Only 'Geniuses' Can Solve - How Much Water To Cover The Balls?

HUGE thanks to Stephen Lerantges for solving the problem! A glass has an inner diameter of 10 cm and inner height of 10 cm. Each of the 4 balls has a diameter of 5 cm. How much water do you need to exactly cover to the top of the balls?

HUGE thanks to Stephen Lerantges for suggesting and solving it. I got the wrong answer, and learned my mistake by reading Stephen’s solution. Also thanks to Zack, Tomas for independently sending me this problem.

Thanks to all patrons! Special thanks to:
Shrihari Puranik
Kyle

If you like my videos, you can support me at Patreon and get exclusive rewards:

My blog post for this video:

Sources
This is Stephen's solution, which is slightly different but generally the same idea:

If you like my videos, you can support me at Patreon:

Connect on social media. I update each site when I have a new video or blog post, so you can follow me on whichever method is most convenient for you.

If you buy from the links below I may receive a commission for sales. This has no effect on the price for you.

My Books
"The Joy of Game Theory" shows how you can use math to out-think your competition. (rated 3.8/5 stars on 31 reviews)

"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" is a handbook that explains the many ways we are biased about decision-making and offers techniques to make smart decisions. (rated 5/5 stars on 2 reviews)

"Math Puzzles Volume 1" features classic brain teasers and riddles with complete solutions for problems in counting, geometry, probability, and game theory. Volume 1 is rated 4.4/5 stars on 13 reviews.

"Math Puzzles Volume 2" is a sequel book with more great problems. (rated 5/5 stars on 3 reviews)

"Math Puzzles Volume 3" is the third in the series. (rated 3.8/5 stars on 4 reviews)

"40 Paradoxes in Logic, Probability, and Game Theory" contains thought-provoking and counter-intuitive results. (rated 4.3/5 stars on 12 reviews)

"The Best Mental Math Tricks" teaches how you can look like a math genius by solving problems in your head (rated 4.7/5 stars on 4 reviews)

"Multiply Numbers By Drawing Lines" This book is a reference guide for my video that has over 1 million views on a geometric method to multiply numbers. (rated 5/5 stars on 3 reviews)

[edderiofer]

A possibly-easier way: It's easy to see that the four spheres' centres form a regular tetrahedron. A regular tetrahedron is also formed when you connect up four of the vertices of a cube. So constructing the cube that exactly exscribes the tetrahedron, you have that each of its face diagonals have length 5, so the cube has side 2.5sqrt(2) (and so does the vertical distance between the centres). From there, proceed as before.

[andrewperrin]

I tried testing this with ping pong balls and ended up drowning 6 people.

[donaldasayers]

Centres of spheres forms a tetrahedron. inscribe a tetrahedron in a cube, edges of tetrahedron are face diagonals of the cube. Diagonals are 5cm so height =5/sqr2, to which you add 5 for the two half radii. The rest follows easy.

[iwersonsch]

We rotate the picture so that the spheres at the bottom span from (0, -5) to (0, 5), and the spheres at the top span from (-5, 0) to (5, 0) when viewn from above. That gives us immediate information that the centers of the bottom spheres must be at (0, -2.5) and (0, 2.5), while the centers of the top spheres must be at (-2.5, 0) and (2.5, 0).
We know that the distance of the centers of two touching spheres is twice their radius, i.e. 5. This distance is the square root taken from the distances in each of the three dimensions squared. Regarding a bottom vs. a top ball, we get 5=sqrt(2.5^2+2.5^2+h^2), where h is the height of the top sphere. This gives us that h^2=12.5, i.e. h=5/sqrt(2).
So the height that the top spheres are off the ground is 5/sqrt(2) centimeters, resulting in their tops being at 5+5/sqrt(2) centimeters of height.
Now using the formulas for the volume of a sphere and for the area of a circle (which leads us to the volume of a cylinder), we can calculate the volume in the glass up to the height of the top of the higher balls, then subtract 4 times the volume of a ball, and we should get the solution.

=
~408.59 cm³
And now I take the risk of posting this before watching the solution. Wish me luck

[BigDBrian]

There is an assumption. It's that the balls are sufficiently dense to stay in position. If they were to float... well, you'd have to fill it to the ceiling of the house it's in ;)

[sealand]

It took me less than a second to figure out the answer to his question: "Can you figure it out?"

I feel so smart!

[stmounts]

I worked it out as follows....it helps if you have a model tetrahedron to look at ;)
The centres of the 4 balls (diameter 2r) form the vertices of a regular tetrahedron whose properties are well know, but I'll explain a bit further anyway.
The edges of the tetrahedron are the same length as the ball diameter (2r) since the balls are touching, also the top and bottom edges are parallel to the bottom of the cylinder.
Each face of the tetrahedron is an equilateral triangle with each side = 2r and an
'altitude' = r x sqrt(3).
Now consider the triangle formed by taking a vertical slice upward parallel to the bottom edge of the tetrahedron (i.e. with top vertice at the midpoint of the upper edge of the tetrahedron.)
We get an isosceles triangle with bottom side = 2r and remaining sides = r x sqrt(3).
The 'altitude' of this triangle = r x sqrt(2) (from pythag. on the included right angle triangle with hypotenuse = r x sqrt(3) and top side = r)
Thus the water height = r( 2 + sqrt(2) ) and the rest follows easily!

[Crivella]

Taking L=5cm (balls diameter)
The four balls form a figure made up by 4 equilateral triangle as a surface. We can easly calculate the height of this solid by first calculatingthe height of 1 equilatera triangle (sqrt(3)/2 * L) and than we get the height of the solid ( sqrt(2) / 2 * L)
To this we add the eight of the balls (The solid we just considered has the vertexes in the center of the balls) so
H_tot = (1 + sqrt(2)/2) * L/2
So the cilinder volumes is
pi * (1 + sqrt(2)/2) * L^3
To this we subtract the volume of the spheres
4 * (4/3 * pi * (L/2)^3)
So finally
V_water = pi * L^3 * (1 +sqrt(2)/2 - 2/3) =~ 408.58 cm^3

[Anonymous-jono]

Just out of curiosity, but... how did you get this wrong?

[eithan]

My answer before watching the solution in the video and before reading any comments is:
(125/3 + 125/sqrt(2))*PI cubed cm of water.
This is how I got it:

If we connect the centers of the top 2 spheres with the center of the bottom left sphere (as seen from the perspective of the illustration in the video) we get an equilateral triangle, and we get another one by connecting to the center of the bottom right sphere.
Now we place a point in the middle of the line segment which connect the centers of the top two spheres (lets call it A) and connect to the bottom vertex of the left triangle which is exactly its height, lets call that point B, and the same on the right triangle, lets call that point C. Now the height of triangle ABC + 5 is the height from bottom of the glass to the top of the balls. The rest is just calculating:

Length of AB = BC = 5*sqrt(3)/2
Length of BC = 5
By Pythagorean theorem the height of triangle ABC = sqrt((5*sqrt(3)/2)^2 - (5/2)^2) = 5/sqrt(2)
Bottom to top = 5 + 5/sqrt(2)

Volume of the 4 balls = 4 * (4/3)*(5/2)^3*PI = (250/3)*PI
Volume of the water without balls = (5 + 5/sqrt(2)) * 25 * PI = 250*PI

(5 + 5/sqrt(2)) * 25 * PI - (250/3)*PI = (125/3 + 125/sqrt(2))*PI

Edit: Saw that many found the solution by looking at a tetrahedron formed by the centers of the 4 balls, that is basically my approach, only that I calculated the height of this tetrahedron.

[Horinius]

Another approach is analytical geometry.
Consider the centers of the lower layer spheres lying on x-axis, and those of upper layer spheres on y-axis at a height h, the axle of the cylinder on z-axis, we have
lower layer: (-2.5, 0, 0) and (2.5, 0, 0)
upper layer: (0, -2.5, h) and (0, 2.5, h)
The distance between any two centers has to be 5, which means we have to solve
Distance [(2.5, 0, 0), (0, 2.5, h)] = 5
==> |2.5 - 0|² + |0 - 2.5|² + |0 - h|² = 25
==> h = 2.5 sqrt(2)
And the rest is in your video.

[sambobag]

Make a equal-sided triangle from centers of 2 bottom and one top balls, it's side is 2*2.5=5, its height is 5*sqrt(3)/2
Center of upper ball is 5-2.5=2.5 cm out of cylinder's axis - this is the projection of the triangle's height to base plain, so projection of the triangle's height to vertical axis is sqrt(25*3/4-25/4)=5/sqrt(2) - this is height of top ball's center above bottom ones, so top of top balls is 2*2.5+5/sqrt(2)=5+5/sqrt(2)

[Vcimdarf]

Another way to get the same answer is, consider the balls to have centers A, B, C, D with A and B at the top. Let E at midpoint between C and D. Triangle ABE is vertical and has height 2.5 sqrt(2) by pythagorean theorem, where AB has length 5 and AE, BE have length 2.5 sqrt (3). The length of AE and BE can be found from ACD and BCD equilateral triangles.

[eytansuchard]

The easiest way: The centers of the balls make a tetrahedron. The distance between two juxtaposed edges of a tetrahedron = L/sqrt(2) where L is the edge length, in our case L=5. So the height will be that distance plus one radius up and one radius down so h = 5 / sqrt(2) + 5 = 5(1+1/sqrt(2)). The volume without the balls, at that height is 5*(1+1/sqrt(2))*Pi*5*5=125Pi + 125Pi/sqrt(2). We have to subtract the 4 balls. We subtract 4 *(4/3) Pi * 2.5 * 2.5 *2.5 = (16/3) * Pi * 6.25 * 2.5 = (250/3) Pi. Now we subtract the two volumes, 125Pi + 125Pi/sqrt(2) - (250/3) Pi and we have Volume = (125-250/3 + 125/sqrt(2))Pi.

[jameshowland]

As the spheres are tangent we can work out that the distance between any 2 centers of a sphere is 5(double the radius). As the diagram is symmetrical similar we can work out the height by using pythagoras where a=b so that 2a^2=5^2. When we solve this we find a to equal root 12.5= 2.5 * root 2. Then all we do is add this to 5 to get a value of the height of the water. Then just multiply this by cylinders cross section and deduct the volume of the spheres and then you get your answer.

[kenhaley]

Got it! I did it a little differently by realizing that the centers of the spheres form a regular tetrahedron with edge length 5. Cutting the tetrahedron in half, perpendicular to the top horizontal edge exposes a triangle whose height is 2.5(sqrt(2)). This is easily calculated by splitting that triangle vertically into 2 right triangles. The hypotenuse forms the altitude of one of the original tetrahedron's faces (an equilateral triangle) which is 2.5(sqrt(3)). The short leg of that triangle is 2.5, so the height of it is 2.5(sqrt(2)). The rest of the solution is the same as Presh's.

[DouglasKubler]

I connect the centers of a top and bottom sphere. It is 5 cm long. By symmetry(*) it is at an angle of 45 degrees to the horizontal. The legs of the 45 right triangle are 2.5sqrt(2). The rest follows.
Symmetry - top and bottom are arbitrary labels, if you flip the complex upside down the angle doesn't change, hence 45 degrees.

[DamienConcordel]

Another solution:
1-Notice indeed that the four spheres' centers form a perfect tetrahedron: the bottom two spheres fit perfectly in the width of the glass, so they touch and the distance between their centers is exactly one sphere diameter. Same for the top two. Also, the top two are sitting on top of the bottom ones, neatly tucked into the space left above them. Because of the exact symmetry of the setup of the bottom spheres, each of the top ones is touching the edge of the glass, the other top one, AND each of the two bottom ones. Hence the perfect regular tetrahedron with a side length of 5.

2-From there, it's just triangle calculations: You need the vertical distance between the top and bottom edges. This is a two-step calculation. First, you need the height of one of the faces (for example, one of the faces with the bottom edge as a base). This is important for the next step, as you will need a vertical length as a result, and this height will be in a vertical plane with the top edge. As all faces of the tetrahedron are equilateral triangles, this means the height of the one face is 5/2 sqrt(3).
The same calculation applies to the other face with the bottom edge as a base.

3-From there we have one exactly vertical triangle made up of the top edge of the tetrahedron and the two heights we have just calculated. This is an isoceles triangle, with a base (the top edge) of 5 and the other two sides of 5/2 sqrt(3). From there, all it takes is Pythagoras on half the triangle to find the distance we're looking for. So the top and bottom spheres are 5/2sqrt(2) units apart.

4-Remember this is just the height separating the centers of the bottom spheres from those of the bottom spheres. So we need to add in two radii: the total height of the spheres setup is therefore 5+5/2sqrt(2). That is the HEIGHT of water to pour into the glass to cover the spheres.

5-use the cylinder volume and sphere volume formulae to figure out the VOLUME of water to pour: with h=5+5/2sqrt(2),
pi r²h - 4*4/3 pi r³ = 408.58 mL

[HenryFabianGT]

*Nice approach! A much easier way here:*
- Construct a regular tetrahedron from each sphere center
- Calculate the height of that tetrahedron
- Use that result to calculate the total height of the arrangement of spheres (don't forget to add the radii)
- Using the total height, calculate the total volume (bottom to top), and subtract the volume of the 4 spheres. DONE!

[robxyz]

There is an error in your calculation at 5:13.  The height 2.5*sqrt(2) assumes the three circles are tangent in 2 dimensions (if you expand Pythagorean theorem using that as a side, the Hypotenuse = 5, which is not consistent with the diagram - radii overlap).  You need to imagine the balls from the side view then rotate the view again to see that the balls are skewed at a 30 degree angle.  so your new height is Volume = ~371.2ml