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Problem A. 556. (February 2012)

A. 556. Prove that for arbitrary real numbers a_1,\ldots,a_n there exist a real t such that

\sum_{i=1}^n \big|\sin(t-a_i)\big| \le \ctg\frac\pi{2n}.

(5 pont)

Deadline expired on March 12, 2012.

Solution. Define the function f(x) = \sum_{i=1}^n \big|\sin(x-a_i)\big| and let M = \min\big( f(a_1), \ldots, f(a_n) \big). We will show that M \le \ctg\frac\pi{2n}. Then it follows that at least one of the choices t=a1, ..., t=an proves the statement.


The role of a1,...,an is symmetric and the function |sin x| is periodic by \pi, so without loss of generality we may assume 0<a_1\le a_2\le\ldots\le a_n=\pi. Define a0=0 too; then f(a0)=f(an).

If a_1=\ldots=a_n=\pi, then f(a_1)=\ldots=f(a_n)=0, and the statement is trivial. In the rest of the solution we assume a1<\pi as well; then 0\le a_1-a_0,a_2-a_1,\ldots,a_n-a_{n-1}<\pi.

By the periodicity of |sin x|,

\int_0^\pi f
= \sum_{i=1}^n \int_0^\pi \big|\sin(x-a_i)\big| \,\mathrm{d}x
= \sum_{i=1}^n \int_0^\pi \big|\sin x\big| \,\mathrm{d}x
= n \cdot 2 = 2n. (1)

Now we prove that

\int_{a_{k-1}}^{a_k} f = \big(f(a_{k-1})+f(a_k)\big) \cdot \tg\frac{a_k-a_{k-1}}2

for all 1\lek\len.

We prove (2) termwise. For each index 1\lei\len,

  \int_{a_{k-1}}^{a_k} \sin(x-a_i) \,\mathrm{d}x
  = \cos(a_{k-1}-a_i) - \cos(a_k-a_i)
  = 2 \sin\frac{(a_{k-1}-a_i)+(a_k-a_i)}2 \sin\frac{a_k-a_{k-1}}2 =

  = \bigg(2 \sin\Big(\frac{a_{k-1}+a_k}2-a_i\Big)
  \cos\frac{a_k-a_{k-1}}2\bigg) \tg\frac{a_k-a_{k-1}}2
  = \Big( \sin(a_{k-1}-a_i) + \sin(a_k-a_i) \Big) \tg\frac{a_k-a_{k-1}}2.

In the interval [ak-1,ak] the function sin (x-ai) has constant sign: it is nonnegative for i\lek-1, and nonpositive for i\gek. Multiplying by (-1) for i<k and summing up we obtain (2).

Combining (1) and (2), and applying Jensen's inequality to the tangent function (which is convex in [0,\pi/2), we get

  2n = \int_0^\pi f = \sum_{k=1}^n \int_{a_{k-1}}^{a_k} f
  = \sum_{k=1}^n \big(f(a_{k-1})+f(a_k)\big) \cdot \tg\frac{a_k-a_{k-1}}2 \ge

  \ge 2M \cdot \sum_{k=1}^n \tg\frac{a_k-a_{k-1}}2
  \ge 2M \cdot n \tg\bigg( \frac1n \sum_{k=1}^n \frac{a_k-a_{k-1}}2 \bigg)
  = 2nM \cdot \tg \frac\pi{2n},

  M \le \ctg \frac\pi{2n}.


6 students sent a solution.
5 points:Gyarmati Máté, Janzer Olivér, Omer Cerrahoglu.
4 points:Machó Bónis.
0 point:2 students.

Problems in Mathematics of KöMaL, February 2012