Difference between revisions of "2018 USAMO Problems/Problem 2"
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==Problem 2== | ==Problem 2== | ||
− | Find all functions <math>f:(0,\infty) \ | + | Find all functions <math>f:(0,\infty) \to (0,\infty)</math> such that |
<cmath>f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right) + f\left(z+\frac{1}{x}\right) = 1</cmath> for all <math>x,y,z >0</math> with <math>xyz =1.</math> | <cmath>f\left(x+\frac{1}{y}\right)+f\left(y+\frac{1}{z}\right) + f\left(z+\frac{1}{x}\right) = 1</cmath> for all <math>x,y,z >0</math> with <math>xyz =1.</math> | ||
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==Solution== | ==Solution== | ||
− | + | Obviously, the output of <math>f</math> lies in the interval <math>(0,1)</math>. Define <math>g:(0,1)\to(0,1)</math> as <math>g(x)=f\left(\frac1x-1\right)</math>. Then for any <math>a,b,c\in(0,1)</math> such that <math>a+b+c=1</math>, we have <math>g(a)=f\left(\frac1a-1\right)=f\left(\frac{1-a}a\right)=f\left(\frac{b+c}a\right)</math>. We can transform <math>g(b)</math> and <math>g(c)</math> similarly: | |
− | + | <cmath>g(a)+g(b)+g(c)=f\left(\frac ca+\frac ba\right)+f\left(\frac ab+\frac cb\right)+f\left(\frac bc+\frac ac\right)</cmath> | |
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− | + | Let <math>x=\frac ca</math>, <math>y=\frac ab</math>, <math>z=\frac bc</math>. We can see that the above expression is equal to <math>1</math>. That is, for any <math>a,b,c\in(0,1)</math> such that <math>a+b+c=1</math>, <math>g(a)+g(b)+g(c)=1</math>. | |
− | + | (To motivate this, one can start by writing <math>x=\frac ab</math>, <math>y=\frac bc</math>, <math>z=\frac ca</math>, and normalizing such that <math>a+b+c=1</math>.) | |
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− | + | For convenience, we define <math>h:\left(-\frac13,\frac23\right)\to\left(-\frac13,\frac23\right)</math> as <math>h(x)=g\left(x+\frac13\right)-\frac13</math>, so that for any <math>a,b,c\in\left(-\frac13,\frac23\right)</math> such that <math>a+b+c=0</math>, we have | |
+ | |||
+ | <cmath>h(a)+h(b)+h(c)=g\left(a+\frac13\right)-\frac13+g\left(b+\frac13\right)-\frac13+g\left(c+\frac13\right)-\frac13=1-1=0.</cmath> | ||
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+ | Obviously, <math>h(0)=0</math>. If <math>|a|<\frac13</math>, then <math>h(a)+h(-a)+h(0)=0</math> and thus <math>h(-a)=-h(a)</math>. Furthermore, if <math>a,b</math> are in the domain and <math>|a+b|<\frac13</math>, then <math>h(a)+h(b)+h(-(a+b))=0</math> and thus <math>h(a+b)=h(a)+h(b)</math>. | ||
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+ | At this point, we should realize that <math>h</math> should be of the form <math>h(x)=kx</math>. We first prove this for some rational numbers. If <math>n</math> is a positive integer and <math>x</math> is a real number such that <math>|nx|<\frac13</math>, then we can repeatedly apply <math>h(a+b)=h(a)+h(b)</math> to obtain <math>h(nx)=nh(x)</math>. Let <math>k=6h\left(\frac16\right)</math>, then for any rational number <math>r=\frac pq\in\left(0,\frac13\right)</math> where <math>p,q</math> are positive integers, we have <math>h(r)=6p*h\left(\frac1{6q}\right)=\frac{6p}q*h\left(\frac16\right)=kr</math>. | ||
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+ | Next, we prove it for all real numbers in the interval <math>\left(0,\frac13\right)</math>. For the sake of contradiction, assume that there is some <math>x\in\left(0,\frac13\right)</math> such that <math>h(x)\ne kx</math>. Let <math>E=h(x)-kx</math>, then obviously <math>0<|E|<1</math>. The idea is to "amplify" this error until it becomes so big as to contradict the bounds on the output of <math>h</math>. Let <math>N=\left\lceil\frac1{|E|}\right\rceil</math>, so that <math>N\ge2</math> and <math>|NE|\ge1</math>. Pick any rational <math>r\in\left(\frac{N-1}Nx,x\right)</math>, so that <cmath>0<x-r<N(x-r)<x<\frac13.</cmath> | ||
+ | All numbers and sums are safely inside the bounds of <math>\left(-\frac13,\frac13\right)</math>. Thus | ||
+ | <cmath>h(N(x-r))=Nh(x-r)=N(h(x)+h(-r))=N(h(x)-h(r))=kN(x-r)+NE,</cmath> | ||
+ | but picking any rational number <math>s\in\left(N(x-r),\frac13\right)</math> gives us <math>|kN(x-r)|<|ks|</math>, and since <math>ks=h(s)\in\left(-\frac13,\frac23\right)</math>, we have <math>kN(x-r)\in\left(-\frac13,\frac23\right)</math> as well, but since <math>NE\ge1</math>, this means that <math>h(N(x-r))=kN(x-r)+NE\notin\left(-\frac13,\frac23\right)</math>, giving us the desired contradiction. | ||
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+ | We now know that <math>h(x)=kx</math> for all <math>0<x<\frac13</math>. Since <math>h(-x)=-h(x)</math> for <math>|x|<\frac13</math>, we obtain <math>h(x)=kx</math> for all <math>|x|<\frac13</math>. For <math>x\in\left(\frac13,\frac23\right)</math>, we have <math>h(x)+h\left(-\frac x2\right)+h\left(-\frac x2\right)=0</math>, and thus <math>h(x)=kx</math> as well. So <math>h(x)=kx</math> for all <math>x</math> in the domain. Since <math>h(x)</math> is bounded by <math>-\frac13</math> and <math>\frac23</math>, we have <math>-\frac12\le k\le1</math>. It remains to work backwards to find <math>f(x)</math>. | ||
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+ | <cmath>\begin{align*} | ||
+ | h(x) &= kx \\ | ||
+ | g(x) &= kx+\frac{1-k}3 \\ | ||
+ | f(x) &= \frac k{1+x}+\frac{1-k}3\quad\left(-\frac12\le k\le1\right). | ||
+ | \end{align*}</cmath> | ||
+ | - wzs26843545602 | ||
+ | |||
+ | {{USAMO newbox|year=2018|num-b=1|num-a=3}} |
Latest revision as of 09:40, 27 August 2023
Problem 2
Find all functions such that
for all with
Solution
Obviously, the output of lies in the interval . Define as . Then for any such that , we have . We can transform and similarly:
Let , , . We can see that the above expression is equal to . That is, for any such that , .
(To motivate this, one can start by writing , , , and normalizing such that .)
For convenience, we define as , so that for any such that , we have
Obviously, . If , then and thus . Furthermore, if are in the domain and , then and thus .
At this point, we should realize that should be of the form . We first prove this for some rational numbers. If is a positive integer and is a real number such that , then we can repeatedly apply to obtain . Let , then for any rational number where are positive integers, we have .
Next, we prove it for all real numbers in the interval . For the sake of contradiction, assume that there is some such that . Let , then obviously . The idea is to "amplify" this error until it becomes so big as to contradict the bounds on the output of . Let , so that and . Pick any rational , so that All numbers and sums are safely inside the bounds of . Thus but picking any rational number gives us , and since , we have as well, but since , this means that , giving us the desired contradiction.
We now know that for all . Since for , we obtain for all . For , we have , and thus as well. So for all in the domain. Since is bounded by and , we have . It remains to work backwards to find .
- wzs26843545602
2018 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |