IMO 1967

Key idea: the circles cover the parallelogram exactly when the circumradius $R$ of triangle $ABD$ is at most $1$. Translating $R\le1$ with the law of sines and the law of cosines gives the stated inequality.

The diagonal $BD$ cuts the parallelogram into triangles $ABD$ and $CDB$, which are congruent by a half-turn about the center. That half-turn swaps the circles about $A$ and $C$ and swaps those about $B$ and $D$. Let $O$ be the circumcenter of $ABD$. It lies inside the triangle, since the triangle is acute.

If $R\le1$, the circles cover. The segments $OA,OB,OD$ cut $ABD$ into three triangles, and each of these is cut by the perpendicular from $O$ into two right triangles, such as $O,A,$ and the midpoint $N$ of $AB$. Every point $X$ of triangle $OAN$ satisfies $AX\le\max(AO,AN)\le R\le1$, because the distance from $A$ is largest at a vertex of that triangle. So triangle $ABD$ is covered by the circles about $A,B,D$. By the half-turn, triangle $CDB$ is covered by those about $C,D,B$.

If $R>1$, they do not. The point $O$ is at distance $R>1$ from $A,B,D$. It is also farther than $R$ from $C$. Indeed, $C$ and $A$ are on opposite sides of $BD$, and $\angle BCD=\alpha$. A point on the circle on that side would see $BD$ under $180^\circ-\alpha$, and since $\alpha<90^\circ$ we have $\alpha<180^\circ-\alpha$. So $C$ is outside the circumcircle and $OC>R>1$. The point $O$ is not covered.

Translating $R\le1$. By the law of sines in $ABD$, $BD=2R\sin\alpha$, so $R\le1\iff BD^2\le4\sin^2\alpha$. By the law of cosines, $BD^2=a^2+1-2a\cos\alpha$. So the condition is $$a^2-2a\cos\alpha+\cos^2\alpha\le4\sin^2\alpha-1+\cos^2\alpha=3\sin^2\alpha,$$ that is $|a-\cos\alpha|\le\sqrt3\sin\alpha$. The angle of $ABD$ at $B$ is acute, which means the foot of the perpendicular from $D$ to line $AB$ lies on the same side of $B$ as $A$: $\cos\alpha<a$. So the lower half of the double inequality holds automatically, and the condition is $$a\le\cos\alpha+\sqrt3\sin\alpha .$$

Key idea: look at the edge opposite the long one. Both faces through it are short triangles, so their heights over it are small. The volume is at most one sixth of the edge times the two heights.

Let $AB$ be the edge longer than $1$, so the other five are at most $1$. Let $c=CD\le1$.

Heights over $CD$. In triangle $ACD$, drop the perpendicular from $A$ to line $CD$, with foot $F$. One of $FC,FD$ is at least $\frac c2$ (if $F$ is between $C$ and $D$, they add up to $c$; if not, one of them is at least $c$). Say $FC\ge\frac c2$. Then $$h_1=AF=\sqrt{AC^2-FC^2}\le\sqrt{1-\tfrac{c^2}4}.$$ In the same way, the height $h_2$ of $B$ over line $CD$ in triangle $BCD$ is at most $\sqrt{1-c^2/4}$.

The volume. The base $BCD$ has area $\frac12c\,h_2$, and the height of $A$ over the plane $BCD$ is at most $h_1$. So $$V\le\frac13\cdot\frac12c\,h_2\cdot h_1\le\frac16\,c\Bigl(1-\frac{c^2}{4}\Bigr).$$ The function $f(c)=c-\frac{c^3}4$ has derivative $1-\frac{3c^2}4>0$ on $(0,1]$, so $f(c)\le f(1)=\frac34$. Hence $V\le\frac16\cdot\frac34=\frac18$.

Equality holds for two equilateral triangles of side $1$ on a common edge, in perpendicular planes. The long edge is then $\sqrt{3/2}>1$.

Key idea: each difference factors, and the product becomes two blocks of $n$ consecutive integers. The prime takes care of the extra factor $n+1$.

Since $c_u-c_v=u^2+u-v^2-v=(u-v)(u+v+1)$, $$\prod_{i=1}^n(c_{m+i}-c_k)=\prod_{i=1}^n(m-k+i)\cdot\prod_{i=1}^n(m+k+1+i).$$ The divisor is $c_1c_2\cdots c_n=\prod i(i+1)=n!\,(n+1)!$.

• First block. $\prod_{i=1}^n(m-k+i)$ is a product of $n$ consecutive integers. It is divisible by $n!$. (If it contains $0$, it is $0$. If all factors are negative, change signs. If all are positive, the quotient is a binomial coefficient.)
• Second block. Let $p=m+k+1$, a prime with $p>n+1$. The product $p(p+1)\cdots(p+n)$ of $n+1$ consecutive positive integers is divisible by $(n+1)!$. Since $p$ is a prime larger than $n+1$, it shares no factor with $(n+1)!$. So $(n+1)!$ divides $(p+1)\cdots(p+n)=\prod_{i=1}^n(m+k+1+i)$.

Multiplying, $n!\,(n+1)!$ divides the product.

Write $\alpha,\beta,\gamma$ for the angles of $A_1B_1C_1$. Call a triangle $ABC$ allowed if it is similar to $A_1B_1C_1$ (in this vertex order) and $A_0,B_0,C_0$ lie on the sides $BC,CA,AB$ (endpoints allowed). All allowed triangles have the same shape, so the one with the longest side $AB$ has the largest area.

Key idea: each vertex of an allowed triangle moves on a fixed circle, and the three circles share a point $M$. A segment through a common point of two circles, with its ends on the circles, is longest when it is parallel to the line of centers. That happens exactly when $A,B,C$ are the points opposite $M$ on the circles.

Step 1: the point $M$. The angles $p=180^\circ-\alpha$, $q=180^\circ-\beta$, $r=180^\circ-\gamma$ add up to $360^\circ$, and each is more than $90^\circ$, so more than the matching angle of the acute triangle $A_0B_0C_0$. (This is where both triangles being acute is used.) We show there is a point $M$ inside $A_0B_0C_0$ with $\angle B_0MC_0=p$, $\angle C_0MA_0=q$, $\angle A_0MB_0=r$.

Write $x,y,z$ for the angles of $A_0B_0C_0$ at $A_0,B_0,C_0$. The points $Q$ on $A_0$’s side of $B_0C_0$ with $\angle B_0QC_0=p$ form an arc from $B_0$ to $C_0$; label them by $\varphi=\angle QB_0C_0\in(0^\circ,180^\circ-p)$. Such a $Q$ is inside the triangle exactly when $\varphi<y$ and $180^\circ-p-\varphi<z$. That is an interval $J$ of values of $\varphi$, and it is not empty because $p>x$. Inside the triangle, $\angle C_0QA_0+\angle A_0QB_0=360^\circ-p$, so we only need $g=\angle A_0QB_0$ to equal $r$.

• At the lower end of $J$, $Q$ tends to $C_0$ (then $g\to z<r$) or to a point of side $C_0A_0$ (then $\angle C_0QA_0\to180^\circ$, so $g\to180^\circ-p<r$, because $q<180^\circ$).
• At the upper end, $Q$ tends to $B_0$ (then $\angle C_0QA_0\to y<q$, so $g\to360^\circ-p-y>r$) or to a point of side $A_0B_0$ (then $g\to180^\circ>r$).

By continuity $g=r$ somewhere in $J$. That point is $M$.

Step 2: three fixed circles. Let $\omega_A,\omega_B,\omega_C$ be the circles $MB_0C_0$, $MC_0A_0$, $MA_0B_0$, with centers $O_A,O_B,O_C$. For every allowed triangle, $A\in\omega_A$. If $A$ is $B_0$ or $C_0$ this is clear. Otherwise $B_0,C_0$ lie on the rays $AC,AB$, so $\angle B_0AC_0=\alpha$. The line $B_0C_0$ cuts off the corner $A$ from the side $BC$, which contains $A_0$, and $M$ is on $A_0$’s side. So $A$ and $M$ are on opposite sides of $B_0C_0$ with $\angle B_0AC_0+\angle B_0MC_0=180^\circ$, and $A,B_0,M,C_0$ are concyclic. In the same way $B\in\omega_B$ and $C\in\omega_C$.

Step 3: the bound. Drop perpendiculars from $O_A$ and $O_B$ to the line $AB$. Since $A$ and $C_0$ are on $\omega_A$, the first foot is the midpoint of $AC_0$; likewise the second foot is the midpoint of $C_0B$. These midpoints are $\frac12AB$ apart, so $$AB=2\cdot\big(\text{length of the projection of }O_AO_B\text{ on line }AB\big)\le2\,O_AO_B,$$ with equality exactly when $AB\parallel O_AO_B$. (If $A=C_0$, the tangent-chord angle shows that line $AB$ touches $\omega_A$ at $C_0$, and the same formula holds.)

Step 4: the triangle that reaches the bound. Let $A^*,B^*,C^*$ be the points diametrically opposite $M$ on $\omega_A,\omega_B,\omega_C$. By Thales, $\angle MB_0A^*=\angle MC_0A^*=90^\circ$. So the three lines through $A_0,B_0,C_0$ perpendicular to $MA_0,MB_0,MC_0$ form the triangle $A^*B^*C^*$.

• Its angles. The chord $B_0C_0$ is seen from $M$ at the obtuse angle $p$, so the center $O_A$, and with it $A^*=2O_A-M$, lies on the far side of $B_0C_0$ from $M$. So $MB_0A^*C_0$ is a convex quadrilateral with right angles at $B_0$ and $C_0$, and its angle at $A^*$ is $180^\circ-p=\alpha$. In the same way the angles at $B^*$ and $C^*$ are $\beta$ and $\gamma$.
• The points are inside the sides. From the convex quadrilaterals $MA_0B^*C_0$ and $MA_0C^*B_0$, $B^*$ lies on $C_0$’s side of line $MA_0$ and $C^*$ on $B_0$’s side. Since $M$ is inside $A_0B_0C_0$, line $MA_0$ separates $B_0$ from $C_0$. So $A_0$ lies strictly between $B^*$ and $C^*$. The same holds for $B_0$ and $C_0$.

So $A^*B^*C^*$ is allowed. Its side $A^*B^*$ is perpendicular to $MC_0$, the common chord of $\omega_A$ and $\omega_B$, so it is parallel to $O_AO_B$, and $A^*B^*=2\,O_AO_B$.

Conclusion. Every allowed triangle has $AB\le A^*B^*$, so its area is at most that of $A^*B^*C^*$. Equality forces $AB\parallel O_AO_B$. That fixes the line $AB$ through $C_0$, hence $A=A^*$, $B=B^*$, and then $C=C^*$. So the maximum is reached by $A^*B^*C^*$ only.

Construction. On $B_0C_0$, on the side of $A_0$, draw the arc of points that see $B_0C_0$ at the angle $180^\circ-\angle A_1$. Do the same on $C_0A_0$ with $180^\circ-\angle B_1$. The arcs meet inside the triangle at $M$ (the third arc passes through it as well). Draw the lines through $A_0,B_0,C_0$ perpendicular to $MA_0,MB_0,MC_0$.

Key idea: for large odd $n$, the numbers of largest absolute value dominate the sum. They must cancel in pairs $\pm M$. Peel them off and repeat.

For even $n$, $c_n$ is a sum of non-negative terms, not all zero, so $c_n>0$. Hence $c_n=0$ for infinitely many odd $n$.

Claim: the non-zero $a_i$ split into pairs $\{x,-x\}$. Let $M>0$ be the largest $|a_i|$. Suppose $M$ appears $u$ times and $-M$ appears $v$ times. For odd $n$, $$\frac{c_n}{M^n}=u-v+\sum_{|a_i|<M}\Bigl(\frac{a_i}{M}\Bigr)^n .$$ The last sum tends to $0$ as $n\to\infty$. Along the infinitely many odd $n$ with $c_n=0$, this forces $u-v=0$. For odd $n$, the $u$ copies of $M$ and the $u$ copies of $-M$ cancel, so removing them does not change $c_n$ for any odd $n$. The remaining numbers again have $c_n=0$ for infinitely many odd $n$. Repeat until only zeros remain.

So the numbers are zeros together with pairs $x,-x$. Then every odd power sum is $0$, and every even one is positive. The answer is: $c_n=0$ exactly for odd $n$.

Key idea: the remaining medals follow a linear recurrence. Its solution has a term $(\frac67)^{k}$, and whole numbers force that term to vanish.

Let $r_k$ be the number of medals left before day $k$, so $r_1=m$. On day $k$, $k+\frac17(r_k-k)$ medals are given out, which leaves $$r_{k+1}=\frac67(r_k-k).$$ Solving. Try $r_k=\alpha k+\beta$: $\alpha(k+1)+\beta=\frac67(\alpha k+\beta-k)$ gives $\alpha=-6$, $\beta=42$. So the general solution is $$r_k=42-6k+C\Bigl(\frac67\Bigr)^{k-1}.$$ The last day. On day $n$ all remaining medals, $n$ of them, are given, so $r_n=n$: $$C\Bigl(\frac67\Bigr)^{n-1}=7n-42,\qquad m=r_1=36+C=36+\frac{7^n(n-6)}{6^{n-1}} .$$ Since $m$ is a whole number and $7^n$ is coprime to $6$, we need $6^{n-1}\mid n-6$. For $n\ge2$, $6^{n-1}>|n-6|$ (check $n=2$: $6>4$; after that the left side grows much faster). So $n-6=0$.

Hence $n=6$ and $m=36$. Check: each day $6$ medals are given ($1+\frac{35}{7}$, then $2+\frac{28}{7}$, and so on), and on day $6$ the last $6$ remain.