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After you posted mean things online about Aquaman, he found out where you lived and kidnapped you (using his awesome aqua-powers), taking you to his super-secret aqua-lair. He has placed you in a room with a ceiling less than $10$ feet tall, a perfectly level floor, and, at the end of the room, a perfectly vertical wall with a mark exactly $5$ feet above floor level. There is a drain in the floor so that water drained into the room will not accumulate (at all). The room is otherwise featureless. Behind the marked wall is a reservoir of water. You have access to a compass (which can only draw circles around a given center), a plumb bob (which can only indicate a perfectly vertical line), and an awl (which can only poke holes through the wall). All the walls, ceiling, and floor are all opaque and you cannot see the reservoir. Notice that the tools given do not allow you to make any measurement.

Aquaman has challenged you

If you can drain the reservoir such that its water level is exactly $10$ feet above the floor, I will return you to your house and we can forget this ever happened. Otherwise, I will get a real superhero to come and deal with you - and you don't want that!

How can you use the three tools given to drain the reservoir and to know exactly when its water level is $10$ feet above the floor?

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  • $\begingroup$ The water remaining in the reservoir should be 10 feet high? Or do we need to drain water equivalent to the room's volume? $\endgroup$
    – CodeNewbie
    Commented Jul 18, 2015 at 16:33
  • $\begingroup$ @CodeNewbie Yes, the water remaining in the reservoir should be $10$ feet high (when measured starting from the floor). $\endgroup$
    – Milo Brandt
    Commented Jul 18, 2015 at 16:35
  • $\begingroup$ What are the dimensions of the reservoir then? What is the initial level of water? $\endgroup$
    – CodeNewbie
    Commented Jul 18, 2015 at 16:37
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    $\begingroup$ @CodeNewbie The water is initially more than $10$ feet high. No further information is necessary to solve this. $\endgroup$
    – Milo Brandt
    Commented Jul 18, 2015 at 16:42
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    $\begingroup$ Assuming the floor is at least 10 feet x 10 feet, you have a pen and can keep the span of the compass... Pick a point A at the 5 feet mark and use the compass to find two points B and C on the lower edge of the wall which are equidistant from it. Keeping the same radius, you draw two circles with centers at these points and find a point D which is at distance 5 feet from the wall. Using the bob you also mark the midpoint E between these two points. Now you can measure distances of 5 feet and sqrt(2)*5 feet. Intersect S(D, 5) with S(B, sqrt(2)*5) at point F. Intersect S(D,5) with S(F,sqrt(2)*5) $\endgroup$
    – Puzzle Prime
    Commented Jul 18, 2015 at 18:05

4 Answers 4

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Set the compass to some size, less than the distance from the 5ft mark to the ceiling. Use the bob to find two points on the circle, one directly above the other. Then use the awl to punch holes at these points. Wait until the streams from both holes fall on the same spot. You're done! Plug the holes in a hurry and mock Aquaman some more.

Explanation: the velocity of a stream from a given hole will be $v=\sqrt{2gd}$ where d is the distance to the top of the water. The time to fall from a height h is given by $t=\sqrt{\frac{2h}{g}}$. This means that the horizontal distance traveled by a stream before fitting the ground will be $vt=2\sqrt{hd}$.

Given that $h+d=l$ where l is the height of water above the floor, the only time $h_1d_1=h_2d_2$ is when $h_1=d_2, h_2=d_1$, which means $l=h_1+h_2$. Because the holes are symmetric about the 5ft (height) line, $h_1+h_2=10ft, l=10ft$

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  • $\begingroup$ Great solution! I'm sure the OP didn't think about this! $\endgroup$
    – hjhjhj57
    Commented Jul 18, 2015 at 22:35
  • $\begingroup$ This is a lovely explanation of the solution I was going for! One note I might make is that we don't actually need to punch holes directly above and below the mark (it's kind of hard to line up a bob with anything) - the circle and the bob are both symmetric about a plane $5$ feet from the floor, so $h_1+h_2=10$ feet even if we don't line things up exactly. $\endgroup$
    – Milo Brandt
    Commented Jul 19, 2015 at 0:28
  • $\begingroup$ Good point! I'll edit that in. $\endgroup$
    – frodoskywalker
    Commented Jul 19, 2015 at 0:31
  • $\begingroup$ @Meelo , I had given the same answer "1 hour" before frodoskywalker , so I wonder why you accepted this answer ? Maybe , my answer lacked the equations ? $\endgroup$
    – Prem
    Commented Jul 19, 2015 at 8:07
  • $\begingroup$ Unless the wall is infinitesimally thin, the water will be slowed by friction and hence exit at a velocity that is lower than $\sqrt{2gd}$ by an unknown amount. $\endgroup$
    – A. P.
    Commented Mar 27, 2018 at 9:07
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Use the compass to draw a very small circle around the 5 feet MARK.
Now use the bob to make it pass through the MARK, and get two points on the circle, one on top and the other on bottom of the circle.
Use the awl to poke 3 holes at center+top+bottom.

The water will flow out and fall on the floor at 3 moving points. When the reservoir has 10 feet of water, then the water from the center hole will fall at the maximum Distance, while the water from other two holes will fall very near but at a lesser Distance.

Plug the holes now.
Done.

Alternate solution:
Make only 2 holes on the circle, at top and bottom.
When both the water flows fall at the same point, the water from the center hole (which we did not make) would have fallen at the maximum Distance.
So plug the 2 holes now.
Done.

References :
https://en.wikipedia.org/wiki/Torricelli%27s_law
https://physics.stackexchange.com/questions/6341/how-far-will-water-squirt-out-from-a-hole-in-a-can

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    $\begingroup$ This is really close to the intended solution, but doesn't quite work as well - in an ideal world, this only tells us the waterlevel to plus or minus the radius of the circle. In a less ideal world, we'll have issues with the streams interfering with each other and won't be able to tell. (Hint: Your solution works without puncturing a center hole and can be done with any sized circle around the center. What will be true of the other two holes when the water level is $10$ feet?) $\endgroup$
    – Milo Brandt
    Commented Jul 18, 2015 at 17:45
  • $\begingroup$ @Meelo, thanks , I added an alternate solution based on your comment. $\endgroup$
    – Prem
    Commented Jul 18, 2015 at 17:49
  • $\begingroup$ FYI for viewers : for some inexplicable reason, @Meelo has chosen to "accept" a later answer, even though I have given two solutions (including the expected solution) but that is his prerogative. $\endgroup$
    – Prem
    Commented Jul 22, 2015 at 5:29
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If you drill a hole at the $5$ ft height, and the current water height is $H$, then the water lands at a distance $2\sqrt{5(H-5)}$ from the wall. When $H=10$ ft, this is a distance of $10$ ft. So, we just need to make a mark $10$ ft from where the wall meets the floor, drill a hole at $5$ ft, then wait till the water falls on that mark.

But how do we make the mark? First, we show how to do this using a straightedge and compass. Since every straightedge/compass construction can be done with a compass alone, this means we can make the mark with the tools we have.

The construction (using a straightedge) isn't hard to figure out, but for completeness, here it is. Draw a point $P_1$ on the wall, on the 5ft line. Use the plumb bob to draw a point $P_2$ directly below it on the floor. Use the compass to make two points $P_3$ and $P_4$, also on the edge between the wall and the floor, both $5$ ft from $P_2$. Now, draw two circles, one with center $P_3$ passing through $P_4$, and another with center $P_4$ passing $P_3$. These intersect at $P_5$. Draw the line $L$ through $P_2$ and $P_5$, and a circle $C$ centered at $P_2$ through $P_3$ and $P_4$. Say that $C$ and $L$ intersect at $P_6$. Finally, draw a circle $\tilde C$ centered at $P_6$ passing through $P_2$. Then $\tilde C$ intersects $L$ at the desired mark, $10$ ft from $P_2$. Now drill at $P_1$, wait for the stream to hit the mark, and you're done!

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  • $\begingroup$ Posted a similar approach in the comments of Meelo's problem... I'm not sure we can draw lines, so used a circle with radius sqrt(2)*5 instead. For this solution we still need to have a pen though. $\endgroup$
    – Puzzle Prime
    Commented Jul 18, 2015 at 18:14
  • $\begingroup$ (1) Drawing lines is not allowed with only compass+bob+awl. (2) While it is true that Compass is enough to get all constructions with compass+straightedge, this applies to infinite plane, but here we are limited to a wall or a floor, so that theorem is not sufficient. If the wall is not sufficiently wide, you may not be able to draw P3 & P4. $\endgroup$
    – Prem
    Commented Jul 18, 2015 at 18:23
  • $\begingroup$ Actually even if the floor is too narrow, we still can find the 10 feet mark, using the side walls. The assumptions we need are that we have a pen and the floor is at least 10 feet long. $\endgroup$
    – Puzzle Prime
    Commented Jul 18, 2015 at 18:48
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    $\begingroup$ @ArturKirkoryan We definitely have a writing utensil, since the compass can draw circles $\endgroup$
    – Mike Earnest
    Commented Jul 18, 2015 at 18:51
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Use the awl to make a hole anywhere in the wall to let water out, and let the water drain. Since the reservoir's water level started higher than 10 feet, eventually it will drain to the level of the hole (which must be less than 10 feet high). Therefore at some time in between, the water level was exactly at 10 feet. Aquaman did not specify that the reservoir's water level needed to stay at 10 feet high for any particular duration of time. =)

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