IMO 1983

Key idea: $x f(x)$ is always a fixed point, and the set of fixed points is closed under squaring and taking reciprocals. Decay at infinity then allows only the fixed point $1$.

Let $c=f(1)$.

• $x=1$: $f(f(y))=cy$ for all $y$.
• $x=y=1$: $f(c)=c$. Then $f(f(c))=f(c)=c$, but also $f(f(c))=c\cdot c$. So $c^2=c$ and $c=1$.

Fixed points. With $y=x$: $f(xf(x))=xf(x)$, so $u=xf(x)$ is a fixed point for every $x$. If $u$ is a fixed point, then:

• $f(u^2)=f(u\,f(u))=u\,f(u)=u^2$, so $u^{2^k}$ is a fixed point for every $k$;
• with $x=\frac1u$, $y=u$: $f\left(\tfrac1u\cdot u\right)=u\,f\left(\tfrac1u\right)$, so $f\left(\tfrac1u\right)=\tfrac1u$ is a fixed point.

If some fixed point $u\ne1$ existed, then one of $u,\frac1u$ would be a fixed point $v>1$, and $f\left(v^{2^k}\right)=v^{2^k}\to\infty$, against $f(x)\to0$. So the only fixed point is $1$, and $xf(x)=1$ for all $x$.

Check. $f(x)=\frac1x$ gives $f\left(\frac xy\right)=\frac yx=y\,f(x)$, and $\frac1x\to0$.

Key idea: with $S$ the meeting point of the common tangents, $SA^2=SM_1\cdot SO_2=SM_2\cdot SO_1$. That gives two pairs of similar triangles, which move each angle at $A$ to an angle at a center.

If the radii are equal, the common tangents are parallel to $O_1O_2$, so $M_1=O_1$ and $M_2=O_2$, and the two angles are the same angle. From now on let the radii be $r_1<r_2$, and let $k=\frac{r_2}{r_1}$. The two common tangents of intersecting circles meet at the external center of similitude $S$, on line $O_1O_2$, beyond $O_1$. The homothety $h$ with center $S$ and ratio $k$ maps $C_1$ to $C_2$, $P_1\to P_2$ and $O_1\to O_2$.

Where $M_1$, $M_2$ are. The chord $P_1Q_1$ is perpendicular to $SO_1$ (by symmetry in the line of centers), so its midpoint $M_1$ is on segment $SO_1$, and in the right triangle $SP_1O_1$ the foot of the altitude gives $SM_1\cdot SO_1=SP_1^2$. Likewise $SM_2\cdot SO_2=SP_2^2$. With $SP_2=k\,SP_1$ and $SO_2=k\,SO_1$: $$SM_1\cdot SO_2=k\,SP_1^2,\qquad SM_2\cdot SO_1=\frac{k^2SP_1^2}{k}=k\,SP_1^2 .$$

$SA^2=k\,SP_1^2$. The image $h(A)$ lies on $C_2$ and on line $SA$, and $h(A)\ne A$ because $k\ne1$. So line $SA$ meets $C_2$ at $A$ and $h(A)$, and the power of $S$ with respect to $C_2$ is $$SA\cdot S\,h(A)=k\,SA^2=SP_2^2=k^2SP_1^2 .$$ So $SA^2=k\,SP_1^2=SM_1\cdot SO_2=SM_2\cdot SO_1$.

Similar triangles. From $\frac{SA}{SM_1}=\frac{SO_2}{SA}$ and the common angle at $S$, triangles $SAM_1$ and $SO_2A$ are similar, so $\angle SAM_1=\angle SO_2A$. In the same way $\angle SAM_2=\angle SO_1A$.

Finish. The points $S,O_1,O_2$ lie in this order on a line, so in triangle $AO_1O_2$ the exterior angle at $O_1$ gives $\angle SO_1A=\angle O_1AO_2+\angle SO_2A$. Hence $\angle SAM_2-\angle SAM_1=\angle O_1AO_2>0$. Both $M_1$ and $M_2$ lie on the ray from $S$ through $O_1$, so the rays $AM_1$ and $AM_2$ are on the same side of $AS$, and $$\angle M_1AM_2=\angle SAM_2-\angle SAM_1=\angle O_1AO_2 .$$

Key idea: modulo $a$, only the term $xbc$ matters, so remainders pin down $x$ and $y$; the size bound decides whether $z\ge0$.

Let $N=2abc-ab-bc-ca$.

$N$ is not representable. Suppose $xbc+yca+zab=N$ with $x,y,z\ge0$. Modulo $a$: $xbc\equiv-bc$, and $bc$ is prime to $a$, so $x\equiv-1\pmod a$, hence $x+1\ge a$. Likewise $y+1\ge b$ and $z+1\ge c$. But the equation says $$(x+1)bc+(y+1)ca+(z+1)ab=2abc,$$ while the left side is at least $abc+abc+abc=3abc$. Contradiction.

Every $n>N$ is representable. Choose $x\in\{0,\dots,a-1\}$ with $xbc\equiv n\pmod a$, and $y\in\{0,\dots,b-1\}$ with $yca\equiv n\pmod b$ (possible since $bc$ is prime to $a$ and $ca$ to $b$). Then $m=n-xbc-yca$ is divisible by $a$ (as $yca\equiv0$) and by $b$ (as $xbc\equiv0$), so by $ab$: $m=zab$. And $$m\ge n-(a-1)bc-(b-1)ca>N-(a-1)bc-(b-1)ca=-ab ,$$ so $z>-1$, that is, $z\ge0$. Hence $n=xbc+yca+zab$.

Key idea: the points at one third of each side form a triangle whose sides are perpendicular to the sides of $ABC$. Two of them share a set, and then any point of the matching side either completes a right triangle or the other set fills that side and gets one.

Let the side be $3$. Take $P$ on $BC$ with $BP=1$, $Q$ on $CA$ with $CQ=1$, $R$ on $AB$ with $AR=1$. In triangle $BPR$, $BP=1$, $BR=2$ and $\angle B=60^\circ$, so $PR^2=1+4-2=3$ and $BP^2+PR^2=BR^2$: the angle at $P$ is right, so $RP\perp BC$. In the same way $PQ\perp CA$ and $QR\perp AB$.

Two of $P,Q,R$ are in the same set; by the symmetry of the figure under turning, say $P,R\in S_1$.

1. If some $X\ne P$ on side $BC$ is in $S_1$, then $R,P,X$ is a right triangle in $S_1$ (right angle at $P$). Done.
2. Otherwise all of $BC$ except $P$ is in $S_2$.
   • If $Q\in S_2$: its foot $F$ on $BC$ has $CF=\frac12\ne2=CP$, so $F\in S_2$, and $Q,F,B$ is a right triangle in $S_2$.
   • If $Q\in S_1$: then $Q,R\in S_1$ and $QR\perp AB$. If some $Y\ne R$ on $AB$ is in $S_1$, then $Q,R,Y$ is a right triangle in $S_1$. Otherwise all of $AB$ except $R$ is in $S_2$, including $A$. The midpoint $D$ of $BC$ is not $P$, so $D\in S_2$, and $C\in S_2$. Then $A,D,C$ is a right triangle in $S_2$ (right angle at $D$).

In every case one set contains the vertices of a right triangle.

Key idea: numbers whose base-$3$ digits are only $0$ and $1$ add without carries, so $x+z=2y$ forces $x=z$.

Let $T$ be the set of numbers whose base-$3$ representation, with $11$ digits, uses only $0$ and $1$. It has $2^{11}=2048$ elements, from $0$ up to $1+3+\dots+3^{10}=\frac{3^{11}-1}2=88573$.

No three in progression. Suppose $x,y,z\in T$ with $x+z=2y$. In base $3$, digit by digit, $x_i+z_i\le2$, so adding $x$ and $z$ involves no carries, and the digits of $x+z$ are $x_i+z_i$. The digits of $2y$ are $2y_i\in\{0,2\}$, also without carries. So $x_i+z_i=2y_i$ for every $i$, which with $x_i,z_i\in\{0,1\}$ forces $x_i=z_i=y_i$. So $x=y=z$.

The answer. Add $1$ to every element of $T$; this keeps the property (a progression shifted is still a progression). Now all $2048$ numbers are positive and at most $88574\le10^5$. Any $1983$ of them work. So the answer is yes.

Key idea: the substitution from the touch points of the incircle turns the inequality into $\frac{y^2}{z}+\frac{z^2}{x}+\frac{x^2}{y}\ge x+y+z$, which is AM–GM term by term.

For a triangle there are positive $x,y,z$ (the tangent lengths from the vertices to the incircle) with $a=y+z$, $b=z+x$, $c=x+y$. Expanding (a routine check), $$a^2b(a-b)+b^2c(b-c)+c^2a(c-a)=2\left(xy^3+yz^3+zx^3-xyz(x+y+z)\right).$$ Divide by $2\,x\,y\,z>0$. We need $$\frac{y^2}{z}+\frac{z^2}{x}+\frac{x^2}{y}\ge x+y+z .$$ By AM–GM, $\frac{y^2}z+z\ge2y$, $\frac{z^2}x+x\ge2z$ and $\frac{x^2}y+y\ge2x$. Adding these gives the claim.

Equality needs $y=z$, $z=x$ and $x=y$, so $x=y=z$, which means $a=b=c$. And for an equilateral triangle every term is $0$.