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2011 USAMO Problems/Problem 3

Revision as of 15:25, 29 October 2012 by Math154 (talk | contribs) (→‎Solution 2)

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.

Solutions

Solution 1

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 $BDF$.

Solution 2

We work in the complex plane, where lowercase letters denote point affixes. Let $P$ denote hexagon $ABCDEF$. Since $AB=DE$, the condition $AB\not\parallel DE$ is equivalent to $a-b+d-e\ne 0$.

Construct a "phantom hexagon" $P'=A'B'C'D'E'F'$ as follows: let $A'C'E'$ be a triangle with $\angle{A'C'E'}=\angle{F}$, $\angle{C'E'A'}=\angle{B}$, and $\angle{E'A'C'}=\angle{F}$ (this is possible since $\angle{B}+\angle{D}+\angle{F}=180^\circ$ by the angle conditions), and reflect $A',C',E'$ over its sides to get points $D',F',B'$, respectively. By rotation and reflection if necessary, we assume $A'B'\parallel AB$ and $P',P$ have the same orientation (clockwise or counterclockwise), i.e. $\frac{b-a}{b'-a'}\in\mathbb{R}^+$. It's easy to verify that $\angle{X'}=\angle{X}$ for $X\in\{A,B,C,D,E,F\}$ and opposite sides of $P'$ have equal lengths. As the corresponding sides of $P$ and $P'$ must then be parallel, there exist positive reals $r,s,t$ such that $r=\frac{a-b}{a'-b'}=\frac{d-e}{d'-e'}$, $s=\frac{b-c}{b'-c'}=\frac{e-f}{e'-f'}$, and $t=\frac{c-d}{c'-d'}=\frac{f-a}{f'-a'}$. But then $0\ne a-b+d-e=r(a'-b'+d'-e')$, etc., so \begin{align*} 0 &=(a-b+d-e)+(b-c+e-f)+(c-d+f-a) \\ &=r(a'-b'+d'-e')+s(b'-c'+e'-f')+t(c'-d'+f'-a') \\ &=(r-t)(a'-b'+d'-e')+(s-t)(b'-c'+e'-f'). \end{align*} If $r-t=s-t=0$, then $P$ must be similar to $P'$ and the conclusion is obvious. Otherwise, since $a'-b'+d'-e'\ne0$ and $b'-c'+e'-f'\ne0$, we must have $r-t\ne0$ and $s-t\ne0$. Construct parallelograms $A'XE'D'$ and $C'YD'E'$; if $U$ is the reflection of $A'$ over $B'X$ and $V$ is the reflection of $C'$ over $D'Y$, then by simple angle chasing we can show that $\angle{UA'C'}=180^\circ-\angle{A'E'C'}$ and $\angle{VC'A'}=180^\circ-\angle{C'E'A'}$. But $(r-t)(s-t)\ne0$ means $u-a'=b'-a'+e'-d'$ and $v-c'=b'-c'+e'-f'$ must be linearly dependent (note that $A'B'=A'X$ and $C'D'=C'V$), so we must have $\angle{UA'C'}+\angle{VC'A'}=180^\circ\implies \angle{A'E'C'}=90^\circ$. But then $C'D'\parallel A'F'$, which is impossible, so we're done.

Alternatively, WLOG assume the $(A'C'E')$ is the unit circle, and compute $b'=a'+c'-\frac{a'c'}{e'}$, etc.

See Also

2011 USAMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
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