Difference between revisions of "1991 OIM Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | From | + | From property c: <math>f(1)=1-f(0)=1</math> |
From property c: <math>f(1/2)=f(1-1/2)=1-f(1/2)</math> which means that <math>f(1/2)=1/2</math> | From property c: <math>f(1/2)=f(1-1/2)=1-f(1/2)</math> which means that <math>f(1/2)=1/2</math> | ||
| + | |||
| + | We don really need to calculate <math>f(1/2)</math> to solve this problem but it's good to have it so that we can see what happens in the range where <math>\frac{1}{3} \le x \le \frac{2}{3}</math> | ||
From property b: <math>f(1/3) = f(1)/2=1/2</math> | From property b: <math>f(1/3) = f(1)/2=1/2</math> | ||
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From property c: <math>f(2/3) = 1-f(1/3)=1-1/2=1/2</math> | From property c: <math>f(2/3) = 1-f(1/3)=1-1/2=1/2</math> | ||
| − | Here 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. | + | 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 <math>1/3 \le x \le 2/3</math> the function can neither increase nor decrease and <math>f(1/3)=f(2/3)=1/2</math>, then in the interval <math>1/3 \le x \le 2/3</math>, <math>f(x)=1/2</math> | Therefore, since in the interval <math>1/3 \le x \le 2/3</math> the function can neither increase nor decrease and <math>f(1/3)=f(2/3)=1/2</math>, then in the interval <math>1/3 \le x \le 2/3</math>, <math>f(x)=1/2</math> | ||
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From property b: <math>f(7/3^3)=\frac{f(7/3^2)}{2}=3/8</math> and <math>f(8/3^3)=\frac{f(8/3^2)}{2}=3/8</math> | From property b: <math>f(7/3^3)=\frac{f(7/3^2)}{2}=3/8</math> and <math>f(8/3^3)=\frac{f(8/3^2)}{2}=3/8</math> | ||
| + | From property c: <math>f(19/3^3)=1-f(8/3^2)=1-3/8=5/2^3</math> and <math>f(20/3^3)=1-f(7/3^2)=1-3/8=5/2^3</math> | ||
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| + | From property b: <math>f(19/3^4)=\frac{f(19/3^3)}{2}=5/2^4</math> and <math>f(20/3^4)=\frac{f(20/3^3)}{2}=5/2^4</math> | ||
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| + | From property b: <math>f(19/3^5)=\frac{f(19/3^4)}{2}=5/2^5</math> and <math>f(20/3^5)=\frac{f(20/3^4)}{2}=5/2^5</math> | ||
| + | |||
| + | From property b: <math>f(19/3^6)=\frac{f(19/3^5)}{2}=5/2^6</math> and <math>f(20/3^6)=\frac{f(20/3^5)}{2}=5/2^6</math> | ||
| + | |||
| + | From property b: <math>f(19/3^7)=\frac{f(19/3^6)}{2}=5/2^7</math> and <math>f(20/3^7)=\frac{f(20/3^6)}{2}=5/2^7</math> | ||
| + | |||
| + | Since <math>\frac{19}{3^7} < \frac{18}{1991} < \frac{20}{3^7}</math> and <math>f(19/3^7)=f(20/3^7)</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 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 | ||
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== See also == | == See also == | ||
| + | [[OIM Problems and Solutions]] | ||
| + | |||
https://www.oma.org.ar/enunciados/ibe6.htm | https://www.oma.org.ar/enunciados/ibe6.htm | ||
Latest revision as of 08:40, 23 December 2023
Problem
Let 
be an increasing function defined for every real number 
, 
, such that:
a. 
b. 
c. 
Find 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
From property c: 
From property c: 
which means that 
We don really need to calculate 
to solve this problem but it's good to have it so that we can see what happens in the range where 
From property b: 
From property c: 
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 
the function can neither increase nor decrease and 
, then in the interval , 
From property b: 
and 
From property c: 
and 
From property b: 
and 
From property c: 
and 
From property b: 
and 
From property b: 
and 
From property b: 
and 
From property b: 
and 
Since 
and 
,
Then 
- 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.
