Bottles in a Wine Rack

Ron Honsberger, the author of several bestsellers published by the MAA, begins one of his latest problem collections with an exciting problem:

Across the horizontal bottom of a rectangular wine rack there is room for more than three bottles (A, B, C) but not enough for a fourth bottle. All the bottles that are put into this rack are the same size. Naturally, bottles A and C are laid against the sides of the rack and the second layer, consisting of just two bottles (D, E), holds B in place somewhere between A and C. Now we can lay in a third row of bottles (F, G, H), with F and H resting against the sides of the rack. Then a fourth layer is held to just two bottles (I, J).

Now, if the bottles are not evenly spaced in the bottom row, the second, third and fourth rows can slope considerably, tiling at different angles for different spacings. Prove, however, that, whatever the spacing in the bottom row, the fifth row (K, L, M) is always perfectly horizontal!

(The applet below lets you experiment with the problem. Bottle B is moveable and the radius of the bottles can be changed somewhat with the help of the scroll bar at the bottom of the applet.)

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Bottles in a Wine Rack

What if applet does not run?



  1. R. Honsberger, Mathematical Diamonds, MAA, 2003

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The problem has been borrowed from a remarkable Which Way Did the Bicycle Go?, another MAA's publication. It was discovered by Charles Payan of the Laboratoire de Structures Dèscretes et Didactique in France, one of the creators of the CABRI package, while experimenting with this program. The solution is due to Hung Dinh.

Bottles in a Wine Rack, proof

We wish to show that KL is horizontal. Since AK is vertical, suffice it to show that angle FKL (or AKL) is right. (The angle HML is treated similarly.)

The distance between the centers of touching bottles is twice their radius. So that I is at the same distance from F, K, and L, and is therefore the circumcenter of triangle FKL. If I were to lie on FL, then FL would be the diameter of the circumcircle of the triangle, making angle FKL right.

On the other hand, we do know that in ΔBCH angle C is right, so that I lies midway between B and H. The four quadrilaterals around G are all rhombuses and so is their union BFLH. Importantly, the lines BE, EH, DG, GJ, FI, IL are equal and parallel. We concluded that indeed I is the midpoint of IL, as needed.

The solution is simple and elegant, but has a flaw. Or rather, the flaw is in the statement of the problem. Experimenting with the applet, you may have noticed that it is not always the case that both F and H snuggle against a side of the rack. When the common radius of the bottles is on the lower end of the admissible length, this condition is violated and the top layer bottles K, L, M are no longer all horizontal. Remarkably, either K and L or L and M are still horizontal. This requires an additional proof.

If, for example, the circle H is away from the rack (this is achieved with circle B near circle A), then BF, EI, HL are sill parallel, although E no longer lies on BH. Vectors BF, EI, HL are equal and define a translation under which circles, B, E, H are mapped onto circles F, I, L. Circle C is mapped by that translation to some circle C'. Circle C is tangent to both circles E and H. Circle C' therefore is tangent to circles I and L. Circles B and C being horizontal are symmetric with respect to the vertical line through the center of circle E. After the translation, the same holds for circles F and C': they are symmetric in the vertical axis through I. In particular their centers are on the same level horizontally. Because of the symmetry of circles F and C', the tangent circles K and L are also symmetric in the same vertical axis. If so, they are also horizontal. Circle M is a little down.

I must add that the original formulation in Which Way Did the Bicycle Go? is a little different from Honsberger's. Originally, the bottles have radius 1 and the distance between the centers of A and B (and those of B and C) is less than 4. This, however, does not save the day: it is still possible that only 2 circles out of K, L, M be horizontal.

When there are 4 bottles at the bottom, the rack contains 13 bottles. This is reflected in the caption - Thirteen Bottles of Wine - under which the problem appears in Which Way Did the Bicycle Go?. Elsewhere there is a tool to experiment with more than 13 bottles.

It may be observed that the center of circle G is located on the vertical axis of the rack regardless of the radius and location of bottle B. Furthermore, it was shown by A. Brown that, in a pyramid that starts with any number of circles at the bottom, the center of the top bottle always projects at the midpoint of the base of the rack. We have a dynamic illustration of this result.


  1. A. Brown, A Circle-Stacking Theorem, Math Magazine, v. 76, no 4 (Oct., 2003), pp. 301-302
  2. R. Honsberger, Mathematical Diamonds, MAA, 2003
  3. J. Konhauser, D. Velleman, S. Wagon, Which Way Did the Bicycle Go?, MAA, 1996

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Copyright © 1996-2018 Alexander Bogomolny