IMO 1982

Key idea: $f$ is superadditive, so $f(3n)\ge n$. The value at $9999$ forces equality up to $9999$, and superadditivity squeezes $f(1982)$ between $660$ and $660.67$.

The condition says $f(m+n)\ge f(m)+f(n)$, with excess at most $1$.

• $0=f(2)\ge2f(1)$, so $f(1)=0$.
• $f(3)\in\{f(2)+f(1),\,f(2)+f(1)+1\}=\{0,1\}$ and $f(3)>0$, so $f(3)=1$.
• $f(3(n+1))\ge f(3n)+f(3)$, so by induction $f(3n)\ge n$.

$f(3n)=n$ for $n\le3333$. If $f(3n)\ge n+1$ for some $n<3333$, then $$f(9999)\ge f(3n)+f(9999-3n)\ge(n+1)+(3333-n)=3334,$$ a contradiction. So $f(3n)=n$ for these $n$, in particular $f(5946)=1982$ and $f(1980)=660$.

Squeeze. From superadditivity, $f(5946)\ge3f(1982)$, so $f(1982)\le\frac{1982}3<661$. And $f(1982)\ge f(1980)+f(2)=660$. So $f(1982)=660$.

(Such a function exists: $f(n)=\lfloor n/3\rfloor$ satisfies every condition.)

Key idea: the reflected points lie on the incircle and form a triangle whose sides are parallel to those of the medial triangle. Two triangles with parallel sides and different sizes are related by a homothety, and its center is the common point.

Let $I$ and $r$ be the incenter and inradius. The incircle is symmetric in each bisector $A_iI$, so each $S_i$ is on the incircle. Describe points of the incircle by the direction angle of the ray from $I$: say $T_i$ has angle $\theta_i$.

Positions of the $S_i$. The bisector $A_1I$ bisects the angle $T_2IT_3$ (the two touch points on the sides from $A_1$ are mirror images in it), so it has direction $\frac{\theta_2+\theta_3}2$. Reflecting in it sends the angle $\theta$ to $\theta_2+\theta_3-\theta$. So $S_1$ has angle $\theta_2+\theta_3-\theta_1$, and likewise $S_2$ has $\theta_1+\theta_3-\theta_2$ and $S_3$ has $\theta_1+\theta_2-\theta_3$.

Parallel sides. A chord between angles $\psi$ and $\psi'$ is perpendicular to the direction $\frac{\psi+\psi'}2$. For $S_1S_2$ this is $\theta_3$: the chord $S_1S_2$ is perpendicular to $IT_3$, so it is parallel to side $A_1A_2$, which is tangent at $T_3$. (The points $S_1$, $S_2$ are different: otherwise $2\theta_1\equiv2\theta_2$, making $T_1$, $T_2$ equal or opposite, and two sides parallel.) The midline $M_1M_2$ is also parallel to $A_1A_2$. In the same way $S_2S_3\parallel M_2M_3$ and $S_3S_1\parallel M_3M_1$.

Homothety. Two triangles with corresponding sides parallel are related by a translation or by a homothety. The medial triangle has circumradius $\frac R2$, while $S_1S_2S_3$ has circumradius $r$. Euler’s inequality $R\ge2r$, with equality only for an equilateral triangle, gives $\frac R2>r$. So the map is not a translation but a homothety, and the lines joining corresponding vertices, $M_iS_i$, pass through its center. (Each $M_i\ne S_i$: the incircle meets side $a_i$ only at $T_i$, and $S_i=T_i$ would put $T_i$ on the bisector from $A_i$, making the triangle isosceles.)

Key idea: AM–GM gives a telescoping lower bound $4(x_0-x_n)$. Either $x_n$ gets small, or the terms themselves stay large.

Let $S_n=\sum_{i=0}^{n-1}\frac{x_i^2}{x_{i+1}}$.

(a) By AM–GM, $\frac{x_i^2}{x_{i+1}}+4x_{i+1}\ge2\sqrt{4x_i^2}=4x_i$. Summing for $i=0,\dots,n-1$, $$S_n\ge4\sum_{i=0}^{n-1}x_i-4\sum_{i=1}^{n}x_i=4(x_0-x_n)=4-4x_n .$$

• If some $x_n\le\frac1{4000}$, then $S_n\ge4-0.001=3.999$.
• Otherwise all $x_i>\frac1{4000}$, and since $x_{i+1}\le x_i$, each term $\frac{x_i^2}{x_{i+1}}\ge x_i>\frac1{4000}$. Then $S_{16000}>4$.

(b) Take $x_n=2^{-n}$. Then $\frac{x_i^2}{x_{i+1}}=\frac{2^{-2i}}{2^{-i-1}}=2^{1-i}$, and $$S_n=\sum_{i=0}^{n-1}2^{1-i}=4\left(1-2^{-n}\right)<4 .$$

Key idea: a linear map of order $3$ preserves the form and moves every nonzero solution, so solutions come in triples. Modulo $9$ the form takes only a few values, and $2891$ is not among them.

Let $F(x,y)=x^3-3xy^2+y^3$ and $\tau(x,y)=(y-x,-x)$.

$\tau$ preserves $F$. $$F(y-x,-x)=(y-x)^3-3(y-x)x^2-x^3=y^3-3y^2x+3yx^2-x^3-3x^2y+3x^3-x^3=F(x,y).$$ $\tau$ has order $3$: $\tau^2(x,y)=(-y,x-y)$ and $\tau^3(x,y)=(x,y)$. Its only fixed point is $(0,0)$ (from $y-x=x$ and $-x=y$). So if $F(x,y)=n>0$, then $(x,y)\ne(0,0)$, and $(x,y)$, $\tau(x,y)$, $\tau^2(x,y)$ are three different solutions (an orbit of size $3$, since its size divides $3$ and is not $1$).

No solutions for $2891$. Modulo $9$: $(x+3s)^3\equiv x^3$ and $3(x+3s)y^2\equiv3xy^2$, and the same for $y$. So $F\bmod9$ depends only on $x,y\bmod3$. Since $2891\equiv2\pmod 9$, and $F\equiv x^3+y^3\equiv x+y\pmod3$, we need $x+y\equiv2\pmod3$, that is, $(x,y)\equiv(0,2),(2,0)$ or $(1,1)\pmod3$. These give $$F(0,2)=8,\qquad F(2,0)=8,\qquad F(1,1)=-1\equiv8\pmod9 .$$ Never $2$. So $F(x,y)=2891$ has no integer solutions.

Key idea: with coordinates, collinearity is one determinant, and it reduces to $3r^2=1$.

Take the hexagon with center at the origin and vertices on the unit circle: $$A=(1,0),\ B=\left(\tfrac12,\tfrac{\sqrt3}2\right),\ C=\left(-\tfrac12,\tfrac{\sqrt3}2\right),\ E=\left(-\tfrac12,-\tfrac{\sqrt3}2\right).$$ Then $$M=A+r(C-A)=\left(1-\tfrac32r,\ \tfrac{\sqrt3}2r\right),\qquad N=C+r(E-C)=\left(-\tfrac12,\ \tfrac{\sqrt3}2-\sqrt3\,r\right).$$ Relative to $B$: $M-B=\left(\tfrac12-\tfrac32r,\ \tfrac{\sqrt3}2(r-1)\right)$ and $N-B=\left(-1,\ -\sqrt3\,r\right)$. They are parallel when $$\left(\tfrac12-\tfrac32r\right)(-\sqrt3\,r)-\tfrac{\sqrt3}2(r-1)(-1)=\frac{\sqrt3}2\left(3r^2-1\right)=0 .$$ With $0<r<1$, this gives $r=\frac1{\sqrt3}$.

Key idea: find a moment when the path is near a corner $W$, while some point $P$ on the two sides away from $W$ is approached both before and after that moment. Going from near $P$ to near $W$ and back costs at least $2\cdot 99$.

Let the path have length $\ell$ and parametrize it by arc length, $L(t)$ for $0\le t\le\ell$, from $A_0$ to $A_n$. The path does not meet itself and $A_0\ne A_n$, so each of its points has exactly one parameter. The part between $L(s)$ and $L(t)$ is $L([s,t])$, of length $|t-s|$.

For a corner $V$ let $f(V)$ and $g(V)$ be the first and the last time $t$ with $|L(t)-V|\le\frac12$. They exist: the set of such times is closed and, by the hypothesis, not empty.

Choosing the corner. Let $V_1$ be a corner with the smallest $f$, and $V_2$ a corner with the largest $g$. Pick a corner $W$ different from both (there are four corners), and let $\tau=f(W)$. Then

• $f(V_1)\le f(W)=\tau$: the path is near the corner $V_1\ne W$ at some time $\le\tau$;
• $g(V_2)\ge g(W)\ge\tau$: the path is near the corner $V_2\ne W$ at some time $\ge\tau$.

The far sides. Let $K$ be the union of the two sides of the square that do not contain $W$. It is a connected broken line through the other three corners, and each of its points is at distance at least $100$ from $W$. Put $$K_-=\{P\in K:\ |P-L(t)|\le\tfrac12\ \text{for some } t\le\tau\},\qquad K_+=\{P\in K:\ |P-L(t)|\le\tfrac12\ \text{for some } t\ge\tau\}.$$ Each is closed (the points of $K$ within $\frac12$ of a compact piece of the path). By the hypothesis $K_-\cup K_+=K$. Also $V_1\in K_-$ and $V_2\in K_+$, so neither is empty. A connected set cannot be split into two disjoint nonempty closed sets, so some point $P$ lies in both.

The two points. Choose $t_1\le\tau\le t_2$ with $|L(t_1)-P|\le\frac12$ and $|L(t_2)-P|\le\frac12$, and let $X=L(t_1)$, $Y=L(t_2)$. Then $XY\le1$. Since $|L(\tau)-W|\le\frac12$ and $PW\ge100$, $$|X-L(\tau)|\ge100-\tfrac12-\tfrac12=99,\qquad |L(\tau)-Y|\ge99.$$ The part of $L$ from $X$ to $Y$ passes through $L(\tau)$, and a path is at least as long as the straight segments joining its points in order. So that part has length at least $99+99=198$.