This problem is from Korea Winter Program Practice Test 1 Problem 1(i think), and I was somehow asked to post this. All solutions are welcome

wlog f(1)=1 f(2*3^0.5n)=2*3^0.5f(n) (perfect triangle) f(kx)=kf(x) (12 12 k makes radius the incircle 1 k is smaller one) lim f(x) = 0 x->0+ f is continuos (x x+dh root((x+dh)^2-x^2)forms triangle) r<k1^(an)*k2^(bn) an bn is integer sequence Dirichlet approximation (k1=2*3^0.5 k2=k) lim f(k1^(an)*k2^(bn))=f(r)=r n->oo f(x)=f(1)x (sorry for my bad english)

seojinkim wrote: f(kx)=kf(x) (12 12 k makes radius the incircle 1 k is smaller one) Sorry but I dont understand this step ....

I mean we can make triangle which length is 12 ,12 , k and radius of incircle is 1 then 12=(2*3^0.5)^2 so 12 12 f(k) must form triangle which radius of incircle is 1 there are two roots of k and if we choose the smaller one we can get f(k)=k(we can proove f(x)<2*3^0.5x and smaller root is almost 0 and larger root is almost 24)

But then you just proved $f(k)=k$, and not what you claimed $f(kx)=kf(x)$

pco wrote: But then you just proved $f(k)=k$, and not what you claimed $f(kx)=kf(x)$ Oh I made a mistake ! Think of 12x, 12x, kx which radius of incircle is x and do the same thing...

@above What about $f(kx)=k'f(x)$, where $k'$ is the other value such that the triangle with sides $12x,12x,k'x$ have a incircle with radius $x$? P.S. I think just a little modification to the former solution can make it correct. Note that $2a>b\Longrightarrow 2f(a)>f(b)$, so one can prove that $(1-\epsilon) x<\frac{f(x)}{f(1)}<(1+\epsilon )x$ using this and $f(2\sqrt{3}x)=2\sqrt{3}f(x)$

First, we can WLOG scale $f$ so that $f(1)=1$. Easy to see that $f(2\sqrt{3}x)=2\sqrt{3}f(x)$ for all positive real $x$. Also, we have $f(x)<2f(1)$ for all $1<x<2$. This means $\lim_{x\to 0}{f(x)}$ exists and is equal to $0$. For any positive reals $a,b$, consider the sequence of positive reals $x_1,x_2,\dotsc $ that all terms are less than $a+b$ and $\lim_{n\to \infty}{x_n}=a+b$, we get that There exists subsequence $y_1,y_2,\dotsc $ that $\lim_{n\to \infty}{f(y_n)}$ exists and is equal to $|f(a)-f(b)|$, or There exists subsequence $z_1,z_2,\dotsc $ that $\lim_{n\to \infty}{f(z_n)}$ exists and is equal to $f(a)+f(b)$. Given positive reals $a,b,c$ that $a+b=2c$ and sequence of positive reals $x_1,x_2,\dotsc$ that all terms are less than $a+b$ and $\lim_{n\to \infty}{x_n}=2c$. We have two cases There exists subsequence $y_1,y_2,\dotsc $ that $\lim_{n\to \infty}{f(y_n)}$ exists and is equal to $|f(c)-f(c)|=0$. Note that there exists positive integer $k$ that $(2\sqrt{3})^k(a+b)>2$. Consider the triangle with side lengths $(2\sqrt{3})^ky_m,(2\sqrt{3})^ky_m,1$. But there not exists triangle with side lengths $2(\sqrt{3})^kf(y_m),2(\sqrt{3})^kf(y_m),1$. There exists subsequence $z_1,z_2,\dotsc $ that $\lim_{n\to \infty}{f(z_n)}$ exists and is equal to $2f(c)$. Note that $\lim_{n\to \infty}{z_n}=2c=a+b$. Since $2c>\max \{ a,b\}$, we get that $2f(c)>\max \{ f(a),f(b)\} > |f(a)-f(b)|$. So, $\lim_{n\to \infty}{f(z'_n)}$ can't equal to $|f(a)-f(b)|$ for any subsequence $z'_1,z'_2,\dotsc$ of $z_1,z_2,\dotsc$. Therefore, there exists subsequence $z_1',z_2',\dotsc$ that $\lim_{n\to \infty}{f(z'_n)}$ exists and is equal to $f(a)+f(b)$. In summary, the above two cases give $f(a)+f(b)=2f(c)=2f((a+b)/2)$. This gives $f(a)+f(b)=f(c)+f(d)$ for all positive reals $a,b,c,d$ that $a+b=c+d$. Not hard to deduce that $f$ must be linear. Checking with $f(1)=1$ and $f(2\sqrt{3}x)=2\sqrt{3}f(x)$ gives us $f(x)=x$. Hence, the answer is $f(x)=cx$ for any positive constant $c$.

Here is a slightly different solution. Let $S$ be the set of real numbers $x$ so that $xf(a)=f(ax)$ for all $a\in\mathbb{R}^+$. We see that $1\in S$, $x\in S\implies \frac{1}{x}\in S$, and $x,y\in S \implies xy\in S$. Lemma 1. If $x\in S$ and $x>1/2$, then $$\frac{1}{2}\sqrt{\frac{2x-1}{2x+1}}\in S.$$

We now show the following: Lemma 2. $S$ is dense in $\mathbb{R}^+$.

We now show that $f$ is nondecreasing.

Now let $x$ be an arbitrary positive real, and let $\{y_n\}$ be a sequence of elements of $S$ that approach $x$ from below and $\{z_n\}$ be a sequence of elements of $S$ that approach $x$ from above. We have $f(y_n)=y_nf(1)$ and $f(z_n)=z_nf(1)$, so, $$y_nf(1)\leq f(x)\leq z_nf(1)$$ for all $n$. The squeeze theorem gives that $$xf(1)=\lim_{n\to\infty} y_nf(1)=f(x)=\lim_{n\to\infty} z_nf(1)=xf(1),$$ so $f(x)=xf(1)$ for all $x$, finishing the proof.

pieater314159 wrote: Here is a slightly different solution. Let $S$ be the set of real numbers $x$ so that $xf(a)=f(ax)$ for all $a\in\mathbb{R}^+$. We see that $1\in S$, $x\in S\implies \frac{1}{x}\in S$, and $x,y\in S \implies xy\in S$. Lemma 1. If $x\in S$ and $x>1/2$, then $$\frac{1}{2}\sqrt{\frac{2x-1}{2x+1}}\in S.$$

We now show the following: Lemma 2. $S$ is dense in $\mathbb{R}^+$.

Here's a slightly different way to prove that $f$ is linear if $S$ is dense. Assume that $S$ is dense in $\mathbb{R}^{+}$, and consider some $t\in \mathbb{R}^{+}$ with $f(t)\neq f(1)\cdot t$. Case 1. If $f(t)>f(1)\cdot t$, then take some $u,v \in S$ with \[ \frac{f(t)}{f(1)}>u+v>t>u-v>0 \]gives $f(t)<f(u)+f(v)=(u+v)f(1)<f(t)$, a contradiction. Case 2. If $f(t)<f(1) \cdot t$, then take some $u,v \in S$ with \[ \frac{f(t)}{f(1)}<u-v<t<u+v \]gives $f(t)>f(u)-f(v)=f(1)(u-v)>f(t)$, a contradiction. Hence $f(t)=f(1) \cdot t$ for all $t \in \mathbb{R}^{+}$.

Maybe the same as above, but just gonna post it for fun I claim that the only solutions are $f(x) = kx$, $k \in \mathbb{R}$. It's quite immediate to see that such $f$ works, by similarity. Now, we'll prove that these are the only ones. Denote $S$ as the subgroup of $\mathbb{R}$, which is a set of constants $c \in \mathbb{R}^+$ such that \[ f(cx) = cf(x) \]Now, denote that $S$ is multiplicative, as if $a \in S$ and $B \in S$, then we must have \[ f(abx) = af(bx) = a(bf(x)) = ab f(x) \]Therefore, $ab \in S$. Moreover, if $a \in S$, then \[ f(x) = f \left( a \cdot \frac{1}{a} x \right) = a \cdot f \left( \frac{1}{a} x \right) \]This results \[ f \left( \frac{1}{a} x \right) = \frac{1}{a} \cdot f(x) \]We'll provide a lemma: $\textbf{Lemma 01.}$ If $x,y \in S$, such that $2x > y$, then we must have \[ \frac{1}{2} y \sqrt{ \frac{2x - y}{2x + y} } \in S \]as well. $\textit{Proof.}$ Consider a triangle of side $x, x, y$, now we have \[ r = \frac{\frac{1}{2} y \sqrt{x^2 - \frac{1}{4}y^2}}{\frac{2x + y}{2}} = \frac{1}{2} y \sqrt{\frac{2x - y}{2x + y}} \]Therefore, since $x,y \in S$, then we must have $r \in S$. $\textbf{Lemma 02.}$ $S$ is dense in $\mathbb{R}^+$. $\textit{Proof.}$ Notice that one can have $f(x) = \frac{1}{2} \sqrt{\frac{2x - 1}{2x + 1}} \in S$, as $1 \in S$. Now, notice that as $(2 \sqrt{3} )^n \in S$ for every $n \in S$. Then we can have a sequence of real numbers $x_1 < x_2 < x_3 < \dots < \frac{1}{2}$ by putting $x_i = (2\sqrt{3})^n$. Now consider $y_n = \frac{x_{n+1}}{x_n}$ approaches $1$ as $n$ tends to infinity, and each element is greater than $1$ by the way we construct $\{ x_i \}$. Now, we can extract numbers in $S$ in the form of $1 + \varepsilon$ where $\varepsilon$ is something arbitrarily small. Now, we consider three cases of interval $(a,b)$: 1. If $a < 1, b > 1$, then $1 \in S$ satisfy as $a < 1 < b$. 2. If $1 < a < b$, then we could take $\varepsilon$ sufficiently small and for some integer $n$ we can have $a < (1 + \varepsilon)^n < b$, and therefore we have found an element of $S$ in this interval. 3. If $ a < b < 1$, notice that since $1 < \frac{1}{b} < \frac{1}{a}$. By the second case, there exists a number $u \in S$ such that $\frac{1}{b} < u < \frac{1}{a}$. Since $u \in S$, then we must have $\frac{1}{u} \in S$ and this satisfy $a < \frac{1}{u} < b$. $\textbf{Lemma 03.}$ $f$ is linear over $\mathbb{R}^+$. $\textit{Proof.}$ We'll now prove that for any positive real number $c$, we must have $f(c) = cf(1)$. Fixed the value of $c$, and let $a,b \in S$ vary. Now we know by triangle inequality, that $c < a + b$, and in fact $f(c) < f(a) + f(b) = (a + b) f(1)$. Take $a + b = c + \varepsilon$ where $\varepsilon$ is something arbitrarily small. Therefore, by taking limit of $a+b$ that tends to $c$, we have \[ f(c) \le c f(1) \]Similarly, take $a,b \in S$ such that $a \le c \le b$ and $a + c > b$, this gives us $f(c) > (b - a)f(1)$. Take $(b - a) = c - \varepsilon$ where $\varepsilon$ is something really small. Therefore, by taking limit of $b - a$ that tends to $c$, we have \[ f(c) \ge cf(1) \]Therefore, we conclude that $f(c) = cf(1)$ for all $c \in \mathbb{R}^+$. Therefore, we have proved that $f(x) = kx$ for all $k \in \mathbb{R}^+$.

What the heck. The answer is $f(x) = kx$. Without loss of generality scale $f$ so that $f(1) = 1$; we will show that $f$ is the identity. We begin with the following geometric claim. Claim. For a triangle with side lengths $a$, $b$, $c$ and inradius $r$, we have $\min\{a,b,c\} > r/2$. Proof. Without loss of generality let $\min\{a,b,c\} = a$, and for sake of contradiction say $a \leq r/2$. Then note $90 + \tfrac12 \angle A = \angle BIC \leq 90^\circ$, or $\angle A \leq 0^\circ$, contradiction. $\square$ The claim implies $f(r) < f(a)/2$ for all $r < a/2$. Setting $a=b=c$ also gives us $f(2\sqrt{3} r) = 2\sqrt{3} f(r)$ for all $r$. Now note the sets

, so for any real number $r$ and $\varepsilon > 0$, there is an element $r_- \in S_-$ and $r_+ \in S_+$ such that $r_- - r > \varepsilon$ and $0 < r - r_+ < \varepsilon$. But by the properties, $r_+ < f(r) < r_-$. Thus, $f(r) = r$, as desired.

Weird problem, reminds of hamel arguments lol. Consider the set $\mathcal{S}$ of real numbers $x$ s.t. $xf(y) = f(yx)$ for all $y \in \mathcal{R}^{+}$, evidently $1 \in \mathcal{S}$, if $x \in \mathcal{S}$ $\iff \frac{1}{x} \in \mathcal{S}$, and $x,y \in \mathcal{S}$ implies $xy \in \mathcal{S}$ Now we consider the triangles of form $x,x,1$ and $x > \frac{1}{2}$, and $x \in \mathcal{S}$, this implies $\frac{1}{2} \cdot \sqrt{\frac{2x+1}{2x-1}} \in \mathcal{S}$, the key claim is as follows: Key Claim: $\mathcal{S}$ is dense in $\mathcal{R}^{+}$

. So now, consider some $t$ such that $f(t) \neq t \cdot f(1)$ If $f(t) > t \cdot f(1)$, then by using density of $\mathcal{S}$ choose $m,n$ s.t. $\frac{f(t)}{f(1)} > m+n > t > m-n >0$ but this gives $f(t) < f(m)+f(n) = (m+n) \cdot f(1) < f(t)$ similarly we get contradiction otherwise. Hence the only solution is $f(x) = cx$ for $c \in \mathcal{R}^{+}$