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Difference between revisions of "2011 USAMO Problems/Problem 3"
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===Solution 2===
 
===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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We work in the complex plane, where lowercase letters denote their corresponding point's poition. 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>.
  
 
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 the non-parallel condition "transfers" directly from <math>P</math> to <math>P'</math> and
 
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 the non-parallel condition "transfers" directly from <math>P</math> to <math>P'</math> and
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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}}
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[[Category:Olympiad Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 11:07, 31 August 2023

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 $\angle D = \alpha$, $\angle F = \gamma$, and $\angle B = \beta$, $AB=DE=p$, $BC=EF=q$, $CD=FA=r$. Define the vectors: \[\vec{u} = \vec{AB} + \vec{DE}\] \[\vec{v} = \vec{BC} + \vec{EF}\] \[\vec{w} = \vec{CD} + \vec{FA}\] Clearly, $\vec{u}+\vec{v}+\vec{w}=\textbf{0}$.

Let $AB$ intersect $DE$ at $X$. Note that $\angle X = 360^\circ - \angle D - \angle C - \angle B = 360^\circ - \alpha - 3\gamma - \beta = 180^\circ - 2\gamma$. Define the points $M$ and $N$ on lines $AB$ and $DE$ respectively so that $\vec{MX} = \vec{AB}$ and $\vec{XN} = \vec{DE}$. Then $\vec{u} = \vec{MN}$. As $XMN$ is isosceles with $XM = XN = p$, 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 $\vec{EF}$ and $\vec{v}$ is $\alpha$. Since the angle between $\vec{NE}$ and $\vec{EF}$ is $\angle E = 3\beta$, the angle between $\vec{u}$ and $\vec{v}$ is $360^\circ - \gamma - 3\beta - \alpha = 180^\circ - 2\beta$. Similarly, the angle between $\vec{v}$ and $\vec{w}$ is $180^\circ - 2\gamma$, and the angle between $\vec{w}$ and $\vec{u}$ is $180^\circ - 2\alpha$.

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 corresponding angles of $\gamma$, $\alpha$, and $\beta$. It follows by SAS congruency that this triangle is congruent to $FAB$, $BCD$, and $DEF$, so $FD=p$, $BF=q$, and $BD=r$, and $D$, $F$, and $B$ are the reflections of the vertices of triangle $ACE$ about the sides. So $AD$, $BE$, and $CF$ concur at the orthocenter of triangle $ACE$.

Solution 2

We work in the complex plane, where lowercase letters denote their corresponding point's poition. 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 the non-parallel condition "transfers" directly from $P$ to $P'$ and \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 $s-t\ne0$. Now let $x=\frac{a'+d'}{2}$, $y=\frac{c'+f'}{2}$, $z=\frac{e'+b'}{2}$ be the feet of the altitudes in $\triangle{A'C'E'}$; by the non-parallel condition in $P'$, $x,y,z$ are pairwise distinct. But $\frac{z-x}{z-y}=\frac{s-t}{r-t}\in\mathbb{R}$, whence $x,y,z$ are three distinct collinear points, which is clearly impossible. (The points can only be collinear when $\triangle{A'C'E'}$ is a right triangle, but in this case two of $x,y,z$ must coincide.)

Alternatively (for the previous paragraph), WLOG assume that $(A'C'E')$ is the unit circle, and use the fact that $b'=a'+c'-\frac{a'c'}{e'}$, etc. to get simple expressions for $a'-b'+d'-e'$ and $b'-c'+e'-f'$.

Solution 3

We work in the complex plane to give (essentially) a complete characterization when the parallel condition is relaxed.

WLOG assume $a,b,c$ are on the unit circle. It suffices to show that $a,b,c$ uniquely determine $d,e,f$, since we know that if we let $E$ be the reflection of $B$ over $AC$, $D$ be the reflection of $A$ over $CE$, and $F$ be the reflection of $C$ over $AE$, then $ABCDEF$ satisfies the problem conditions. (*)

It's easy to see with the given conditions that \begin{align*} (a-b)(c-d)(e-f) &= (b-c)(d-e)(f-a) \Longleftrightarrow f=\frac{(a-b)(c-d)e+(c-b)(e-d)a}{(a-b)(c-d)+(c-b)(e-d)} \\ \frac{(e-a)(c-b)}{(a-b)(c-d)+(c-b)(e-d)} = \frac{f-e}{d-e} &= \left(\frac{c-b}{a-b}\right)^2 \overline{\left(\frac{a-b}{c-b}\right)} = \frac{c-b}{a-b}\cdot\frac{c}{a} \Longleftrightarrow d=\frac{c[(a-b)c+(c-b)e]+a(a-e)(a-b)}{c[(a-b)+(c-b)]} \\ \frac{(a-b)(c-d)+(c-b)(e-d)}{(a-e)(c-d)} = \frac{b-a}{f-a} &= \left(\frac{e-d}{c-d}\right)^2 \overline{\left(\frac{c-d}{e-d}\right)}. \end{align*} Note that \[\frac{e-d}{c-d}=\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)},\] so plugging into the third equation we have \begin{align*} \frac{a(a-b)(2b-a-c)}{c(c-e)(c-b)-a(a-e)(a-b)} &=\frac{(a-b)+(c-b)\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}}{(a-e)}\\ &=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\overline{\left(\frac{c(c-e)(c-b)-a(a-e)(a-b)}{(a-b)[c(e-c)+a(e-a)]}\right)}\\ &=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\frac{\frac{1}{c}\left(\overline{e}-\frac{1}{c}\right)\frac{b-c}{bc}-\frac{1}{a}\left(\overline{e}-\frac{1}{a}\right)\frac{b-a}{ba}}{\frac{b-a}{ab}\left(\frac{1}{c}\left(\frac{1}{c}-\overline{e}\right)+\frac{1}{a}\left(\frac{1}{a}-\overline{e}\right)\right)}\\ &=\left(\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)}\right)^2\frac{c^3(a\overline{e}-1)(a-b)-a^3(c\overline{e}-1)(c-b)}{c(a-b)[a^2(c\overline{e}-1)+c^2(a\overline{e}-1)]}\\ &=\frac{(a-b)[c(e-c)+a(e-a)]^2}{[c(c-e)(c-b)-a(a-e)(a-b)]^2}\frac{c^3(a\overline{e}-1)(a-b)-a^3(c\overline{e}-1)(c-b)}{c[a^2(c\overline{e}-1)+c^2(a\overline{e}-1)]}. \end{align*} Simplifying, this becomes \begin{align*} &ac(2b-a-c)[c(c-e)(c-b)-a(a-e)(a-b)][a^2(c\overline{e}-1)+c^2(a\overline{e}-1)]\\ &=[c(e-c)+a(e-a)]^2[c^3(a\overline{e}-1)(a-b)-a^3(c\overline{e}-1)(c-b)]. \end{align*} Of course, we can also "conjugate" this equation -- a nice way to do this is to note that if \[x=\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)},\] then \[\frac{a(2b-a-c)}{c(e-c)+a(e-a)}=\frac{x}{\overline{x}}=\overline{\left(\frac{c(e-c)+a(e-a)}{a(2b-a-c)}\right)}=\frac{\frac{1}{c}\left(\overline{e}-\frac{1}{c}\right)+\frac{1}{a}\left(\overline{e}-\frac{1}{a}\right)}{\frac{1}{a}\left(\frac{2}{b}-\frac{1}{a}-\frac{1}{c}\right)},\] whence \[ac(2b-a-c)[2ac-b(a+c)]=b[c(e-c)+a(e-a)][a^2(c\overline{e}-1)+c^2(a\overline{e}-1)].\] If $a+c\ne 0$, then eliminating $\overline{e}$, we get \[e\in\left\{a+c-\frac{ac}{b},a+\frac{2c(c-b)}{a+c},c+\frac{2a(a-b)}{a+c}\right\}.\] The first case corresponds to (*) (since $a,b,c,e$ uniquely determine $d$ and $f$), the second corresponds to $AB\parallel DE$ (or equivalently, since $AB=DE$, $a-b=e-d$), and by symmetry, the third corresponds to $CB\parallel FE$.

Otherwise, if $c=-a$, then we easily find $b^2e=a^4\overline{e}$ from the first of the two equations in $e,\overline{e}$ (we actually don't need this, but it tells us that the locus of working $e$ is a line through the origin). It's easy to compute $d=e+\frac{a(a-b)}{b}$ and $f=e+\frac{a(a+b)}{b}$, so $a-c=2a=f-d\implies c-d=a-f\implies CD\parallel AF$, and we're done.

Comment. It appears that taking $(ABC)$ the unit circle is nicer than, say $e=0$ or $(ACE)$ the unit circle (which may not even be reasonably tractable).

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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