IMO 1996 · Problem 5

Let \(ABCDEF\) be a convex hexagon such that \(AB\) is parallel to \(DE\), \(BC\) is parallel to \(EF\), and \(CD\) is parallel to \(FA\). Let \(R_A,R_C,R_E\) denote the circumradii of triangles \(FAB,BCD,DEF\), respectively, and let \(P\) denote the perimeter of the hexagon. Prove that \[ R_A+R_C+R_E\ge \frac{P}{2}. \]

Official text of the problem. Official problem paper (PDF)

Solution

Write \[ a=AB,\quad b=BC,\quad c=CD,\quad d=DE,\quad e=EF,\quad f=FA. \] Orient the boundary counterclockwise. Convexity and the parallelisms imply that the successive directed sides have directions \[ 0,\quad \alpha,\quad \alpha+\beta,\quad \pi,\quad \pi+\alpha,\quad \pi+\alpha+\beta, \] where \[ \alpha>0,\qquad \beta>0,\qquad \gamma:=\pi-\alpha-\beta>0. \] Indeed, opposite parallel sides have opposite directions. The exterior angles therefore repeat in opposite pairs. These angles are positive because the triangles whose circumradii occur are nondegenerate.

Put \[ x=\sin\alpha,\qquad y=\sin\beta,\qquad z=\sin\gamma, \] so \(x,y,z>0\).

First, projecting \(FB\), \(BD\), and \(DF\) onto lines perpendicular to \(BC\), \(AB\), and \(CD\), respectively, gives \[ FB\ge ax+fy,\qquad BD\ge bx+cz,\qquad DF\ge dz+ey. \] For example, \(\overrightarrow{FB}=\overrightarrow{FA}+\overrightarrow{AB}\), and its two summands have components \(fy\) and \(ax\) in the same direction perpendicular to \(BC\).

The interior angles at \(A,C,E\) are respectively \(\pi-\gamma,\pi-\beta,\pi-\alpha\). Hence the extended sine rule yields \[ 2(R_A+R_C+R_E) =\frac{FB}{z}+\frac{BD}{y}+\frac{DF}{x} \ge S, \] where \[ S:=\frac{ax+fy}{z}+\frac{bx+cz}{y}+\frac{dz+ey}{x}. \] We will prove \(S\ge P\).

Let \(\mathbf u,\mathbf v,\mathbf w\) be unit vectors in the directions of \(AB,BC,CD\). Closure of the hexagon gives \[ (a-d)\mathbf u+(b-e)\mathbf v+(c-f)\mathbf w=\mathbf0. \] Taking components perpendicular to \(\mathbf u\) and \(\mathbf v\), respectively, gives \[ (b-e)x+(c-f)z=0,\qquad -(a-d)x+(c-f)y=0. \] Consequently, for some real number \(t\), \[ a-d=ty,\qquad e-b=tz,\qquad c-f=tx. \tag{1} \]

Set \[ a_0=\min(a,d),\qquad b_0=\min(b,e),\qquad c_0=\min(c,f),\qquad h=|t|. \] Equation (1) implies \[ P=2(a_0+b_0+c_0)+h(x+y+z). \tag{2} \] Also, regrouping the terms of \(S\) by opposite sides gives \[ S=a\frac{x}{z}+d\frac{z}{x} +b\frac{x}{y}+e\frac{y}{x} +c\frac{z}{y}+f\frac{y}{z}. \] If \(t\ge0\), the excesses over the three minima occur in \(a,e,c\); if \(t\le0\), they occur in \(d,b,f\). In either case, substitution from (1) gives \[ \begin{aligned} S={}&a_0\left(\frac{x}{z}+\frac{z}{x}\right) +b_0\left(\frac{x}{y}+\frac{y}{x}\right) +c_0\left(\frac{z}{y}+\frac{y}{z}\right)\\ &+h\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right). \end{aligned} \] Each parenthesis on the first line is at least \(2\). Moreover, \[ \frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}-(x+y+z) = \frac{(xy-yz)^2+(yz-zx)^2+(zx-xy)^2}{2xyz} \ge0. \] Together with (2), these inequalities show that \(S\ge P\). Therefore \[ \boxed{R_A+R_C+R_E\ge\frac{P}{2}}. \]

Equality is attained for a regular hexagon.

The solution was checked against John Scholes' IMO solution archive (Kalva).