Can You Solve the Origin-Inclusion Convex Polygon Problem?

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  • Thread starter Ackbach
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    2016
In summary, the POTW can always be solved, but it may require different strategies. To check if your solution is correct, compare it to the answer key provided by the organizer. If you are having difficulty, seek help from others. There can be multiple solutions to the POTW, as long as they meet the requirements. The use of outside resources may or may not be allowed, depending on the rules set by the organizer.
  • #1
Ackbach
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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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  • #2
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.
 

Related to Can You Solve the Origin-Inclusion Convex Polygon Problem?

1. Can the POTW be solved?

Yes, every problem of the week can be solved. It may require different approaches or strategies, but there is always a solution.

2. How do I know if my solution is correct?

Once you have solved the problem, you can compare your solution to the answer key provided by the POTW organizer. If your solution matches the correct answer, then it is considered correct.

3. What if I can't solve the problem?

If you are having difficulty solving the problem, you can seek help from fellow scientists, online forums, or consult with your mentor or professor. Remember, there is no shame in asking for help.

4. Is there only one solution to the POTW?

No, there can be multiple ways to solve the POTW. As long as your solution is valid and meets the requirements of the problem, it can be considered correct.

5. Can I use outside resources to solve the POTW?

It depends on the rules set by the POTW organizer. Some may allow the use of outside resources, while others may require you to solve the problem independently. Make sure to check the guidelines before attempting to use outside resources.

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