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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?

Milo Brandt
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  • The water remaining in the reservoir should be 10 feet high? Or do we need to drain water equivalent to the room's volume? – CodeNewbie Jul 18 '15 at 16:33
  • @CodeNewbie Yes, the water remaining in the reservoir should be $10$ feet high (when measured starting from the floor). – Milo Brandt Jul 18 '15 at 16:35
  • What are the dimensions of the reservoir then? What is the initial level of water? – CodeNewbie Jul 18 '15 at 16:37
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    @CodeNewbie The water is initially more than $10$ feet high. No further information is necessary to solve this. – Milo Brandt Jul 18 '15 at 16:42
  • I am not a physicist, so don't know much about that, but... Can we calculate easily how the water propels from a hole depending on the pressure which is caused by the water above it? Then put a hole at the 5 feet line on the wall and wait until the water starts propelling on the floor at some certain level, which we have indicated using the given instruments? I am suggesting this because I guess it doesn't really matter how wide the reservoir is for the calculations. – Puzzle Prime Jul 18 '15 at 16:54
  • Is it safe to presume that we can plug holes we've made with our finger? – VictorHenry Jul 18 '15 at 17:01
  • @ArturKirkoryan Bernoulli's principle can be used (though the Wikipedia page is hopelessly opaque). The result you end up with is that if you poke a hole a distance of $d$ from the top, the water will come out horizontally at a speed of $\sqrt{2gd}$ where $g$ is the acceleration due to gravity. (But the intended solution doesn't rely on measuring any velocities or distances and I purposefully phrased the problem to eliminate such approaches. You will, nevertheless, probably need the result I wrote here in a solution, though) – Milo Brandt Jul 18 '15 at 17:25
  • @VictorHenry Yes, one can plug holes with their fingers. – Milo Brandt Jul 18 '15 at 17:27
  • Oh I see - I thought that maybe the water will drop at distance exactly 5 feet from the wall or something (which can be measured using a compass). – Puzzle Prime Jul 18 '15 at 17:31
  • @ArturKirkoryan Well, it would drop exactly $10$ feet from the wall (if my calculations are correct) - but I don't think it's possible to make a mark there with the given tools (if you had a square and a straightedge, you could probably do it though) – Milo Brandt Jul 18 '15 at 17:51
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    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) – Puzzle Prime Jul 18 '15 at 18:05
  • We can easily choose A to be in the middle of the wall. If the floor is less than 10 feet wide (but still at least 10 feet long), then using the compass, instead of finding point F on the floor, we can find a point F on the side wall, which is at 5 feet height and minimal distance to D. Now intersect the two circles S(D,5) and S(F,d(F,E)) on the floor in order to find the 10 feet mark. This way we just need to assume the floor is at least 10 feet long and that we have a pen. – Puzzle Prime Jul 18 '15 at 18:47
  • This question is all sorts of confusing for what I assume are all the wrong reasons. You say that the water level is over 10 feet high, you say that the ceiling is 10 feet high, and you say that the water level needs to be made to be 10 feet high. How can all these possibly be true? – Kingrames Jul 22 '15 at 15:25
  • @Kingrames Imagine that there is a tall room next to a short room, sharing the same floor and separated by a wall. The tall room is full of water - so it can be filled above the short room's ceiling. – Milo Brandt Jul 22 '15 at 22:33

4 Answers4

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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, $h_1+h_2=10ft, l=10ft$

wchargin
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frodoskywalker
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  • Great solution! I'm sure the OP didn't think about this! – hjhjhj57 Jul 18 '15 at 22:35
  • 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. – Milo Brandt Jul 19 '15 at 00:28
  • Good point! I'll edit that in. – frodoskywalker Jul 19 '15 at 00:31
  • @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 ? – Prem Jul 19 '15 at 08:07
  • 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. – A. P. Mar 27 '18 at 09:07
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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

Community
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Prem
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    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?) – Milo Brandt Jul 18 '15 at 17:45
  • @Meelo, thanks , I added an alternate solution based on your comment. – Prem Jul 18 '15 at 17:49
  • 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. – Prem Jul 22 '15 at 05:29
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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!

CodeNewbie
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Mike Earnest
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  • 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. – Puzzle Prime Jul 18 '15 at 18:14
  • (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. – Prem Jul 18 '15 at 18:23
  • 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. – Puzzle Prime Jul 18 '15 at 18:48
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    @ArturKirkoryan We definitely have a writing utensil, since the compass can draw circles – Mike Earnest Jul 18 '15 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. =)

jamesdlin
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