IMO 1965

Key idea: the middle expression is $2\min(|\sin x|,|\cos x|)$.

Since $1\pm\sin2x=(\sin x\pm\cos x)^2$, the roots are $|\sin x+\cos x|$ and $|\sin x-\cos x|$. For real $u,v$, the larger of $|u+v|$ and $|u-v|$ is $|u|+|v|$ and the smaller is $\bigl||u|-|v|\bigr|$. So $$\bigl||u+v|-|u-v|\bigr|=2\min(|u|,|v|).$$ Hence the middle expression is $m(x)=2\min(|\sin x|,|\cos x|)$.

Right inequality. $\min(|\sin x|,|\cos x|)\le\frac{\sqrt2}2$, because the two squares add up to $1$. So $m(x)\le\sqrt2$ always holds.

Left inequality. $2\cos x\le m(x)$.

• If $\cos x\le0$, it holds, since $m\ge0$. On $[0,2\pi]$ this is $\frac\pi2\le x\le\frac{3\pi}2$.
• If $\cos x>0$, it says $\cos x\le|\sin x|$ (the condition $\cos x\le|\cos x|$ is automatic). With $\cos x>0$, this means $x\in[\frac\pi4,\frac\pi2)$ or $x\in(\frac{3\pi}2,\frac{7\pi}4]$.

Together: $\dfrac\pi4\le x\le\dfrac{7\pi}4$.

Key idea: look at the equation of the largest unknown. Its diagonal term wins against everything else.

Suppose $(x_1,x_2,x_3)\ne(0,0,0)$ is a solution. Choose $i$ with $|x_i|$ largest. Replacing the solution by its negative if needed, assume $x_i>0$, so $x_j\le x_i$ for every $j$.

In the $i$-th equation, every off-diagonal coefficient $a_{ij}$ is negative and $x_j\le x_i$, so $a_{ij}x_j\ge a_{ij}x_i$. Therefore $$0=a_{ii}x_i+\sum_{j\ne i}a_{ij}x_j\ \ge\ \Bigl(a_{ii}+\sum_{j\ne i}a_{ij}\Bigr)x_i>0,$$ because the coefficients of each equation have a positive sum and $x_i>0$. This is a contradiction. So the only solution is $x_1=x_2=x_3=0$.

(The same proof works for any number of unknowns.)

Key idea: the cross-sections parallel to both edges are parallelograms whose area is a quadratic in the position. Integrating it gives both pieces.

Every point of the tetrahedron can be written as $(1-t)X+tY$ with $X$ on $AB$, $Y$ on $CD$ and $0\le t\le1$. The planes parallel to $AB$ and $CD$ are exactly the sets of fixed $t$, and the one with parameter $t$ is at distance $td$ from $AB$. Its section is $$\{(1-t)X+tY\}=(1-t)\cdot AB+t\cdot CD,$$ a parallelogram with sides $(1-t)a$ and $tb$ at angle $\omega$. Its area is $t(1-t)\,ab\sin\omega$. Integrating over the distance $d\,dt$: $$\text{vol}=abd\sin\omega\int_0^1t(1-t)\,dt=\frac{abd\sin\omega}{6},$$ which is the familiar formula. The plane $\varepsilon$ has distances in the ratio $k$, so it sits at $t=\lambda=\frac k{k+1}$. The piece next to $AB$ has volume $$abd\sin\omega\int_0^\lambda t(1-t)\,dt=abd\sin\omega\Bigl(\frac{\lambda^2}2-\frac{\lambda^3}3\Bigr)=\frac{abd\sin\omega}{6}\,\lambda^2(3-2\lambda).$$ With $\lambda=\frac k{k+1}$: $$\lambda^2(3-2\lambda)=\frac{k^2(k+3)}{(k+1)^3},\qquad 1-\lambda^2(3-2\lambda)=\frac{(k+1)^3-k^3-3k^2}{(k+1)^3}=\frac{3k+1}{(k+1)^3}.$$ So the two pieces are in the ratio $$\frac{k^2(k+3)}{3k+1},$$ independent of $a,b,d,\omega$. For $k=1$ the plane is halfway and the ratio is $1$, as symmetry demands.

Key idea: all four numbers are roots of one quadratic, so they take at most two values. A short case check finishes.

No $x_i$ is $0$: if $x_1=0$, the other equations give $x_2=x_3=x_4=2$, and then the first equation says $0+8=2$. Let $P=x_1x_2x_3x_4\ne0$. The $i$-th equation is $x_i+\frac P{x_i}=2$, that is $$x_i^2-2x_i+P=0 .$$ So each $x_i$ is a root $r$ or $s$ of $t^2-2t+P$, where $r+s=2$ and $rs=P$.

• All equal, $x_i=r$: then $r+r^3=2$, so $(r-1)(r^2+r+2)=0$ and $r=1$. This gives $(1,1,1,1)$.
• Three equal to $r$, one equal to $s\ne r$: $P=r^3s$ and $P=rs$, so $r^2=1$. $r=1$ forces $s=1=r$, so $r=-1$ and $s=3$. Check: $3+(-1)^3=2$ and $-1+(-1)(-1)\cdot3=2$. This gives $(3,-1,-1,-1)$ in any order.
• Two and two: $P=r^2s^2=rs$, so $rs=1$ with $r+s=2$, which forces $r=s=1$, a contradiction.

The solutions are $(1,1,1,1)$ and the four arrangements of $(3,-1,-1,-1)$.

Key idea: $H$ is a linear function of $M$: it is $P+Q-M$ with $O$ as origin. A linear map sends segments to segments and triangles to triangles.

$H$ in vectors. Put $O$ at the origin. The right angles at $P$ and $Q$ put $O,P,Q,M$ on the circle with diameter $OM$, so the triangle $OPQ$ has circumcenter $\frac12M$. For any triangle $XYZ$ with circumcenter $K$, the orthocenter is $X+Y+Z-2K$. (Check: $(X+Y+Z-2K)-X=(Y-K)+(Z-K)$, and this is perpendicular to $Y-Z=(Y-K)-(Z-K)$ because $|Y-K|=|Z-K|$.) Hence $$H=O+P+Q-M=P+Q-M .$$ $P$ and $Q$ are the projections of $M$ onto the lines $OA$ and $OB$, which are linear in $M$. So $H=L(M)$ for a linear map $L=\pi_A+\pi_B-\mathrm{id}$, where $\pi_A,\pi_B$ are these projections.

Images of $A$ and $B$. $L(A)=A+\pi_B(A)-A=\pi_B(A)=:A_1$, the foot of the perpendicular from $A$ to $OB$. Likewise $L(B)=B_1$, the foot from $B$ to $OA$. Since the angle at $O$ is acute and not right, $A_1$ is a non-zero point of ray $OB$ and $B_1$ a non-zero point of ray $OA$. They are linearly independent, so $L$ is invertible.

(a) $M=(1-u)A+uB$ with $0\le u\le1$ gives $H=(1-u)A_1+uB_1$. The locus is the segment $A_1B_1$.

(b) $L$ maps the interior of triangle $OAB$ onto the interior of triangle $OA_1B_1$. (For every $M\ne O$ in the closed triangle the projections $P,Q$ are non-zero, because the angle is acute, so the triangle $OPQ$ is not degenerate.)

Key idea: diameters pairwise meet. So a point with three diameters has a middle neighbour that can have no other diameter, and removing it allows induction.

Call the diameters edges. The degree of a point is the number of edges at it.

Lemma 1. Two edges with four different endpoints cross. Suppose edges $XY$ and $ZW$ do not meet.

• If the four points are in convex position, then $XY$ and $ZW$ are not the two diagonals (diagonals of a convex quadrilateral cross). So they are opposite sides. The two diagonals cross at a point, and the triangle inequality at that point gives: sum of diagonals $>$ sum of the opposite sides $=2d$. So some diagonal is longer than $d$. Contradiction.
• Otherwise one point, say $W$, lies in the triangle formed by the other three (the four points cannot be collinear, since on a line only one segment can have the maximum length). The distance from $Z$ to a point of a triangle with vertex $Z$ is largest only at the other two vertices, so $ZW<\max(ZX,ZY)\le d$. Contradiction. The same argument works whichever point lies inside.

Lemma 2. If $P$ has three edges $PA,PB,PC$, then one of $A,B,C$ has degree $1$. $A,B,C$ lie on the circle of radius $d$ about $P$, and their pairwise distances are at most $d$. So they lie within an arc of at most $60^\circ$. Name them so that $B$ is between $A$ and $C$, strictly inside the angle $APC$. Then $BA<d$ and $BC<d$. Suppose $BQ$ is an edge with $Q\ne P$. Then $Q\ne A,C$, so by Lemma 1, $BQ$ meets both segments $PA$ and $PC$. It cannot pass through $P$ (that would make $BQ=BP+PQ>d$). But segment $BQ$ starts inside the convex angle $APC$. Once it leaves the angle through one side, it cannot come back to cross the other side. Contradiction. So $B$ has degree $1$.

Induction on $n$. For $n=3$ there are at most $3$ segments in all. Let $n\ge4$.

• If some point $V$ has degree at most $1$, remove it. If the remaining set still has the largest distance $d$, its edges are the old edges not at $V$, and by induction there are at most $n-1$ of them, so there were at most $n$. If the largest distance drops, every old edge used $V$, so there was at most $1$.
• If every degree is at least $2$ and some point has degree at least $3$, Lemma 2 gives a point of degree $1$, a contradiction. So every degree is exactly $2$, and the number of edges is $\frac{2n}2=n$.

Either way there are at most $n$ diameters. (A regular polygon with an odd number of vertices has exactly $n$.)