IMO 1968

Key idea: a doubled angle gives the relation $b^2=a(a+c)$ between the sides. With consecutive integers, only one assignment has a whole-number solution.

Lemma. In a triangle with sides $a,b,c$ opposite $A,B,C$: if $B=2A$ and $a\ne c$, then $b^2=a(a+c)$. By the law of sines, $\frac ba=\frac{\sin2A}{\sin A}=2\cos A=\frac{b^2+c^2-a^2}{bc}$. So $b^2c=a(b^2+c^2-a^2)$, which rearranges to $b^2(c-a)=a(c-a)(c+a)$. Dividing by $c-a\ne0$ gives $b^2=a(a+c)$.

The search. The sides $n-1,n,n+1$ are all different, and they form a triangle only when $n\ge3$. If one angle is twice another, call the smaller angle’s opposite side $a$ and the larger one’s $b$. Then $b>a$, and $c$ is the remaining side.

• $(a,b,c)=(n-1,n,n+1)$: $n^2=(n-1)\cdot2n$ gives $n=2$. Not a triangle.
• $(a,b,c)=(n-1,n+1,n)$: $(n+1)^2=(n-1)(2n-1)$ gives $n^2=5n$, so $n=5$.
• $(a,b,c)=(n,n+1,n-1)$: $(n+1)^2=n(2n-1)$ gives $n^2-3n-1=0$, which has no integer root.

So the only candidate has sides $4,5,6$, with the angle opposite $6$ twice the angle opposite $4$.

It works. The angle opposite $4$ has cosine $\frac{25+36-16}{60}=\frac34$, and the angle opposite $6$ has cosine $\frac{16+25-36}{40}=\frac18$. Since $2\bigl(\frac34\bigr)^2-1=\frac18$, the second angle is double the first.

Key idea: the product of the digits is between $0$ and $x$. This traps $x$ between two roots of quadratics.

Let $p(x)$ be the product of the digits. If $x$ has $k$ digits with leading digit $d$, then $p(x)\le d\cdot9^{k-1}\le d\cdot10^{k-1}\le x$. So $$x^2-10x-22\le x\iff x^2-11x-22\le0\iff x\le\frac{11+\sqrt{209}}2\approx12.7 .$$ Also $p(x)\ge0$, so $x^2-10x-22\ge0$, which for positive $x$ means $x\ge5+\sqrt{47}\approx11.9$. The only whole number left is $x=12$. Check: $1\cdot2=2$ and $144-120-22=2$.

Key idea: adding the equations gives $\sum g(x_i)=0$ for $g(t)=f(t)-t$, a quadratic with discriminant $\Delta$. The sign of $\Delta$ controls the sign of $g$.

Let $f(t)=at^2+bt+c$ and $g(t)=f(t)-t=at^2+(b-1)t+c$, whose discriminant is $\Delta$. The system says $x_2=f(x_1),\dots,x_n=f(x_{n-1}),x_1=f(x_n)$. The right-hand sides are $x_1,\dots,x_n$ in a different order, so adding all equations gives $$\sum_{i=1}^n g(x_i)=0 .$$ (a) $\Delta<0$. Then $g$ has no real root and has the sign of $a$ everywhere. A sum of $n$ numbers of the same strict sign is not $0$. No solution.

(b) $\Delta=0$. Then $g(t)=a(t-r)^2$ with $r=\frac{1-b}{2a}$. The sum $a\sum(x_i-r)^2=0$ forces every $x_i=r$. And $x_i=r$ for all $i$ is a solution, since $f(r)=r$. Exactly one solution.

(c) $\Delta>0$. Then $g$ has two different roots $r_1,r_2$, which are fixed points of $f$. The constant tuples $(r_1,\dots,r_1)$ and $(r_2,\dots,r_2)$ are two different solutions. So there is more than one.

Key idea: if both ends of the longest edge fail, adding the two failures contradicts the triangle inequality in the two faces through that edge.

Let $AB$ be a longest edge. At vertex $A$ the edges are $AB,AC,AD$. Since $AB$ is the longest, the only way they can fail to form a triangle is $AB\ge AC+AD$. Likewise at $B$, failure means $AB\ge BC+BD$. If both fail, then adding, $$2AB\ge(AC+CB)+(AD+DB)>AB+AB,$$ using the strict triangle inequality in faces $ABC$ and $ABD$. This is impossible. So at $A$ or at $B$ the three edges form a triangle.

Key idea: with $g=f-\frac12$, the equation says $g(x+a)^2+g(x)^2=\frac14$. Applying it twice brings back $g(x)^2$.

(a) Every real $y$ is $x+a$ for $x=y-a$, so $f(y)\ge\frac12$ for all $y$. Put $g=f-\frac12\ge0$. Since $f-f^2=\frac14-g^2$, the equation reads $$g(x+a)=\sqrt{\tfrac14-g(x)^2},\qquad\text{so}\qquad g(x+a)^2=\tfrac14-g(x)^2 .$$ Applying this twice: $g(x+2a)^2=\frac14-g(x+a)^2=g(x)^2$. Both values are non-negative, so $g(x+2a)=g(x)$, and $f$ has period $2a$.

(b) For $a=1$, take $g(x)=\frac12\bigl|\cos\frac{\pi x}{2}\bigr|$, that is $$f(x)=\frac12+\frac12\Bigl|\cos\frac{\pi x}2\Bigr| .$$ Then $g(x+1)=\frac12\bigl|\sin\frac{\pi x}2\bigr|=\sqrt{\frac14-\frac14\cos^2\frac{\pi x}{2}}=\sqrt{\frac14-g(x)^2}$, as required, and $f$ is not constant. (A step function also works: $f=1$ on $[2j,2j+1)$ and $f=\frac12$ on $[2j+1,2j+2)$ for every integer $j$.)

Key idea: each term is a difference of two floors, so the sum telescopes.

Identity. For real $x$, $\lfloor x\rfloor+\lfloor x+\frac12\rfloor=\lfloor2x\rfloor$. Write $x=j+t$ with $j$ an integer and $0\le t<1$. If $t<\frac12$, both sides are $2j$. If $t\ge\frac12$, both sides are $2j+1$.

The sum. The $k$-th term is $\bigl\lfloor\frac{n+2^k}{2^{k+1}}\bigr\rfloor=\bigl\lfloor\frac n{2^{k+1}}+\frac12\bigr\rfloor$. By the identity with $x=\frac n{2^{k+1}}$, this is $$\Bigl\lfloor\frac n{2^k}\Bigr\rfloor-\Bigl\lfloor\frac n{2^{k+1}}\Bigr\rfloor .$$ Summing over $k=0,1,2,\dots$ telescopes. Once $2^{k}>n$ the floors are $0$, so the sum is $\lfloor n/1\rfloor=n$.

(A way to see it: the $k$-th term counts the numbers from $1$ to $n$ that are divisible by $2^k$ but not by $2^{k+1}$. Every number is counted once.)