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Difference between revisions of "1991 OIM Problems/Problem 3"
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Then <math>f(18/1991)=f(19/3^7)=f(20/3^7)=5/2^7</math>
 
Then <math>f(18/1991)=f(19/3^7)=f(20/3^7)=5/2^7</math>
  
−
* Note.  I actually competed at this event when I was in High School representing Puerto Rico.  While I can solve it now, at the event I was not able to solve it and only got partial points.
+
* Note.  I actually competed at this event in Argentina when I was in High School representing Puerto Rico.  While I can solve it now, at the event I was not able to solve it and only got partial points.
  
 
~Tomas Diaz. orders@tomasdiaz.com
 
~Tomas Diaz. orders@tomasdiaz.com

Revision as of 21:50, 13 December 2023

Problem

Let $f$ be an increasing function defined for every real number $x$, $0 \le x \le 1$, such that:

a. $f(0) = 0$

b. $f(x/3) = f(x)/2$

c. $f(1-x) = 1 - f(x)$

Find $f(18/1991)$

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

From property c: $f(1)=1-f(0)=1$

From property c: $f(1/2)=f(1-1/2)=1-f(1/2)$ which means that $f(1/2)=1/2$

We don really need to calculate $f(1/2)$ to solve this problem but it's good to have it so that we can see what happens in the range where $\frac{1}{3} \le x \le \frac{2}{3}$

From property b: $f(1/3) = f(1)/2=1/2$

From property c: $f(2/3) = 1-f(1/3)=1-1/2=1/2$

Here it is good to note the different between an increasing function an a strictly increasing function. In a strictly increasing function the function needs to be strictly increasing on all intervals. However, in an increasing function an interval that has the same constant value is allowed. In other words, it will not decrease in that interval.

Therefore, since in the interval $1/3 \le x \le 2/3$ the function can neither increase nor decrease and $f(1/3)=f(2/3)=1/2$, then in the interval $1/3 \le x \le 2/3$, $f(x)=1/2$

From property b: $f(1/9)=\frac{f(1/3)}{2}=1/4$ and $f(2/9)=\frac{f(2/3)}{2}=1/4$

From property c: $f(8/9)=1-f(1/9)=1-1/4=3/4$ and $f(7/9)=1-f(2/9)=1-1/4=3/4$

From property b: $f(7/3^3)=\frac{f(7/3^2)}{2}=3/8$ and $f(8/3^3)=\frac{f(8/3^2)}{2}=3/8$

From property c: $f(19/3^3)=1-f(8/3^2)=1-3/8=5/2^3$ and $f(20/3^3)=1-f(7/3^2)=1-3/8=5/2^3$

From property b: $f(19/3^4)=\frac{f(19/3^3)}{2}=5/2^4$ and $f(20/3^4)=\frac{f(20/3^3)}{2}=5/2^4$

From property b: $f(19/3^5)=\frac{f(19/3^4)}{2}=5/2^5$ and $f(20/3^5)=\frac{f(20/3^4)}{2}=5/2^5$

From property b: $f(19/3^6)=\frac{f(19/3^5)}{2}=5/2^6$ and $f(20/3^6)=\frac{f(20/3^5)}{2}=5/2^6$

From property b: $f(19/3^7)=\frac{f(19/3^6)}{2}=5/2^7$ and $f(20/3^7)=\frac{f(20/3^6)}{2}=5/2^7$

Since $\frac{19}{3^7} < \frac{18}{1991} < \frac{20}{3^7}$ and $f(19/3^7)=f(20/3^7)$,

Then $f(18/1991)=f(19/3^7)=f(20/3^7)=5/2^7$

  • Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. While I can solve it now, at the event I was not able to solve it and only got partial points.

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

https://www.oma.org.ar/enunciados/ibe6.htm

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