MHB Can You Solve the Origin-Inclusion Convex Polygon Problem?

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Here is this week's POTW:

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Let $(a_1, b_1), (a_2, b_2), \ldots, (a_n, b_n)$ be the vertices of a convex polygon which contains the origin in its interior. Prove that there exist positive real numbers $x$ and $y$ such that
$$
(a_1, b_1)x^{a_1} y^{b_1} + (a_2, b_2)x^{a_2}y^{b_2} + \cdots
+ (a_n, b_n)x^{a_n}y^{b_n} = (0,0).
$$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Re: Problem Of The Week # 228 - Aug 11, 2016

This was Problem B-6 in the 1996 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

We will prove the claim assuming only that the convex hull of the points $(a_{i}, b_{i})$ contains the origin in its interior. Let $u = \ln(x), v = \ln(y)$ so that the left-hand side of the given equation is
$$
(a_1, b_1) \exp(a_1 u + b_1 v) + (a_2, b_2) \exp(a_2 u + b_2 v) +
\cdots + (a_n, b_n) \exp(a_n u + b_n v).
$$
Now note that (1) is the gradient of the function
$$
f(u,v) = \exp(a_1 u + b_1 v) +\exp(a_2 u + b_2 v) + \cdots + exp(a_n u + b_n v),
$$
and so it suffices to show $f$ has a critical point. We will in fact show $f$ has a global minimum.

Clearly we have
$$
f(u,v) \geq \exp\left( \max_i (a_i u + b_i v) \right).
$$
Note that this maximum is positive for $(u,v) \neq (0,0)$: if we had $a_i u + b_i v < 0$ for all $i$, then the subset $ur + vs < 0$ of the $rs$-plane would be a half-plane containing all of the points $(a_i, b_i)$, whose convex hull would then not contain the origin, a contradiction.

The function $\max_{i} (a_{i}u + b_{i}v)$ is clearly continuous on the unit circle $u^{2} + v^{2} = 1$, which is compact. Hence it has a global minimum $M > 0$, and so for all $u,v$,
$$
\max_{i} (a_{i} u + b_{i} v) \geq M \sqrt{u^{2} + v^{2}}.
$$
In particular, $f \geq n+1$ on the disk of radius $\sqrt{(n+1)/M}$. Since $f(0,0) = n$, the infimum of $f$ is the same over the entire $uv$-plane as over this disk, which again is compact. Hence $f$ attains its infimal value at some point in the disk, which is the desired global minimum.
 
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