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

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The discussion revolves around the "Origin-Inclusion Convex Polygon Problem," which asks to prove that for a convex polygon containing the origin, there exist positive real numbers x and y such that a specific weighted sum of the polygon's vertices equals the origin. This problem was featured as Problem B-6 in the 1996 William Lowell Putnam Mathematical Competition. Despite its significance, no responses were provided for this week's Problem of the Week (POTW). The solution to the problem is credited to Kiran Kedlaya and his associates. The thread emphasizes the challenge and mathematical importance of the 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.