It’s been a while since I gave cute problems, so here are a few with a theme: solving mathematical problems with physics (I’m *not *talking about the other way around, which is why I said “physical mathematics” instead of “mathematical physics”).

- Take a triangle ABC. Prove that the point X inside the triangle which minimizes AX + BX + CX is a point such that the angles AXB, BXC, and CXA are all degrees.
- Consider an ellipse with ratio of the major to minor axes equal to . From either foci, start enlarging a circle until the circle is tangent to the ellipse from the inside. It is obvious that the number of tangent points will be or and that this only depends on . Given an , figure out this count.
- Take a convex polyhedron and any point inside of it. To every face, drop a normal line from the point. Note that it is both possible to land inside the face or outside. Construct a polyhedron where every such normal line drops outside of the corresponding face, or prove such a polyhedron cannot exist.

These are really delightful problems and among my all-time favorite – if you haven’t seen them before, I highly recommend thinking about them before the jump. =)

Here are the solutions:

- Take a table and drill holes so they form vertices of ABC. Now, take three strings and attach one end from all three at a single point X. At the other end of each of the three strings put a weight of mass . Put the weights through the holes. This system will seek to minimize potential energy, which happens when AX + BX + CX is minimized (if you don’t see this immediately, consider the potential energy is just proportional to the negative of the sum of the three taut parts of the string under the table). However, this means the three forces from the three weights must balance at X. Since the forces have the same tension (all ), the forces must be at degrees from each other.
- It doesn’t matter what is – the answer is always . This nonintuitive answer comes easily with some optical thinking: draw a tangent line to one of the tangent points and fire a ray of light from the chosen focus to . Note that because we have an ellipse, the reflection must go through the other focus. However, because the line is also tangent to the circle, the reflection must also go through the center of the circle, which is the focus we started with! Since the reflection must go through both foci, the tangent must have been orthogonal to the major axis, meaning we only had contact point.
- This polyhedron cannot exist. Suppose we put it on a flat table with bottom face . Since by construction the normal line hits outside of , the polyhedron must roll onto another face (yes, or edge, but we can spent basically zero energy to push it onto a face using that edge instead, so the argument still works), but then the same thing happens and we get perpetual motion. As we recall, perpetual motion cannot exist in our world =)

The first problem is fairly canonical – it has been around for a while and I might have seen it first in Titu Andresscu and Razvan Gelca’s “Mathematical Olympiad Challenges” back in the Olympiad days. The second problem met me in middle school – it became one of my favorite problems ever to appear in a contest after I found it one day practicing old ARML/NYCIML problems (sadly I cannot remember which contest this was from). I learned of the third problem from Tadashi Tokieda’s (seems like a really fun guy, by the way) article “Mechanical Ideas in Geometry” in an old *Monthly*.

The prompt for this post is that I recently rediscovered the first problem in Mark Levi’s gorgeous “The Mathematical Mechanic.” Though I was actually a bit sad since I’ve been working on a collection much like his to share them in blog/book form at some point, I think it is really a great thing for the world that he put these problems out there in organized form, so everyone should go read that book at some point and I’ll be happy vicariously through him. Besides, the second and third problems I gave are not in that book, so I can still share them here with you.

Have a nice summer, everyone,

-Yan

Another nice one: on each face of a polyhedron, construct a vector perpendicular to the face with length proportional to the area of the face. Prove that the vectors sum to zero.

The polygonal version of this is pretty easy. This polyhedral version corresponds to the fact that a polyhedron filled with air and currently at rest will stay at rest, with regards to the air pushing outward on each face.

By:

A. Rexon June 24, 2010at 2:27 PM

that’s def. a nice one. I may have used it if I didn’t already have a problem about polyhedra.

By:

yanzhangon June 24, 2010at 3:40 PM

Yan, we already discussed this, but here’s an alternate solution to the third question:

Suppose such a polyhedron and point exists. Among the altitudes from the point to the various faces, choose the shortest. Since the point is inside the polyhedron and the foot of the altitude is outside, the altitude crosses some facet of the polyhedron. But then the altitude to this face is even shorter, which is a contradiction.

This solution (which is fundamentally equivalent to yours; we can think here of what happens to the potential energy of the arrangement as the polyhedron falls from face to face) is strongly reminiscent of the “usual” solution of Sylvester’s Theorem. (I don’t know if HTML works here; if not, can you edit to make this look nice?) This raises the question of whether there is a “physical” proof of Sylvester’s Theorem. Any takers?

By:

JBLon June 26, 2010at 3:17 AM

Here is yet another physical proof of the fist problem, which a physicist friend of mine came up with today:

Put 3 vertical poles at the points A,B and C and take another nonstationed vertical pole for the point X. Take now a rope, tie it at, say, A, pass it by X, then around B, then by X again, then around C, then by X again and finally around A. Pull the rope tight at A. Since the length of the rope inside ABC is 2(AX +BX +CX) and since it’s tight, this length is minimal and X is at the desired position. Now again as in the original proof, tension forces at X are equal in magnitude and sum to zero.

And a (not really)physical proof of the polyhedron problem: expand a spherical balloon centered at X inside the polyhedron until it touches the polyhedron. That point of contact is clearly inside and is the foot of the perpendicular radius from X to the corresponding face.

By:

Gretaon June 27, 2010at 2:18 AM

Wow, I really like the balloon thing

By:

yanzhangon June 27, 2010at 5:24 AM

Tanya Khovanova points me to the following, related to the third problem: http://mathworld.wolfram.com/UnistablePolyhedron.html

She remarks, “I believe that Conway suggested an idea how to make it with 18 faces” (rather than the 19 shown in the linked page).

By:

JBLon June 29, 2010at 1:17 AM

The first problem does not apply for obtuse triangles.

By:

yanon August 27, 2010at 12:42 AM

The relevant condition is actually that the triangle contain no angle larger than 120 degrees, not that the triangle be acute.

By:

JBLon August 27, 2010at 2:36 AM

You can fill a hollow polyhedron with sand, or float it in the ocean, and it will still reach an equilibrium. I wonder if this can be made into a puzzle.

By:

Tkon October 11, 2010at 12:47 AM

I’m a bit confused by the third one. I could see the solution given being correct if it were specified that the point selected must be the centroid (and indeed, given such information, the solution given would be the first thing to pop into my head), but how can we get from “the existence of a point such that a line normal to each face through it will not pass through the associated face” to “the polyhedron will tip over” when the point has no relationship to the polyhedron’s center of gravity?

By:

Davidon April 1, 2011at 8:26 AM

David, Yan seems to have left out a sentence from the explanation: you should take the entire mass of the polyhedron to be concentrated in the special point. (This makes it the center of gravity.)

By:

JBLon April 1, 2011at 2:00 PM