We are still waiting for a good solution for Problem 2014-15.

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We are still waiting for a good solution for Problem 2014-15.

For a (simple) graph \(G\), let \(o(G)\) be the number of odd-sized sets of pairwise non-adjacent vertices and let \(e(G)\) be the number of even-sized sets of pairwise non-adjacent vertices. Prove that if we can delete \(k\) vertices from \(G\) to destroy every cycle, then \[ | o(G)-e(G)|\le 2^{k}.\]

The best solution was submitted by Minjae Park (박민재, 수리과학과 2011학번). Congratulations!

Here is his solution.

An alternative solution was submitted by 김경석 (+3, 경기과학고 3학년). One incorrect solution was received (BHJ).

Let \[p(x)=x^n+x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_1 x_1 + a_0\] be a polynomial. Prove that if \(p(z)=0\) for a complex number \(z\), then \[ |z| \le 1+ \sqrt{\max (|a_0|,|a_1|,|a_2|,\ldots,|a_{n-2}|)}.\]

For a (simple) graph \(G\), let \(o(G)\) be the number of odd-sized sets of pairwise non-adjacent vertices and let \(e(G)\) be the number of even-sized sets of pairwise non-adjacent vertices. Prove that if we can delete \(k\) vertices from \(G\) to destroy every cycle, then \[ | o(G)-e(G)|\le 2^{k}.\]

Let \(\theta\) be a fixed constant. Characterize all functions \(f:\mathcal R\to \mathcal R\) such that \(f”(x)\) exists for all real \(x\) and for all real \(x,y\), \[ f(y)=f(x)+(y-x)f'(x)+ \frac{(y-x)^2}{2} f”(\theta y + (1-\theta) x).\]

Prove or disprove that for all positive integers \(m\) and \(n\), \[ f(m,n)=\frac{2^{3(m+n)-\frac12} }{{\pi}} \int_0^{\pi/2} \sin^{ 2n – \frac12 }\theta \cdot \cos^{2m+\frac12}\theta \, d\theta\] is an integer.

The best solution was submitted by 김경석 (경기과학고등학교 3학년). Congratulations!

Here is his solution.

Alternative solutions were submitted by 이병학 (2013학번, +2), 박훈민 (2013학번, +2), 배형진 (공항중학교 3학년, +2). One incorrect solution was submitted (LSC).

Prove or disprove that for all positive integers \(m\) and \(n\), \[ f(m,n)=\frac{2^{3(m+n)-\frac12} }{{\pi}} \int_0^{\pi/2} \sin^{ 2n – \frac12 }\theta \cdot \cos^{2m+\frac12}\theta \, d\theta\] is an integer.

(A typo is fixed on Saturday.)