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1989 IMO Problems/Problem 2

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) $ABC$ 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 $\sum\frac{1}{2}ar\left(\frac{a+b+c}{b+c-a}\right)$.

Now we can prove that $A_0B_0C_0 = 2 \cdot$ area of hexagon $AC_1BA_1CB_1$ by simplifying. We have $A_0B_0C_0 -2 \cdot$ area of hexagon $AC_1BA_1CB_1$ $=-[ABC]+\sum\left(\frac{1}{2}ar\left(\frac{a+b+c}{b+c-a}\right)-\frac{1}{2}a^2\left(\frac{2r}{b+c-a}\right)\right)$ $=-[ABC]+\sum\left(\frac{1}{2}ar\left(\frac{a+b+c-2a}{b+c-a}\right)\right)$ $=-[ABC]+\sum\left(\frac{1}{2}ar\right)$ $=0$

as desired.

As for the inequality, notice that it is equivalent to $[ABC]+\sum\frac{1}{2}ar\left(\frac{a+b+c}{b+c-a}\right)\ge 4[ABC]$ $\Leftrightarrow \sum\frac{1}{2}ar\left(\frac{a+b+c}{b+c-a}\right)\ge \frac{3}{2}r(a+b+c)$ $\Leftrightarrow \sum\frac{a}{b+c-a}\ge 3$

Letting $a=x+y,b=y+z,c=z+x$ for $x,y,z$ positive reals, the inequality becomes $\sum\frac{x+y}{2z}\ge 3$ $\Leftrightarrow \sum\frac{x+y+z}{z}\ge 9$

which is true by AM-HM.