Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.3 National olympiad Prove it Romania

Let $A_1$ , $B_1$ and $C_1$ be the feet of the altitudes in the acute triangle $ABC$ . On the line segments $B_1C_1$ , $C_1A_1$ , $A_1B_1$ one considers the points $X$ , $Y$ , and $Z$ , respectively, such that
$\frac{C_1X}{XB_1} = \frac{b \cos C}{c \cos B}, \quad \frac{A_1Y}{YC_1} = \frac{c \cos A}{a \cos C} \quad \text{and} \quad \frac{B_1Z}{ZA_1} = \frac{a \cos B}{b \cos A}.$
Prove that the lines $AX$ , $BY$ and $CZ$ are concurrent.

Solution

Since $b \cos C = A_1C$ and $c \cos B = BA_1$ , we deduce that $\frac{C_1X}{XB_1} = \frac{CA_1}{A_1B}$ or $\frac{C_1X}{C_1B_1} = \frac{CA_1}{CB}$ or, equivalently, $\frac{C_1X}{CA_1} = \frac{C_1B_1}{CB}$ . (1)

Triangles $AC_1B_1$ and $ACB$ being similar, we obtain $\frac{C_1B_1}{CB} = \frac{AC_1}{AC}$ , hence $\frac{C_1X}{CA_1} = \frac{AC_1}{AC}$ . Also, $\angle AC_1X = \angle ACA_1$ , so it follows that triangles $AC_1X$ and $ACA_1$ are similar, and we deduce that $AX \perp B_1C_1$ .

On the other hand, the tangent at $A$ to the circumcircle of the triangle $ABC$ , centered at $O$ , is parallel to $B_1C_1$ , and hence $O$ lies on $AX$ .

Similarly, $O$ lies on $BY$ and $CZ$ as well, therefore the lines $AX$ , $BY$ and $CZ$ are concurrent.

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