Vietnamese MO

Geometry Difficulty 6.8 National Olympiad Prove it Vietnam

Let $ABC$ be an acute triangle with circumcenter $O$ . Let $A'$ be the center of the circle passing through $C$ and tangent to $AB$ at $A$ , let $B'$ be the center of the circle passing through $A$ and tangent to $BC$ at $B$ , let $C'$ be the center of the circle passing through $B$ and tangent to $CA$ at $C$ .

a) Prove that the area of triangle $A'B'C'$ is not less than the area of triangle $ABC$ .

b) Let $X, Y, Z$ be the projections of $O$ onto lines $A'B', B'C', C'A'$ . Given that the circumcircle of triangle $XYZ$ intersects lines $A'B', B'C', C'A'$ again at $X', Y', Z'$ ( $X' \neq X, Y' \neq Y, Z' \neq Z$ ), prove that lines $AX', BY', CZ'$ are concurrent.

Solution

a) Let $(A'), (B'), (C')$ respectively represent the circle passing through point $C$ and touching the line $AB$ at point $A$ , the circle passing through point $B$ and touching the line $BC$ at point $B$ , and the circle passing through point $C$ and touching the line $CA$ at point $C$ .

Let $K$ be the second intersection point of two circles $(A')$ and $(B')$ . We have
$(AK, AB) \equiv (BK, BC) \equiv (CK, CA) \pmod{\pi}.$
Therefore, point $K$ also belongs to circle $(C')$ . Now, denote by $D, E, F$ respectively the foot of the perpendicular drawn from point $K$ to lines $BC, CA$ and $AB$ . According to Erdos inequality, we have
$KA + KB + KC \geq 2(KD + KE + KF).$

Let $\angle KBC = \angle KAB = \angle KCA = \omega$ . Because
$\sin \omega = \frac{KD}{KB} = \frac{KE}{KC} = \frac{KF}{KA} = \frac{KD + KE + KF}{KB + KC + KA} \le \frac{1}{2}$
so $\omega \le 30^\circ$ .

The triangles $KAA'$ , $KBB'$ , $KCC'$ are isosceles triangles at $A'$ , $B'$ , $C'$ with vertex angle equal to $2\omega \le 60^\circ$ .

Put
$(\overrightarrow{KA}, \overrightarrow{KA'}) \equiv (\overrightarrow{KB}, \overrightarrow{KB'}) \equiv (\overrightarrow{KC}, \overrightarrow{KC'}) \equiv \phi \pmod{2\pi}$
$\text{and } k = \frac{KA'}{KA} = \frac{KB'}{KB} = \frac{KC'}{KC} \ge 1.$
We denote by $f$ the rotational homothety with center $K$ , angle $\phi$ and coefficient $k$ .

Since $A'$ , $B'$ , $C'$ are images of $A$ , $B$ , $C$ by $f$ respectively, $\triangle A'B'C' \sim \triangle ABC$ . We deduce that $\frac{S(A'B'C')}{S(ABC)} = k^2 \ge 1$ , or $S(A'B'C') \ge S(ABC)$ . The equality occurs if and only if $ABC$ is an equilateral triangle.

b) Since $A'$ , $C'$ are images of $A$ , $C$ through $f$ and $C'B' \perp BK$ , $C'O \perp BC$ respectively, we get
$(C'K, C'A') \equiv (CK, CA) \equiv (BK, BC) \equiv (C'B', C'O) \pmod{\pi}.$
Therefore $C'O$ and $C'K$ are isogonal in angle $A'C'B'$ . By similar argument, we have $O$ and $K$ are isogonal conjugate points in triangle $A'B'C'$ . So $KX' \perp A'B'$ , $KY' \perp B'C'$ and $KZ' \perp C'A'$ . We also have $AK \perp A'B'$ , $BK \perp B'C'$ and $CK \perp C'A'$ so we deduce that three lines $AX'$ , $BY'$ and $CZ'$ concur at $K$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.