1989 IMO Problems/Problem 2

$ABC$ is a triangle, the bisector of angle $A$ meets the circumcircle of triangle $ABC$ in $A_1$ , points $B_1$ and $C_1$ are defined similarly. Let $AA_1$ meet the lines that bisect the two external angles at $B$ and $C$ in $A_0$ . Define $B_0$ and $C_0$ similarly. Prove that the area of triangle $A_0B_0C_0 = 2 \cdot$ area of hexagon $AC_1BA_1CB_1 \geq 4 \cdot$ area of triangle $ABC$ .

Solution

Notice that since $A_1C$ and $A_1B$ substend the same angle in the circle, and so are equal. Thus $A_1BC$ is iscoceles, and similarly for triangles $AB_1C$ and $ABC_1$ . Also, since $AA_1BC$ is a cyclic quadrilatera, $\angle A_1BC=\angle A_1AC=\frac{A}{2}$ , and similarly for the other triangles. Thus, the area of triangle $A_1BC=\frac{1}{4}a^2\tan\frac{A}{2}=\frac{1}{4}a^2\left(\frac{2r}{b+c-a}\right)$ and similarly for the other triangles. Thus, the area of the hexagon is equal to $[ABC]+\sum\frac{1}{4}a^2\left(\frac{2r}{b+c-a}\right)$ .

Now we shall find the area of triangle $A_0B_0C_0$ . It is obvious that the points $A_1,B,C_1$ are collinear since the angle the make at $B$ is $\frac{1}{2}(180-B)+\frac{1}{2}(180-B)+B=180$ , and similarly for the other points. Thus, it suffices to find the area of each of the triangles $A_0BC$ . It is well known that $A_0$ is the center of the excircle $O_a$ with radius $R_a=r\left(\frac{a+b+c}{b+c-a}\right)$ . Thus the area of triangle $A_0BC$ is $\frac{1}{2}ar\left(\frac{a+b+c}{b+c-a}\right)$ and so the area of triangel $A_0B_0C_0$ is $\sum\frac{1}{2}ar\left(\frac{a+b+c}{b+c-a}\right)$ .