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Difference between revisions of "2011 USAMO Problems/Problem 3"

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In hexagon <math>ABCDEF</math>, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy <math>\angle A = 3\angle D</math>, <math>\angle C = 3\angle F</math>, and <math>\angle E = 3\angle B</math>. Furthermore <math>AB=DE</math>, <math>BC=EF</math>, and <math>CD=FA</math>. Prove that diagonals <math>\overline{AD}</math>, <math>\overline{BE}</math>, and <math>\overline{CF}</math> are concurrent.
 
In hexagon <math>ABCDEF</math>, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy <math>\angle A = 3\angle D</math>, <math>\angle C = 3\angle F</math>, and <math>\angle E = 3\angle B</math>. Furthermore <math>AB=DE</math>, <math>BC=EF</math>, and <math>CD=FA</math>. Prove that diagonals <math>\overline{AD}</math>, <math>\overline{BE}</math>, and <math>\overline{CF}</math> are concurrent.
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==Solution==
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==Solutions==
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===Solution 1===
 
Let <math>\angle A = \alpha</math>, <math>\angle C = \gamma</math>, and <math>\angle E = \beta</math>, <math>AB=DE=p</math>, <math>BC=EF=q</math>, <math>CD=FA=r</math>, <math>AB</math> intersect <math>DE</math> at <math>X</math>, <math>BC</math> intersect <math>EF</math> at <math>Y</math>, and <math>CD</math> intersect <math>FA</math> at <math>Z</math>.  Define the vectors: <cmath>\vec{u} = \vec{AB} + \vec{DE}</cmath> <cmath>\vec{v} = \vec{BC} + \vec{EF}</cmath> <cmath>\vec{w} = \vec{CD} + \vec{FA}</cmath> Clearly, <math>\vec{u}+\vec{v}+\vec{w}=\vec{0}</math>.
 
Let <math>\angle A = \alpha</math>, <math>\angle C = \gamma</math>, and <math>\angle E = \beta</math>, <math>AB=DE=p</math>, <math>BC=EF=q</math>, <math>CD=FA=r</math>, <math>AB</math> intersect <math>DE</math> at <math>X</math>, <math>BC</math> intersect <math>EF</math> at <math>Y</math>, and <math>CD</math> intersect <math>FA</math> at <math>Z</math>.  Define the vectors: <cmath>\vec{u} = \vec{AB} + \vec{DE}</cmath> <cmath>\vec{v} = \vec{BC} + \vec{EF}</cmath> <cmath>\vec{w} = \vec{CD} + \vec{FA}</cmath> Clearly, <math>\vec{u}+\vec{v}+\vec{w}=\vec{0}</math>.
  
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And since <math>\vec{u}+\vec{v}+\vec{w}=\vec{0}</math>, we can arrange the three vectors to form a triangle, so the triangle with sides of lengths <math>2p \cos \gamma</math>, <math>2q \cos \alpha</math>, and <math>2r \cos \beta</math> has opposite angles of <math>180^\circ - 2\gamma</math>, <math>180^\circ - 2\alpha</math>, and <math>180^\circ - 2\beta</math>, respectively.  So by the law of sines: <cmath> \frac{2p \cos \gamma}{\sin 2\gamma} = \frac{2q \cos \alpha}{\sin 2\alpha} = \frac{2r \cos \beta}{\sin 2\beta} </cmath> <cmath> \frac{p}{\sin \gamma} = \frac{q}{\sin \alpha} = \frac{r}{\sin \beta}, </cmath> and the triangle with sides of length <math>p</math>, <math>q</math>, and <math>r</math> has corrosponding angles of <math>\gamma</math>, <math>\alpha</math>, and <math>\beta</math>.  But then triangles <math>FAB</math>, <math>CDB</math>, and <math>FDE</math>.  So <math>FD=p</math>, <math>BF=q</math>, and <math>BD=r</math>, and <math>A</math>, <math>C</math>, and <math>E</math> are the reflections of the vertices of triangle <math>BDF</math> about the sides.  So <math>AD</math>, <math>BE</math>, and <math>CF</math> concur at the orthocenter of triangle <math>BDF</math>.
 
And since <math>\vec{u}+\vec{v}+\vec{w}=\vec{0}</math>, we can arrange the three vectors to form a triangle, so the triangle with sides of lengths <math>2p \cos \gamma</math>, <math>2q \cos \alpha</math>, and <math>2r \cos \beta</math> has opposite angles of <math>180^\circ - 2\gamma</math>, <math>180^\circ - 2\alpha</math>, and <math>180^\circ - 2\beta</math>, respectively.  So by the law of sines: <cmath> \frac{2p \cos \gamma}{\sin 2\gamma} = \frac{2q \cos \alpha}{\sin 2\alpha} = \frac{2r \cos \beta}{\sin 2\beta} </cmath> <cmath> \frac{p}{\sin \gamma} = \frac{q}{\sin \alpha} = \frac{r}{\sin \beta}, </cmath> and the triangle with sides of length <math>p</math>, <math>q</math>, and <math>r</math> has corrosponding angles of <math>\gamma</math>, <math>\alpha</math>, and <math>\beta</math>.  But then triangles <math>FAB</math>, <math>CDB</math>, and <math>FDE</math>.  So <math>FD=p</math>, <math>BF=q</math>, and <math>BD=r</math>, and <math>A</math>, <math>C</math>, and <math>E</math> are the reflections of the vertices of triangle <math>BDF</math> about the sides.  So <math>AD</math>, <math>BE</math>, and <math>CF</math> concur at the orthocenter of triangle <math>BDF</math>.
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===Solution 2===
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We work in the complex plane, where lowercase letters denote point affixes. Let <math>P</math> denote hexagon <math>ABCDEF</math>. Since <math>AB=DE</math>, the condition <math>AB\not\parallel DE</math> is equivalent to <math>a-b+d-e\ne 0</math>.
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Construct a "phantom hexagon" <math>P'=A'B'C'D'E'F'</math> as follows: let <math>A'C'E'</math> be a triangle with <math>\angle{A'C'E'}=\angle{F}</math>, <math>\angle{C'E'A'}=\angle{B}</math>, and <math>\angle{E'A'C'}=\angle{F}</math> (this is possible since <math>\angle{B}+\angle{D}+\angle{F}=180^\circ</math> by the angle conditions), and reflect <math>A',C',E'</math> over its sides to get points <math>D',F',B'</math>, respectively. By rotation and reflection if necessary, we assume <math>A'B'\parallel AB</math> and <math>P',P</math> have the same orientation (clockwise or counterclockwise), i.e. <math>\frac{b-a}{b'-a'}\in\mathbb{R}^+</math>. It's easy to verify that <math>\angle{X'}=\angle{X}</math> for <math>X\in\{A,B,C,D,E,F\}</math> and opposite sides of <math>P'</math> have equal lengths. As the corresponding sides of <math>P</math> and <math>P'</math> must then be parallel, there exist positive reals <math>r,s,t</math> such that <math>r=\frac{a-b}{a'-b'}=\frac{d-e}{d'-e'}</math>, <math>s=\frac{b-c}{b'-c'}=\frac{e-f}{e'-f'}</math>, and <math>t=\frac{c-d}{c'-d'}=\frac{f-a}{f'-a'}</math>. But then <math>0\ne a-b+d-e=r(a'-b'+d'-e')</math>, etc., so
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<cmath>\begin{align*}
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0
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&=(a-b+d-e)+(b-c+e-f)+(c-d+f-a) \\
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&=r(a'-b'+d'-e')+s(b'-c'+e'-f')+t(c'-d'+f'-a') \\
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&=(r-t)(a'-b'+d'-e')+(s-t)(b'-c'+e'-f').
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\end{align*}</cmath>
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If <math>r-t=s-t=0</math>, then <math>P</math> must be similar to <math>P'</math> and the conclusion is obvious. Otherwise, since <math>a'-b'+d'-e'\ne0</math> and <math>b'-c'+e'-f'\ne0</math>, we must have <math>r-t\ne0</math> and <math>s-t\ne0</math>. Construct parallelograms <math>A'XE'D'</math> and <math>C'YD'E'</math>; if <math>U</math> is the reflection of <math>A'</math> over <math>B'X</math> and <math>V</math> is the reflection of <math>C'</math> over <math>D'Y</math>, then by simple angle chasing we can show that <math>\angle{UA'C'}=180^\circ-\angle{A'E'C'}</math> and <math>\angle{VC'A'}=180^\circ-\angle{C'E'A'}</math>. But <math>(r-t)(s-t)\ne0</math> means <math>u-a'=b'-a'+e'-d'</math> and <math>v-c'=b'-c'+e'-f'</math> must be linearly dependent (note that <math>A'B'=A'X</math> and <math>C'D'=C'V</math>), so we must have <math>\angle{UA'C'}+\angle{VC'A'}=180^\circ\implies \angle{A'E'C'}=90^\circ</math>. But then C'D'\parallel A'F'<math>, which is impossible, so we're done.
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Alternatively, use the projection formula with unit circle </math>(A'C'E')$.
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==See Also==
 
==See Also==
 
{{USAMO newbox|year=2011|num-b=2|num-a=4}}
 
{{USAMO newbox|year=2011|num-b=2|num-a=4}}

Revision as of 14:20, 29 October 2012

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, use the projection formula with unit circle$ (Error compiling LaTeX. Unknown error_msg)(A'C'E')$.

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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