2011 USAMO Problems/Problem 3

In hexagon $ABCDEF$ , which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy $\angle A = 3\angle D$ , $\angle C = 3\angle F$ , and $\angle E = 3\angle B$ . Furthermore $AB=DE$ , $BC=EF$ , and $CD=FA$ . Prove that diagonals $\overline{AD}$ , $\overline{BE}$ , and $\overline{CF}$ are concurrent.

Solution

Let $\angle A = \alpha$ , $\angle C = \gamma$ , and $\angle E = \beta$ , $AB=DE=p$ , $BC=EF=q$ , $CD=FA=r$ , $AB$ intersect $DE$ at $X$ , $BC$ intersect $EF$ at $Y$ , and $CD$ intersect $FA$ at $Z$ . Define the vectors: $\vec{u} = \vec{AB} + \vec{DE}$ $\vec{v} = \vec{BC} + \vec{EF}$ $\vec{w} = \vec{CD} + \vec{FA}$ Clearly, $\vec{u}+\vec{v}+\vec{w}=\vec{0}$ .

Note that $\angle X = 360^\circ - \angle A - \angle F - \angle E = 360^\circ - \alpha - 3\gamma - \beta = 180^\circ - 2\gamma$ . By sliding the vectors $\vec{AB}$ and $\vec{DE}$ to the vectors $\vec{MX}$ and $\vec{XN}$ respectively, then $\vec{u} = \vec{MN}$ . As $XMN$ is isosceles with $XM = XN$ , the base angles are both $\gamma$ . Thus, $|\vec{u}|=2p \cos \gamma$ . Similarly, $|\vec{v}|=2q \cos \alpha$ and $|\vec{w}| = 2r \cos \beta$ .

Next we will find the angles between $\vec{u}$ , $\vec{v}$ , and $\vec{w}$ . As $\angle MNX = \gamma$ , the angle between the vectors $\vec{u}$ and $\vec{NE}$ is $\gamma$ . Similarly, the angle between $\vec{NE}$ and $\vec{EF}$ is $180^\circ-\beta$ , and the angle between $\vec{EF}$ and $\vec{v}$ is $\alpha$ . Thus, the angle between $\vec{u}$ and $\vec{v}$ is $\gamma + 180^\circ-\beta+\alpha = 360^\circ - 2\beta$ , or just $2\beta$ in the other direction if we take it modulo $360^\circ$ . Similarly, the angle between $\vec{v}$ and $\vec{w}$ is $2 \gamma$ , and the angle between $\vec{w}$ and $\vec{u}$ is $2 \alpha$ .

And since $\vec{u}+\vec{v}+\vec{w}=\vec{0}$ , we can arrange the three vectors to form a triangle, so the triangle with sides of lengths $2p \cos \gamma$ , $2q \cos \alpha$ , and $2r \cos \beta$ has opposite angles of $180^\circ - 2\gamma$ , $180^\circ - 2\alpha$ , and $180^\circ - 2\beta$ , respectively. So by the law of sines: $\frac{2p \cos \gamma}{\sin 2\gamma} = \frac{2q \cos \alpha}{\sin 2\alpha} = \frac{2r \cos \beta}{\sin 2\beta}$ $\frac{p}{\sin \gamma} = \frac{q}{\sin \alpha} = \frac{r}{\sin \beta},$ and the triangle with sides of length $p$ , $q$ , and $r$ has corrosponding angles of $\gamma$ , $\alpha$ , and $\beta$ . But then triangles $FAB$ , $CDB$ , and $FDE$ . So $FD=p$ , $BF=q$ , and $BD=r$ , and $A$ , $C$ , and $E$ are the reflections of the vertices of triangle $BDF$ about the sides. So $AD$ , $BE$ , and $CF$ concur at the orthocenter of triangle $BDF$ .

2011 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
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2011 USAMO Problems/Problem 3

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