IMO 1963

Key idea: squaring twice leaves a single candidate. Substituting it back is clean, because both radicands become perfect squares. That check decides which $p$ work.

The candidate. The left side is non-negative, so any root has $x\ge0$. Move one root across, $\sqrt{x^2-p}=x-2\sqrt{x^2-1}$, and square: $$x^2-p=x^2-4x\sqrt{x^2-1}+4x^2-4\iff 4x\sqrt{x^2-1}=4x^2-4+p .$$ Square again: $16x^4-16x^2=16x^4+8(p-4)x^2+(p-4)^2$, which simplifies to $$x^2(16-8p)=(4-p)^2 .$$ For $p\ge2$ this has no solution (for $p=2$ it reads $0=4$, and for $p>2$ the left side is $\le0$ while the right side is positive). For $p<2$, the only non-negative candidate is $$x_0=\frac{4-p}{\sqrt{8(2-p)}} .$$

The check. Write $D=\sqrt{8(2-p)}$. Then $$x_0^2-1=\frac{(4-p)^2-8(2-p)}{D^2}=\frac{p^2}{D^2},\qquad x_0^2-p=\frac{(4-p)^2-8p(2-p)}{D^2}=\frac{(3p-4)^2}{D^2}.$$ Both are non-negative, so the roots exist, and the equation at $x_0$ becomes $$|3p-4|+2|p|=4-p .$$

• If $0\le p\le\frac43$: the left side is $4-3p+2p=4-p$. True.
• If $\frac43<p<2$: the left side is $5p-4$, which equals $4-p$ only at $p=\frac43$. False.
• If $p<0$: the left side is $4-5p$, which equals $4-p$ only at $p=0$. False.

Answer. For $0\le p\le\frac43$, there is exactly one real root, $x=\dfrac{4-p}{\sqrt{8(2-p)}}$. For example, $p=0$ gives $x=1$. For all other $p$, there is none.

Key idea: $P$ works when $(A-P)\cdot(X-P)=0$ for some $X$ on the segment. This expression is linear in $X$, so it vanishes on the segment exactly when its values at $B$ and $C$ have opposite signs (or one is zero).

The angle at $P$ is right with one side through $A$ and the other through a point $X$ of segment $BC$ exactly when $(A-P)\cdot(X-P)=0$ (for $P\ne A$, $P\ne X$). Equivalently, $P$ lies on the sphere with diameter $AX$. So the locus is the union of these spheres as $X$ runs along $BC$.

Fix $P$ and let $f(X)=(A-P)\cdot(X-P)$. Along the segment, $X=(1-\lambda)B+\lambda C$, and $f$ is a linear function of $\lambda\in[0,1]$. So $f$ has a zero on the segment exactly when $$f(B)\,f(C)\le0 .$$ Now $f(B)\le0$ means the angle $APB$ is right or obtuse, that is, $P$ lies in the closed ball with diameter $AB$. Likewise for $C$. So the condition says:

$P$ lies in one of the two closed balls (on diameters $AB$ and $AC$) and not in the interior of the other.

This is the locus: the solid region between the two spheres, with both spheres included, except for degenerate cases. The point $A$ itself is excluded. A point $P$ on segment $BC$ satisfies the inequality only through the useless choice $X=P$. Such a point is a real solution only when $AP\perp BC$. (Both spheres pass through $A$ and through the foot of the perpendicular from $A$ to line $BC$. In a plane through $A$, $B$ and $C$, the picture is two disks through $A$, and the locus is their “difference” region.)

Key idea: going around an equiangular polygon, each side is the previous direction turned by $2\pi/n$. Summation by parts turns the closing condition into a sum of non-negative terms that must vanish.

All exterior angles are equal, so they are $2\pi/n$ each. With $\omega=e^{2\pi i/n}$ and a suitable rotation, the $k$-th side is the complex number $a_k\omega^k$. The polygon closes, so $$\sum_{k=1}^n a_k\omega^k=0 .$$ Let $S_k=\omega+\omega^2+\dots+\omega^k=\dfrac{\omega(1-\omega^k)}{1-\omega}$. Note $S_n=0$. Put $d_k=a_k-a_{k+1}\ge0$ for $k<n$. Summation by parts gives $$0=\sum_{k=1}^na_k\omega^k=\sum_{k=1}^{n-1}d_kS_k+a_nS_n=\frac{\omega}{1-\omega}\sum_{k=1}^{n-1}d_k(1-\omega^k).$$ So $\sum_{k=1}^{n-1}d_k(1-\omega^k)=0$. Its real part is $$\sum_{k=1}^{n-1}d_k\Bigl(1-\cos\frac{2\pi k}{n}\Bigr)=0 .$$ Every term is non-negative, and $1-\cos\frac{2\pi k}{n}>0$ for $1\le k\le n-1$. So every $d_k=0$, that is $a_1=a_2=\dots=a_n$.

Key idea: the equations say $x_{i+1}=yx_i-x_{i-1}$ around a cycle of length $5$. Run the recurrence from $x_1,x_2$. The two leftover equations have determinant $(y-2)(y^2+y-1)^2$.

Put $x_1=s$ and $x_2=t$. The equations with $yx_2$, $yx_3$ and $yx_4$ on the right give, in turn, $$x_3=yt-s,\qquad x_4=yx_3-t=(y^2-1)t-ys,\qquad x_5=yx_4-x_3=(y^3-2y)t-(y^2-1)s .$$ Two equations remain: $x_5+x_2=yx_1$ and $x_4+x_1=yx_5$. Both are linear in $s,t$, and their determinant works out to $$(y-2)\,(y^2+y-1)^2 .$$ Other $y$. If the determinant is not zero, then $s=t=0$, so all $x_i=0$.

$y=2$. The equations $x_{i-1}+x_{i+1}=2x_i$ say that each $x_i$ is the average of its two neighbours around the cycle. Take $x_i$ to be a largest value: both neighbours must equal it, and so on around the cycle. So all $x_i$ are equal, to any common value, and that does solve the system.

$y^2+y-1=0$, that is $y=\frac{-1\pm\sqrt5}{2}$. Using $y^2=1-y$, the formulas simplify to $$x_4=-y(s+t),\qquad x_5=ys-t .$$ Check the two remaining equations: $x_5+x_2=ys=yx_1$, and $yx_5=y^2s-yt=(1-y)s-yt=s-y(s+t)=x_1+x_4$. Both hold, so every $(s,t)$ gives a solution: $$(x_1,\dots,x_5)=\bigl(s,\ t,\ yt-s,\ -y(s+t),\ ys-t\bigr).$$ (The values $2$, $\frac{-1+\sqrt5}2$ and $\frac{-1-\sqrt5}2$ are $2\cos\frac{2\pi k}5$ for $k=0,1,2$. They are the only $y$ for which the recurrence comes back to its start after $5$ steps.)

Key idea: after one sign flip the sum is $\cos\theta+\cos3\theta+\cos5\theta$ with $\theta=\frac\pi7$. Multiplying by $2\sin\theta$ makes it telescope.

Let $\theta=\frac\pi7$. Since $\cos(\pi-u)=-\cos u$, we have $-\cos2\theta=\cos5\theta$. So the sum is $$S=\cos\theta+\cos3\theta+\cos5\theta .$$ By $2\sin\theta\cos k\theta=\sin(k+1)\theta-\sin(k-1)\theta$: $$2\sin\theta\,S=\sin2\theta+(\sin4\theta-\sin2\theta)+(\sin6\theta-\sin4\theta)=\sin6\theta .$$ And $\sin6\theta=\sin(\pi-\theta)=\sin\theta\ne0$. So $2\sin\theta\,S=\sin\theta$, which gives $S=\frac12$.

Key idea: the second prediction leaves only three choices of its two adjacent pairs. Each choice leaves three blocks, so six orders to test against the first prediction.

We read “finished consecutively” as: next to each other, in the predicted order.

The second prediction $D\,A\,E\,C\,B$ has adjacent pairs $DA,AE,EC,CB$. The disjoint pairs among them are $\{DA,EC\}$, $\{DA,CB\}$ and $\{AE,CB\}$. In each case, glue the two pairs into blocks and arrange the three blocks. The first prediction forbids $A$ first, $B$ second, $C$ third, $D$ fourth, $E$ fifth, and the adjacent pairs $AB,BC,CD,DE$.

• $\{DA,EC\}$ with $B$. The orders: $DAECB$ (the prediction itself, five places right), $DABEC$ (contains $AB$), $ECDAB$ (contains $CD$), $ECBDA$ ($D$ fourth), $BECDA$ (contains $CD$), $BDAEC$ (no place matches the second prediction, but two are needed). None works.
• $\{AE,CB\}$ with $D$. $AECBD$ and $AEDCB$ put $A$ first, $CBAED$ and $CBDAE$ put $B$ second, $DCBAE$ puts $E$ fifth, and $DAECB$ is the prediction itself. None works.
• $\{DA,CB\}$ with $E$. $DACBE$ and $CBDAE$ put $E$ fifth, $CBEDA$ puts $B$ second, $ECBDA$ puts $D$ fourth, $DAECB$ is the prediction itself. That leaves $EDACB$.

Check of $E\,D\,A\,C\,B$. Against $ABCDE$: no one is in the predicted place, and its adjacent pairs $ED,DA,AC,CB$ avoid $AB,BC,CD,DE$. Against $DAECB$: exactly $C$ and $B$ are in place, and the disjoint pairs $DA$ and $CB$ occur. So the order was $E,D,A,C,B$.