Wolfram Blog
Oleksandr Pavlyk

Our First Russian Student Competition

June 5, 2007 — Oleksandr Pavlyk, Kernel Technology

In April, we invited high school and college students in Russia to participate in a Mathematica competition. We gave the students a week to answer a set of seven Mathematica questions. The response was great, with submissions coming in from all across the country.

The first-place winner was Vladimir Dudchenko, an undergraduate student at the Moscow Institute of Physics and Technology (MIPT). He correctly solved all seven competition problems and displayed remarkable ingenuity and skill in his use of Mathematica. In addition to a student copy of Mathematica 6, Vladimir won a new MacBook Pro, a top-of-the-line machine donated by Apple and DPI Computers (Apple’s partner in Russia).


The runner-up winners were Yulia Kalugina, Konstantin Kanishev, Il’ja Lysenkov, and Aleksey Valabuev. Each won a student copy of Mathematica and other prizes from Wolfram Research.

And a special congratulation goes to 14-year-old Askar Safin. Askar was the youngest participant, and he correctly solved two of the problems–a significant achievement for a person of his age.

One of the problems, which turned out to be the most challenging, asked competitors to compute an inertia tensor of the spikey polyhedron, a cumulated icosahedron. You can now visualize the spikey polyhedron (which Michael Trott discussed in a recent blog post) instantly using the new PolyhedronData function in Mathematica.

PolyhedronData[

Spikey Polyhedron

Vladimir’s first-place solution is given here.

We thought the problem’s most elegant solution was the one submitted by Aleksey Valabuev, a student at MIPT. He remarked that the spikey’s inertia tensor with respect to its center of mass must be spherical, i.e. I==I1==I2==I3. Aleksey further noted that
I==(1/3) (I1 + I2 + I3) == (2/3) (1/Vol) Integral[(x^2 + y^2 + z^2)] dV
is manifestly invariant under rotations. He pointed out that it suffices to integrate over a tetrahedron with one vertex at the spikey’s center of mass and a base being one of the spikey’s faces, instead of integrating over the whole spikey. This observation follows because the spikey’s symmetry group acting on this tetrahedron would generate the whole spikey!

In order to carry out the integration over a tetrahedron with base vertices of coordinates {a, b, c}, he represented the point inside the tetrahedron as

r == at + bs + cu

for

0 <= t <= 1
0 <= s <= 1 - t
0 <= u <= 1 - t - s

{a, b, c} = Normal[PolyhedronData[

Of course, Mathematica now lets one look up the inertia tensor of a spikey immediately using PolyhedronData:

PolyhedronData[

{{1/180 (35 + 9 Sqrt[5] + 4 Sqrt[2] (3 + Sqrt[5])), 0, 0}, {0, 1/180 (35 + 9 Sqrt[5] + 4 Sqrt[2] (3 + Sqrt[5])), 0}, {0, 0, 1/180 (35 + 9 Sqrt[5] + 4 Sqrt[2] (3 +Sqrt[5]))}}

We were impressed by the variety of answers we received and congratulate everyone who participated. We’d like to hold other student competitions soon… but more on that later.

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