IMO 2008

Key idea: all six points are at the same distance from the circumcenter $O$. A short vector computation shows that the squared distance is a symmetric expression in $A$, $B$, $C$.

Put the origin at the circumcenter $O$, write $R$ for the circumradius, and use vectors, so $|A|=|B|=|C|=R$. A standard fact: the orthocenter is $H=A+B+C$. (Indeed $(H-A)\cdot(B-C)=(B+C)\cdot(B-C)=|B|^2-|C|^2=0$, and similarly for the other altitudes.)

One point. Let $M_a=\tfrac{B+C}2$. Since $OM_a\perp BC$ and $A_1$ is on line $BC$ with $M_aA_1=M_aH$, $$OA_1^2=OM_a^2+M_aA_1^2=OM_a^2+M_aH^2 .$$ The same holds for $A_2$.

Compute. $H-M_a=A+\tfrac{B+C}2$, so $$OM_a^2+M_aH^2=\frac{|B+C|^2}4+R^2+A\cdot(B+C)+\frac{|B+C|^2}4=R^2+\frac{2R^2+2B\cdot C}2+A\cdot B+A\cdot C .$$ That is $$OA_1^2=2R^2+A\cdot B+B\cdot C+C\cdot A,$$ which is symmetric in $A$, $B$, $C$.

Finish. The same computation for the other two sides gives the same value. So all six points are at the same distance from $O$: they lie on one circle centered at $O$.

Key idea: with the right substitution, the constraint becomes linear in the pair sums, and the left side becomes a perfect square plus $1$.

(a) Let $a=\dfrac x{x-1}$, $b=\dfrac y{y-1}$, $c=\dfrac z{z-1}$. Then $a\ne1$ and $x=\dfrac a{a-1}$, and similarly for $y,z$. The condition $xyz=1$ becomes $abc=(a-1)(b-1)(c-1)$, which expands to $$ab+bc+ca=a+b+c-1 .$$ Let $s=a+b+c$. Then $$a^2+b^2+c^2=s^2-2(s-1)=(s-1)^2+1\ \ge1 .$$

(b) Equality holds when $s=1$, and then $ab+bc+ca=0$. For a rational $r\notin\{0,-1\}$, set $$a=\frac{1+r}{r^2+r+1},\qquad b=\frac{r(1+r)}{r^2+r+1},\qquad c=\frac{-r}{r^2+r+1}.$$ Then $a+b+c=1$, and $ab+bc+ca=0$ (with $b=ra$, the condition $ab+(a+b)c=0$ is exactly what fixed $a$). None of $a,b,c$ equals $1$: $a=1$ needs $r=0$, $b=1$ never happens, and $c=1$ needs $r=-1$. So $x=\frac a{a-1}$, $y=\frac b{b-1}$, $z=\frac c{c-1}$ are rational, different from $1$, satisfy $xyz=1$, and give equality.

Different values of $r$ give different values of $c$, except that each value of $c$ comes from at most two values of $r$. So there are infinitely many such triples.

Key idea: for a prime $p\equiv1\pmod4$, choose the smaller square root of $-1$ mod $p$. The gap $d=p-2n$ satisfies $p\mid d^2+4$, so it can’t be small.

Ingredients. There are infinitely many primes $p\equiv1\pmod4$. (If $p_1,\dots,p_k$ were all, every prime factor of $(2p_1\cdots p_k)^2+1$ would be odd and $\equiv1\pmod4$, but none is on the list.) For each such $p$, $-1$ is a square mod $p$: for example $\big(\tfrac{p-1}2!\big)^2\equiv-1$ by Wilson’s theorem. Replacing $n$ by $p-n$ if needed, there is $n$ with $1\le n\le\frac{p-1}2$ and $p\mid n^2+1$.

The gap. Let $d=p-2n\ge1$. Then $$d^2+4\equiv4n^2+4=4(n^2+1)\equiv0\pmod p,$$ so $d^2+4\ge p=2n+d$. Since $d$ is odd, $d\in\{1,3\}$ would give $p\mid5$ or $p\mid13$. So for $p>13$, $d\ge5$, and then $$d^2\ge2n+d-4>2n,$$ so $d>\sqrt{2n}$ and $p=2n+d>2n+\sqrt{2n}$.

Infinitely many $n$. Each prime $p>13$ with $p\equiv1\pmod4$ gives such an $n$, and $n^2+1\ge p$, so $n\ge\sqrt{p-1}$. As $p$ grows, so does $n$, and we get infinitely many different $n$.

Key idea: a few substitutions show that $f(t)$ is $t$ or $\tfrac1t$ for each $t$. One more substitution shows that the choice can’t change from point to point.

Values. $w=x=y=z=1$: $\dfrac{2f(1)^2}{2f(1)}=1$, so $f(1)=1$. Then $w=x=y=z=t$: $\dfrac{2f(t)^2}{2f(t^2)}=1$, so $f(t^2)=f(t)^2$. Then $w=t$, $x=1$, $y=z=\sqrt t$ (so $wx=yz$): $$\frac{f(t)^2+1}{2f(t)}=\frac{t^2+1}{2t},\quad\text{so}\quad f(t)+\frac1{f(t)}=t+\frac1t .$$ The map $u\mapsto u+\frac1u$ takes each value at $u$ and $\frac1u$ only, so $f(t)\in\{t,\frac1t\}$ for every $t$.

No mixing. Suppose $f(a)=a\ne\frac1a$ and $f(b)=\frac1b\ne b$; so $a\ne1$ and $b\ne1$. Put $w=a$, $x=b$, $y=z=\sqrt{ab}$: $$\frac{a^2+b^{-2}}{2f(ab)}=\frac{a^2+b^2}{2ab}.$$

• If $f(ab)=ab$: $a^2+b^{-2}=a^2+b^2$, so $b=1$. Contradiction.
• If $f(ab)=\frac1{ab}$: $(a^2+b^{-2})a^2b^2=a^2+b^2$, so $a^4b^2=b^2$ and $a=1$. Contradiction.

So either $f(t)=t$ for all $t$, or $f(t)=\frac1t$ for all $t$ (at $t=1$ the two agree).

Check. $f(t)=t$ clearly works. For $f(t)=\frac1t$, the left side is $\dfrac{(w^2+x^2)/(w^2x^2)}{(y^2+z^2)/(y^2z^2)}=\dfrac{w^2+x^2}{y^2+z^2}$ because $wx=yz$.

Key idea: fold the lamps $n+i$ onto the lamps $i$. Every restricted sequence has exactly $2^{k-n}$ unrestricted sequences folding onto it.

The map. Given a sequence counted by $N$, replace every switch of lamp $n+i$ by a switch of lamp $i$. In the original, lamp $i$ is switched an odd number of times and lamp $n+i$ an even number, so in the new sequence lamp $i$ is switched an odd number of times and lamps $n+1,\dots,2n$ never. So the new sequence is counted by $M$.

Fibers. Fix a sequence counted by $M$, where lamp $i$ is switched $c_i$ times ($c_i$ odd, $\sum c_i=k$). Its preimages come from choosing, for each $i$, which of the $c_i$ switches of lamp $i$ should really be switches of lamp $n+i$. Lamp $n+i$ must end off, so we choose an even-sized subset. A set of $c_i\ge1$ elements has $2^{c_i-1}$ even-sized subsets. So the number of preimages is $$\prod_i2^{c_i-1}=2^{k-n}.$$

Hence $N=2^{k-n}M$.

Key idea: the point $K$ of $\omega$ nearest to $AC$ is the external center of similitude of the two incircles. Homotheties at $B$ and at $D$ put $K$ on two lines through that center.

Measure “height” as signed distance from line $AC$, positive on $D$’s side, so $B$ has negative height. Let $\omega_1$, $\omega_2$ (radii $r_1$, $r_2$) touch $AC$ at $T_1$, $T_2$. Let $T_1'$ be the lowest point of $\omega_1$ (opposite $T_1$) and $T_2'$ the highest point of $\omega_2$ (opposite $T_2$). Let $\omega$ have radius $r$ and touch lines $BA$, $BC$, $AD$, $CD$ at $P$, $Q$, $R$, $S$.

Step 1: $\omega$ sits beyond $D$.

• $\omega$ lies on one side of its tangent line $AB$, and it contains $Q$, which is on $C$’s side. By convexity, $C$ and $D$ are on the same side of $AB$.
• $AD$ is the other tangent from $A$. If $\omega$ were on $B$’s side of $AD$, it would lie in the angle $DAB$, and its touching point with line $AB$ would be on ray $AB$. But $P$ is on the opposite ray, beyond $A$. So $\omega$ is on the side of $AD$ away from $B$, and hence away from $C$ (convexity puts $B$, $C$ on one side of $AD$).
• In the same way, $\omega$ is on the side of $CD$ away from $B$, hence away from $A$.

So $\omega$ lies in the angle vertical to angle $ADC$. Every point of that angle other than $D$ is higher than $D$, so $\omega$ lies entirely above $D$, and $R$, $S$ lie on rays $AD$, $CD$ beyond $D$.

Step 2: $AB+AD=CB+CD$. Tangent segments from one point are equal, so $$BA+AD+DR=BA+AR=BA+AP=BP=BQ=BC+CQ=BC+CS=BC+CD+DS,$$ and $DR=DS$. Hence $AB-CB=CD-AD$, and the incircle tangent lengths give $$AT_1=\tfrac12(AB+AC-BC)=\tfrac12(CD+AC-AD)=CT_2 .$$ So $T_1$ and $T_2$ are symmetric about the midpoint of $AC$. They are different points, since $T_1$ is the midpoint only if $AB=BC$.

Step 3: two lines through $K$. Let $K$ be the lowest point of $\omega$.

• Homothety at $B$, $\omega_1\to\omega$. Both circles touch both rays from $B$, so a homothety with center $B$ and positive ratio maps $\omega_1$ to $\omega$. It keeps lines parallel to $AC$ parallel and keeps their order by height, so it maps the lowest point $T_1'$ to the lowest point $K$. So $B$, $T_1'$, $K$ are collinear.
• Homothety at $B$, $\omega_1\to$ excircle. The excircle of $ABC$ opposite $B$ touches $AC$ at the point $E$ with $AE=\tfrac12(BC+CA-AB)=CT_1$. By Step 2, $AT_2=CT_1$, so $E=T_2$. The positive homothety at $B$ taking $\omega_1$ to this excircle maps the lowest point $T_1'$ to the excircle’s lowest point, which is $E=T_2$. So $B$, $T_1'$, $T_2$ are collinear.
• Homothety at $D$, $\omega_2\to\omega$. $\omega_2$ lies in angle $ADC$ and $\omega$ in the vertical angle, both touching both lines. So the homothety with center $D$ and ratio $-r/r_2$ maps $\omega_2$ to $\omega$. It sends height $y$ to $d+\tfrac r{r_2}(d-y)$, where $d$ is the height of $D$. That reverses the order, so the highest point $T_2'$ goes to the lowest point $K$. So $D$, $T_2'$, $K$ are collinear.
• Homothety at $D$, $\omega_2\to$ excircle. The excircle of $ADC$ opposite $D$ touches $AC$ at $E'$ with $CE'=\tfrac12(DA+AC-DC)=AT_2=CT_1$, so $E'=T_1$. The positive homothety at $D$ maps the highest point $T_2'$ of $\omega_2$ to the highest point of that excircle, which is $E'=T_1$. So $D$, $T_2'$, $T_1$ are collinear.

Together: line $T_1'T_2$ passes through $K$ (it is line $BT_1'$), and line $T_1T_2'$ passes through $K$ (it is line $DT_2'$). These lines differ: otherwise $T_1,T_1',T_2,T_2'$ would lie on one line perpendicular to $AC$, forcing $T_1=T_2$.

Step 4: the external center is $K$. The circles $\omega_1$, $\omega_2$ lie on opposite sides of $AC$ and touch it at different points, so they are disjoint. If $r_1=r_2$, the translation taking $\omega_1$ to $\omega_2$ maps $T_1'\mapsto T_2$ and $T_1\mapsto T_2'$, so lines $T_1'T_2$ and $T_1T_2'$ would be parallel. But they are different lines through $K$. So $r_1\ne r_2$. The homothety with positive ratio $r_2/r_1$ taking $\omega_1$ to $\omega_2$ sends lowest point to lowest point ($T_1'\mapsto T_2$) and highest to highest ($T_1\mapsto T_2'$). So its center lies on both lines, and it must be their common point $K$.

That center is where the two external common tangents meet. So they meet at $K$, which is on $\omega$.