This is the conclusion of the original post about Tony’s Cup Drop Paradox. If you haven’t read the puzzle already, please do so first, take a stab (you can submit your response anonymously and see what others said) below:
Ok, assuming you have given it a shot and perhaps even nailed it, here’s my answer along with detailed explanation and proof. Read on…
To recap, the setup is as follows:
The cups in orange and purple are laid out in rows with each adjacent cup at equal distance from another: 1 foot apart. The orange cups are for Player A to pick up (one by one) and drop into the orange Bucket A (one by one). The purple cups are for Player B to pick up (one by one) and drop into the purple Bucket B (one by one). As you can see from this court layout, Player A starts from the left and moves to the right to pick up the first cup (Cup #1), then return to (moving left) Bucket A to drop it in, then moves on to pick up the second cup (Cup #2), then return to (moving left) Bucket A to drop it in, then moves on to pick up the third cup (Cup #3), until the 10th cup is picked up and dropped into Bucket A.
Similarly, Player B has to accomplish the same but starting from the right side of the court, picking up her first cup (Cup #10 since we’re counting along the x-axis, left to right) which is closest to her and travel left toward her Bucket B to drop it off, then return (moving right) to pick up the next cup (Cup #9) and doing the same until all of her 10 cups are dropped into Bucket B.
Since the speed is constant, the only thing that will determine who finishes first (that is, shortest time), depends on the distances covered by the players. If the distances are the same, then they’ll finish at the exact same time resulting in a tie. However, are there distances covered the same? Let’s find out.
I tabulated the distance to each cup, and from each cup to the target bucket for Player A as follows:
You can follow the arrows and the operators to see how the 2-way trip at each cup and the running total is calculated at each cup position. We find the final total to be 110. That is, Player A must travel 110 feet total for her 10 cups.
Now, let’s do the same for Player B. Her table of distance looks as follows:
That is, Player B must travel 101 feet total for her 10 cups. Therefore, the winner is Player B!
Let’s visualize this and see if we can derive some algorithms for the distances covered. If we chart the running total distance at each of the 10 cups, the trends look as follows for the players.
The trend equations are also shown for each player on their charts above. We clearly see that Player B reduces the distance with each subsequent cup and by the last cup travels 9 feet less than Player A!
We can simplify the equations furthers. For Player A, it becomes: y = x(x+1) which is same as y = x^2 + 1x – 5E-13 when x is factored out, and as 5e-13 is 0.0000000000005 (or, 5 x 10^-13 in scientific notation) is such a very, very small number that we can drop that part. You can test this by substituting x for any cup number, and matching the resulting y with the table and chart above for proof. For example, for 10 cups, x=10, y becomes 10 (10+1) = 110 — exactly what we find from the table above. With this formula, we can now determine the distance at any cup position, for any number of cups accurately!
Focusing on Player B‘s equation, we can simplify her trend equation to: y = x (-x +1) + 101 (which is the same as y = -x2 + 1x + 101 written in a cleaner form). Again, plug in any value for x, and check the y output value and match it with the Player B table or chart, and we find an exact match.
Showing both players’ distances in a single chart for direct comparison (as 2-series XY Scatter plot):
There you have it! Surprised? Or did you solve it yourself and backed it up with math and/or logic? If so, congratulations! Either way, I hope you found this puzzle and the formulaic approach to the proof amusing and educational.
For more puzzles and brain-twisters, see the list here https://flyingsalmon.net/tag/puzzle/ and come back often for more as newer ones will be added over time.
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