IMO 1989 · Problem 2

In an acute-angled triangle \(ABC\) the internal bisector of angle \(A\) meets the circumcircle of the triangle again at \(A_1\). Points \(B_1\) and \(C_1\) are defined similarly. Let \(A_0\) be the point of intersection of the line \(AA_1\) with the external bisectors of angles \(B\) and \(C\). Points \(B_0\) and \(C_0\) are defined similarly. Prove that:

(i) The area of the triangle \(A_0B_0C_0\) is twice the area of the hexagon \(AC_1BA_1CB_1\).

(ii) The area of the triangle \(A_0B_0C_0\) is at least four times the area of the triangle \(ABC\).

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

Solution

Write \([P_1\cdots P_k]\) for the area of a polygon, and let \(I\) be the incentre of \(ABC\).

The points \(A_0,B_0,C_0\) are the excentres opposite \(A,B,C\), respectively. Thus \(A_0\) lies beyond \(BC\) on the ray \(AI\), and cyclically. Both \(B_0\) and \(C_0\) lie on the external bisector at \(A\), on opposite sides of \(AB\); consequently \(A\) lies between them. Similarly, \[ A\in B_0C_0,\qquad B\in C_0A_0,\qquad C\in A_0B_0. \] In particular, the excentral triangle contains \(ABC\) and \(I\).

(i) We first prove that \(A_1\) is the midpoint of \(IA_0\).

Put \(\alpha=\angle BAC\) and \(\beta=\angle ABC\). The angle-bisector property shows that \(A_1\) is the midpoint of the arc \(BC\) not containing \(A\). By the inscribed-angle theorem, \[ \angle CBA_1=\angle CAA_1=\frac{\alpha}{2}. \] Since \(A,I,A_1\) occur in that order, we obtain \[ \angle IBA_1=\frac{\beta}{2}+\frac{\alpha}{2}, \qquad \angle BIA_1 =180^\circ-\angle BIA =\frac{\alpha+\beta}{2}. \] Hence \(A_1I=A_1B\).

The internal and external bisectors at \(B\) are perpendicular, so \(IB\perp BA_0\). Therefore the midpoint \(M\) of the hypotenuse \(IA_0\) of the right triangle \(IBA_0\) satisfies \(MI=MB\). Both \(M\) and \(A_1\) lie on \(AI\) and on the perpendicular bisector of \(IB\). These two lines are distinct, since \(I\) belongs only to the first, so their intersection is unique. Thus \(A_1=M\).

Cyclically, we have proved \[ A_1=\operatorname{midpoint}(IA_0),\qquad B_1=\operatorname{midpoint}(IB_0),\qquad C_1=\operatorname{midpoint}(IC_0). \]

The six points occur on the circumcircle in the order \[ A,C_1,B,A_1,C,B_1. \] Their convex hexagon contains \(ABC\), hence contains \(I\). Joining \(I\) to its vertices partitions it into six triangles. The midpoint identities give, for example, \[ [IAC_1]=\frac12[IAC_0], \qquad [IC_1B]=\frac12[IC_0B]. \] Adding these relations and their cyclic counterparts yields \[ \begin{aligned} [AC_1BA_1CB_1] &=\frac12\bigl( [IAC_0]+[IC_0B]+[IBA_0]\\ &\hspace{35mm}+[IA_0C]+[ICB_0]+[IB_0A] \bigr)\\ &=\frac12[A_0B_0C_0], \end{aligned} \] because the six triangles inside the parentheses partition the excentral triangle. This proves (i).

(ii) Put \[ a=BC,\qquad b=CA,\qquad c=AB,\qquad s=\frac{a+b+c}{2},\qquad \Delta=[ABC]. \] Let \(r_a\) be the common perpendicular distance from \(A_0\) to the three side lines of \(ABC\), and define \(r_b,r_c\) cyclically.

Since \(A_0\) lies beyond \(BC\) on the internal bisector from \(A\), \[ [A_0AB]+[A_0AC]=[ABC]+[A_0BC]. \] Expressing these areas using their bases on the side lines gives \[ \frac{cr_a}{2}+\frac{br_a}{2} =\Delta+\frac{ar_a}{2}. \] Consequently, \[ r_a=\frac{\Delta}{s-a},\qquad r_b=\frac{\Delta}{s-b},\qquad r_c=\frac{\Delta}{s-c}. \]

The excentral triangle is partitioned into \(ABC\) and the three triangles \(A_0BC,B_0CA,C_0AB\). Therefore \[ \frac{[A_0B_0C_0]}{\Delta} =1+\frac12\left( \frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c} \right). \] Set \[ x=s-a,\qquad y=s-b,\qquad z=s-c. \] These numbers are strictly positive by the triangle inequalities, and \[ a=y+z,\qquad b=z+x,\qquad c=x+y. \] Thus \[ \begin{aligned} \frac{[A_0B_0C_0]}{\Delta} &=1+\frac12\left( \frac{x}{y}+\frac{y}{x} +\frac{y}{z}+\frac{z}{y} +\frac{z}{x}+\frac{x}{z} \right)\\ &\ge 1+\frac12(2+2+2)=4. \end{aligned} \] Here each paired inequality follows from \[ \frac{u}{v}+\frac{v}{u}-2 =\frac{(u-v)^2}{uv}\ge0 \qquad(u,v>0). \] This proves (ii). Equality holds precisely when \(x=y=z\), equivalently when \(ABC\) is equilateral.

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