1}{2}q = r-q$
• FH has slope $-\frac{1}{3}$, therefore $1-p=\tfrac{1}{3}(r-p)$

Solving these equations, we obtain
\[
p=\frac{3}{4},\quad q=1,\quad r=\frac{3}{2}.
\]Since I used units of 10 minutes, the arrival time is $10r = 10 \cdot \frac{3}{2}$ minutes, or exactly 15 minutes.

Note: I’d like to thank commenter TLK for pointing out an error in a previous version of my solution. I was a bit hasty in my first write-up and presumed that A would carry C all the way home before returning to fetch B!

Extra credit

For the extra credit, the solution is quite a bit more complicated:

1. A+D ride together, and B+C ride together
2. At some point, A drops off D and returns to meet with B.
3. D continues toward home.
4. A meets up with B, returns home with B, and C continues alone.
5. A+B meet up with D, B carries D home and A turns back.
6. A meets up with C and carries C home.

Once again, the optimal thing to have happen is for A,B,C,D to all arrive home at the same time. Here is the corresponding diagram.

This time, we have more equations and more variables.

• FG has slope $1$, therefore $p-\tfrac{1}{2}q = q-p$.
• FH has slope $-\tfrac{1}{6}$, therefore $H = (p+s,1-p-\tfrac{1}{6}s)$ for some $s$.
• GJ has slope $-\tfrac{1}{3}$, therefore $J = (q+r,1-\tfrac{1}{2}q-\tfrac{1}{3}r$.
• GH has slope $-1$, therefore $p+s-q=p+\tfrac{1}{6}s-\tfrac{1}{2}q$
• HJ has slope $1$, therefore $q+r-p-s=p+\tfrac{1}{6}s-\tfrac{1}{2}q-\tfrac{1}{3}r$.
• HK has slope $-\tfrac{1}{2}$, therefore $1-p-\tfrac{1}{6}s = \tfrac{1}{2}(t-p-s)$.
• JK has slope $-1$, therefore $1-\tfrac{1}{2}q-\tfrac{1}{3}r=t-q-r$.

Solving these equations, we obtain:
\[
p= \frac{5}{8},\quad
q= \frac{5}{6},\quad
r= \frac{7}{16},\quad
s= \frac{1}{2},\quad
t= \frac{41}{24}.
\]The arrival time is $10t=\frac{205}{12}$ minutes, or 17.0833 minutes,
or exactly 17 minutes and 5 seconds.

Note: I’d like to thank commenter Sanandan Swaminathan for pointing out an error in a previous version of my solution. Once again, I was a bit hasty in my first write-up and presumed that A would pick up C on their first return trip, when in fact it’s better for A to pick up B! Here is a diagram of the previous (sub-optimal) solution I found, which only achieves 17.8125 minutes, or 17 minutes and 48.75 seconds.