IMO 1979

Key idea: the alternating sum equals $\sum_{k=660}^{1319}\frac1k$, and pairing $k$ with $1979-k$ puts the prime $1979$ in every numerator.

Subtracting twice the even terms from the full harmonic sum, $$\sum_{k=1}^{1319}\frac{(-1)^{k+1}}{k}=\sum_{k=1}^{1319}\frac1k-2\sum_{k=1}^{659}\frac1{2k}=\sum_{k=660}^{1319}\frac1k .$$ These $660$ terms pair up as $k$ and $1979-k$ for $k=660,\dots,989$, and $$\frac1k+\frac1{1979-k}=\frac{1979}{k(1979-k)} .$$ So $\frac pq=1979\cdot\frac{N}{D}$, where $D$ is the product of all the denominators $k(1979-k)$, all less than $1979$. The number $1979$ is prime (it has no prime factor up to $43$, and $44^2>1979$), so $1979\nmid D$. From $pD=1979\,Nq$ we get $1979\mid pD$, hence $1979\mid p$.

Key idea: a top edge of one color limits how many edges of that color its endpoint sends down. Since five edges go down, the other color dominates, and that forces both top edges at each vertex to share a color.

Call the colors red and green, and indices modulo $5$. A fact used twice: any three of the five bottom vertices include two that are neighbors on the bottom pentagon.

Lemma. If the top edge $A_1A_2$ has color $X$, then at most two of the edges $A_1B_j$ have color $X$. Suppose $A_1B_j,A_1B_k,A_1B_l$ all have color $X$. For each of these $B$’s, triangle $A_1A_2B$ has two sides of color $X$, so $A_2B$ has the other color $Y$. Two of $B_j,B_k,B_l$ are neighbors, say $B_jB_k$. Triangle $A_1B_jB_k$ forces $B_jB_k$ to be $Y$. Then triangle $A_2B_jB_k$ is all $Y$. Contradiction.

Each pentagon is one color. Each top vertex sends five edges down, so at least three have one color $Z$. By the lemma, neither top edge at that vertex has color $Z$. So the two top edges at every top vertex have the same color, and going around the pentagon, all five top edges share a color. The same argument works for the bottom.

Top and bottom match. Suppose the top is red and the bottom green. By the lemma, $A_1$ sends at most two red edges down, so at least three green ones. Two of their bottom ends are neighbors, joined by a green bottom edge, which makes a green triangle. Contradiction. So all ten pentagon edges have one color.

Key idea: the two motions are rotations by the same angle about the two centers. The mirror image of $A$ in the perpendicular bisector of the centers is equidistant from both points at every moment.

Put the midpoint of the centers at the origin and the centers at $O_1=-d$ and $O_2=d$, with $d$ real, and let $A=\alpha$. Both points start at $A$, turn in the same sense and return together, so at any moment both have turned through the same angle. With $e=e^{i\omega t}$, $$X=-d+(\alpha+d)e,\qquad Y=d+(\alpha-d)e .$$ Let $P=-\bar\alpha$, the reflection of $A$ in the imaginary axis, which is the perpendicular bisector of $O_1O_2$. Then $$X-P=(\bar\alpha-d)+(\alpha+d)e,\qquad Y-P=(\bar\alpha+d)+(\alpha-d)e .$$ Using $|z+we|^2=|z|^2+|w|^2+2\,\mathrm{Re}(z\bar w\bar e)$ and the fact that $d$ is real: $$|X-P|^2=|\alpha-d|^2+|\alpha+d|^2+2\,\mathrm{Re}\big((\bar\alpha-d)(\bar\alpha+d)\bar e\big),$$ $$|Y-P|^2=|\alpha+d|^2+|\alpha-d|^2+2\,\mathrm{Re}\big((\bar\alpha+d)(\bar\alpha-d)\bar e\big).$$ These are equal. So $PX=PY$ at every moment.

In words: the fixed point is the mirror image of $A$ across the perpendicular bisector of the two centers. (If that bisector passes through $A$, the point is $A$ itself.)

Key idea: by the law of sines the ratio equals $\cos\frac{\hat Q-\hat R}{2}\big/\sin\frac{\hat P}{2}$, where hats denote the angles of triangle $QPR$. The best $R$ makes the triangle isosceles and the angle at $P$ as small as possible.

For $R\ne P$, let $\hat P,\hat Q,\hat R$ be the angles of triangle $QPR$. By the law of sines, $$\frac{QP+PR}{QR}=\frac{\sin\hat R+\sin\hat Q}{\sin\hat P}=\frac{2\sin\frac{\hat Q+\hat R}2\cos\frac{\hat Q-\hat R}2}{2\sin\frac{\hat P}2\cos\frac{\hat P}2}=\frac{\cos\frac{\hat Q-\hat R}2}{\sin\frac{\hat P}2},$$ because $\sin\frac{\hat Q+\hat R}2=\cos\frac{\hat P}2$. (At $R=P$ the ratio is $1$, which is smaller.)

So the ratio is at most $1/\sin\frac{\hat P}2$, with equality when $\hat Q=\hat R$, that is, $PR=PQ$. The angle $\hat P=\angle QPR$ is the angle between ray $PQ$ and a ray in the plane $\pi$. Let $H$ be the foot of the perpendicular from $Q$ to $\pi$.

• If $H\ne P$, the smallest such angle is the angle $\delta$ between $PQ$ and the plane, attained only on the ray $PH$. The maximum ratio is $1/\sin\frac\delta2$, attained only at the point $R$ of ray $PH$ with $PR=PQ$.
• If $H=P$, every angle $\hat P$ is $90^\circ$, the maximum is $\sqrt2$, and it is attained at every $R$ with $PR=PQ$.

Key idea: a combination of the three equations is a sum of nonnegative terms that must vanish, so all the weight sits at one $k$ with $k^2=a$.

Multiply the equations by $a^2$, $-2a$ and $1$ and add: $$\sum_{k=1}^5 k\,(a-k^2)^2\,x_k=a^2\cdot a-2a\cdot a^2+a^3=0 .$$ Each term is $\ge0$, so for each $k$, either $x_k=0$ or $k^2=a$.

• If all $x_k=0$, then $a=0$.
• Otherwise $a=k^2$ for one $k$, and every other $x_j$ is $0$. The first equation gives $kx_k=k^2$, so $x_k=k$. Then $k^3x_k=k^4=a^2$ and $k^5x_k=k^6=a^3$, so all three equations hold.

So $a$ is one of $0,1,4,9,16,25$.

Key idea: look at the frog every two jumps. Two numbers then evolve by a $2\times2$ linear recurrence whose eigenvalues are $2\pm\sqrt2$.

Label the vertices $0,\dots,7$ around the octagon, with $A=0$ and $E=4$. Each jump changes the parity of the label, and $4$ is even, so $E$ can only be reached after an even number of jumps: $a_{2n-1}=0$.

After an even number of jumps, the frog (if not yet stopped) is at $0$, $2$ or $6$. Let $u_n$ be the number of allowed paths of $2n$ jumps ending at $0$, and $v_n$ the number ending at $2$ or $6$. Two jumps pass through an odd vertex, never $E$, so:

• from $0$: back to $0$ in $2$ ways (via $1$ or $7$), to $\{2,6\}$ in $2$ ways;
• from $2$: to $0$ in $1$ way, back to $2$ in $2$ ways, to $E$ in $1$ way (the same for $6$).

So $u_{n+1}=2u_n+v_n$, $v_{n+1}=2u_n+2v_n$, and $a_{2n+2}=v_n$, with $u_0=1$, $v_0=0$.

Eliminate $u$: the second equation gives $2u_n=v_{n+1}-2v_n$, so $$v_{n+2}=2u_{n+1}+2v_{n+1}=2(2u_n+v_n)+2v_{n+1}=2(v_{n+1}-2v_n)+2v_n+2v_{n+1}=4v_{n+1}-2v_n ,$$ whose characteristic roots are $2\pm\sqrt2=x,y$. With $v_0=0$ and $v_1=2$, the solution is $v_n=\frac{x^n-y^n}{\sqrt2}$ (check: $n=1$ gives $\frac{2\sqrt2}{\sqrt2}=2$). Therefore $$a_{2n}=v_{n-1}=\frac{1}{\sqrt2}\left(x^{n-1}-y^{n-1}\right).$$