IMO 2006

Key idea: $P$ sees $BC$ at the same angle as the incenter does, so $P$ lies on the circle $BIC$. The center of that circle is on line $AI$, and the triangle inequality finishes.

Where $P$ is. The four angles $\angle PBA$, $\angle PBC$, $\angle PCA$, $\angle PCB$ add up to $\angle B+\angle C$, and the condition splits them into two groups with equal sums. So $$\angle PBC+\angle PCB=\tfrac12(\angle B+\angle C),\qquad \angle BPC=180^\circ-\tfrac12(\angle B+\angle C)=90^\circ+\tfrac12\angle A .$$ The incenter has the same angle: $\angle BIC=90^\circ+\tfrac12\angle A$. Both $P$ and $I$ are on the same side of $BC$, so $P$ lies on the circle $\omega$ through $B$, $I$, $C$.

The center of $\omega$. Let $M$ be the second point where line $AI$ meets the circumcircle of $ABC$; it is the midpoint of arc $BC$. A classical fact: $MB=MI=MC$. (Angle chase: $\angle MBI=\tfrac12\angle A+\tfrac12\angle B=\angle MIB$.) So $M$ is the center of $\omega$, and $MP=MI$.

Finish. $A$, $I$, $M$ lie on a line in this order, so $AM=AI+IM$. By the triangle inequality, $$AP+PM\ge AM=AI+IM=AI+PM,$$ so $AP\ge AI$. Equality means $P$ lies on segment $AM$. The only point of $\omega$ on that segment is $I$, so equality holds only for $P=I$.

Key idea: each good isosceles triangle can claim two polygon sides of its own, one under each leg. With $2006$ sides, at most $1003$ triangles. A fan of small ear triangles reaches $1003$.

Arcs. A diagonal or side splits the boundary into two arcs. The diagonals don’t cross, so for any two of them the arcs cut off are nested or disjoint. A triangle $XYZ$ of the dissection splits the boundary into three arcs, of $a$, $b$, $c$ sides with $a+b+c=2006$. Equal chords in a regular polygon cut off equal or complementary arcs, and complementary would leave the third arc empty. So in an isosceles triangle with apex $X$, the arcs under the legs $XY$ and $XZ$ have the same length $a$. If the triangle has two good sides, two arcs are odd and one is even (their sum is even), and the equal arcs must be the odd ones. So: a good triangle has legs cutting off two equal odd arcs, and its base cuts off the even remaining arc. Call the arcs under the legs its leg arcs.

Leg arcs of triangles inside. Fix a good triangle $T$ and one leg arc $\Lambda$ of $T$, with $a$ sides, $a$ odd. Look at the good triangles lying in the region cut off by that leg. For each of them, both leg arcs lie inside $\Lambda$ (the other arc of such a triangle contains everything outside $\Lambda$, so it has at least $2006-a\ge1005$ sides, while a leg arc has at most $1001$ sides, since two leg arcs and a nonempty base arc add up to $2006$; so the outside arc is the base arc). Take the maximal ones among all these leg arcs. They are pairwise disjoint, and they come in pairs: if one leg arc of a triangle $T'$ sat inside another triangle’s leg arc, the whole triangle $T'$, and so its other leg arc, would sit inside too. Each pair has two arcs of equal length, so the maximal arcs cover an even number of sides. Since $a$ is odd, at least one side of $\Lambda$ is not under any leg arc of a smaller good triangle. Assign that side to $T$.

Doing this for both legs gives $T$ two sides.

No side is claimed twice. Suppose side $s$ is assigned to $T_1$ and $T_2$, through leg arcs $\Lambda_1$ and $\Lambda_2$ that both contain $s$. They are nested; they can’t be equal, because the two triangles on either side of one diagonal have that diagonal’s two complementary arcs as their arcs. Say $\Lambda_2\subsetneq\Lambda_1$. The triangle $T_2$ lies on the far side of its leg from $\Lambda_2$, which is still inside the region cut off by $T_1$’s leg. So $s$ lies under a leg arc of a smaller good triangle inside $\Lambda_1$, and it was not assigned to $T_1$. Contradiction.

Upper bound. Each good isosceles triangle owns two sides, no side is owned twice, so there are at most $2006/2=1003$ of them.

Construction. Number the vertices $P_0,\dots,P_{2005}$. Draw the $1003$ diagonals $P_{2i}P_{2i+2}$ (indices mod $2006$). They cut off $1003$ triangles $P_{2i}P_{2i+1}P_{2i+2}$, each isosceles with two polygon sides as legs, and sides are good. The middle region is a $1003$-gon, which we split by $1000$ more diagonals from one vertex. In total that is $1003+1000=2003$ diagonals, and $1003$ good isosceles triangles.

Key idea: the left side factors. Then everything depends on two quantities, the spread $d$ and the sum $s$, and a weighted AM–GM finds the best constant.

Factor. The left side $L$ vanishes whenever two of the variables are equal, so it is divisible by $(a-b)(b-c)(c-a)$, and the remaining factor has degree $1$ and is symmetric, so it is a multiple of $a+b+c$. Comparing the coefficient of $a^3b$ (which is $1$) gives $$L=-(a-b)(b-c)(c-a)(a+b+c).$$

Bounds. $|L|$ is symmetric, so assume $a\le b\le c$, and let $d=c-a$, $s=a+b+c$, $Q=a^2+b^2+c^2$.

• By AM–GM, $(b-a)(c-b)\le\big(\tfrac{c-a}2\big)^2$, so $|L|\le\tfrac14d^3|s|$.
• $3Q=s^2+(a-b)^2+(b-c)^2+(c-a)^2$, and $(a-b)^2+(b-c)^2\ge\tfrac12d^2$, so $3Q\ge s^2+\tfrac32d^2$.

AM–GM. Split $\tfrac32d^2$ into three equal parts: $$s^2+\tfrac{d^2}2+\tfrac{d^2}2+\tfrac{d^2}2\ \ge\ 4\sqrt[4]{\tfrac{s^2d^6}8}.$$ So $\tfrac{s^2d^6}8\le\big(\tfrac{3Q}4\big)^4$, which gives $|s|\,d^3\le\tfrac{9\sqrt2}8Q^2$, and then $$|L|\le\frac{9\sqrt2}{32}\,(a^2+b^2+c^2)^2 .$$

Sharp. Equality needs $b$ to be the midpoint of $a$ and $c$, and $s^2=\tfrac{d^2}2$. Take $b=1$, $d=3\sqrt2$: the triple $a=1+\tfrac{3\sqrt2}2$, $b=1$, $c=1-\tfrac{3\sqrt2}2$ gives $L=\tfrac{9\sqrt2}{32}Q^2$ with a plus sign. So the least constant is $M=\dfrac{9\sqrt2}{32}$.

Key idea: factor $y^2-1$. The two factors share only one $2$, so one of them is a big power of two times an odd number. Substituting leaves a tiny equation.

Small and negative $x$. For $x\le-2$ the left side is strictly between $1$ and $2$, not an integer. For $x=-1$ it is $2$, not a square. For $x=0$ it is $4$: $y=\pm2$. For $x=1,2$ it is $11$ and $37$, not squares.

$x\ge3$. By symmetry take $y>0$. Then $y$ is odd and $$(y-1)(y+1)=2^x\big(1+2^{x+1}\big).$$ $\gcd(y-1,y+1)=2$, so one factor is divisible by $2^{x-1}$. Write $y=2^{x-1}m+\varepsilon$ with $\varepsilon=\pm1$ and $m$ odd; $m\ge1$ since $y>0$. Then $$y^2-1=2^{2x-2}m^2+2^x\varepsilon m=2^x\big(1+2^{x+1}\big)\iff 2^{x-2}\big(m^2-8\big)=1-\varepsilon m .$$

• $\varepsilon=1$: the right side is $1-m\le0$, so $m^2\le8$, $m=1$, and then $-7\cdot2^{x-2}=0$: impossible.
• $\varepsilon=-1$: $2^{x-2}(m^2-8)=1+m$. $m=1$ gives a negative left side. For $m\ge5$, the left side is at least $2(m^2-8)>m+1$. So $m=3$, and $2^{x-2}=4$, $x=4$, $y=8\cdot3-1=23$.

Check. $1+2^4+2^9=529=23^2$. So the solutions are $(0,\pm2)$ and $(4,\pm23)$.

Key idea: integer polynomials can only have cycles of length $1$ or $2$ on the integers. Then all the solutions sit on the roots of one polynomial of degree $n$.

Step 1: short cycles. Let $t$ be an integer with $Q(t)=t$. Then $t$ lies on a cycle $x_0\to x_1\to\dots\to x_{m}=x_0$ of $P$, with $x_{i+1}=P(x_i)$ and $m\mid k$. Let $d_i=x_{i+1}-x_i$. Since $x_{i+1}-x_i$ divides $P(x_{i+1})-P(x_i)=x_{i+2}-x_{i+1}$, each $d_i$ divides $d_{i+1}$, and going around the cycle, all $|d_i|$ are equal. If the cycle has more than one point, the $d_i$ are nonzero and add up to $0$, so not all have the same sign: some $d_{i+1}=-d_i$, that is, $x_{i+2}=x_i$. So the cycle has length $1$ or $2$. In every case $P(P(t))=t$.

Step 2a: only fixed points. If every such $t$ has $P(t)=t$, they are roots of $P(x)-x$, which has degree $n$. At most $n$ of them.

Step 2b: a $2$-cycle exists. Say $P(a)=b$, $P(b)=a$ with $a\ne b$. Let $t$ be any integer with $P(P(t))=t$, and let $s=P(t)$. Then $t-a\mid s-b$ and $s-b\mid t-a$, so $|t-a|=|s-b|$. In the same way $|t-b|=|s-a|$. If both held with a plus sign, $t-a=s-b$ and $t-b=s-a$, then subtracting gives $a=b$. So one holds with a minus sign, and either way $t+s=a+b$. Hence every such $t$ is a root of $$P(x)+x-(a+b),$$ which has degree $n$ because $n>1$. At most $n$ of them.

Key idea: the sum of the assigned areas is exactly half the area of the difference set $Q=\{x-y:x,y\in P\}$. Slicing shows $Q$ has at least four times the area of $P$.

Write $|P|$ for area. For a unit vector $u$, let $W(u)=\max_{x\in P}x\cdot u-\min_{x\in P}x\cdot u$ be the width of $P$ in direction $u$.

Step 1: the assigned area. A triangle with side $b$ inside $P$ has its third vertex in $P$, so its area is $\tfrac12|b|$ times that vertex’s distance from line $b$. The polygon lies on one side of that line, so the largest distance is the width across $b$. With $u_b$ the outward unit normal of side $b$, the assigned area is $$S(b)=\tfrac12|b|\,W(u_b).$$

Step 2: the shape of $Q$. $Q$ is a convex polygon, symmetric about the origin. For a unit vector $u$:

• The largest value of $q\cdot u$ on $Q$ is $\max_P x\cdot u-\min_P y\cdot u=W(u)$. So the side of $Q$ facing $u$ (if any) is at distance $W(u)$ from the origin.
• The points of $Q$ where $q\cdot u$ is largest are the differences $x-y$ with $x$ maximizing $x\cdot u$ on $P$ and $y$ minimizing it. Let $\lambda(u)$ be the length of the side of $P$ with outward normal $u$ ($0$ if there is none). Then the side of $Q$ facing $u$ has length $\lambda(u)+\lambda(-u)$.

Step 3: $|Q|=\sum_b|b|\,W(u_b)$. Cut $Q$ into triangles from the origin to its sides. Let $U$ be the set of the normals $u_b$ and their negatives; these are the directions in which $Q$ has a side. Then $$|Q|=\sum_{u\in U}\tfrac12\big(\lambda(u)+\lambda(-u)\big)W(u).$$ In the part with $\lambda(-u)$, rename $-u$ as $u$; since $W(-u)=W(u)$, it equals the part with $\lambda(u)$. So $|Q|=\sum_{u}\lambda(u)W(u)=\sum_b|b|\,W(u_b)$. By Step 1, $$\sum_b S(b)=\tfrac12|Q| .$$

Step 4: $|Q|\ge4|P|$. Put $P$ between heights $\alpha$ and $\beta$, and let $f(y)$ be the length of its horizontal chord at height $y$. By convexity $f$ is continuous, and positive for $\alpha<y<\beta$. Let $A=|P|$. For $0<t<1$ choose heights $y_1(t)$ and $y_2(t)$ so that the part of $P$ below $y_1$ has area $tA$ and the part above $y_2$ has area $tA$. Then $$y_1'=\frac{A}{f(y_1)},\qquad y_2'=-\frac{A}{f(y_2)} .$$ At height $Y=y_1-y_2$, the chord of $Q$ contains every difference of a point on the chord of $P$ at $y_1$ and a point on the chord at $y_2$. These differences form a segment of length $f(y_1)+f(y_2)$. As $t$ runs from $0$ to $1$, $Y(t)$ increases from $\alpha-\beta$ to $\beta-\alpha$, with $Y'=A/f(y_1)+A/f(y_2)$. Adding up chord lengths, $$|Q|\ge\int_0^1\big(f(y_1)+f(y_2)\big)\Big(\frac{A}{f(y_1)}+\frac{A}{f(y_2)}\Big)dt\ge\int_0^1 4A\,dt=4A,$$ because $(p+q)\big(\tfrac1p+\tfrac1q\big)\ge4$ for positive $p,q$.

Conclusion. $\sum_b S(b)=\tfrac12|Q|\ge2|P|$. (Equality holds for centrally symmetric polygons, where $Q$ is $P$ scaled by $2$.)