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

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Next we will find the angles between <math>\vec{u}</math>, <math>\vec{v}</math>, and <math>\vec{w}</math>.  As <math>\angle MNX = \gamma</math>, the angle between the vectors <math>\vec{u}</math> and <math>\vec{NE}</math> is <math>\gamma</math>.  Similarly, the angle between <math>\vec{NE}</math> and <math>\vec{EF}</math> is <math>180^\circ-\beta</math>, and the angle between <math>\vec{EF}</math> and <math>\vec{v}</math> is <math>\alpha</math>.  Thus, the angle between <math>\vec{u}</math> and <math>\vec{v}</math> is <math>\gamma + 180^\circ-\beta+\alpha = 360^\circ - 2\beta</math>, or just <math>2\beta</math> in the other direction if we take it modulo <math>360^\circ</math>.  Similarly, the angle between <math>\vec{v}</math> and <math>\vec{w}</math> is <math>2 \gamma</math>, and the angle between <math>\vec{w}</math> and <math>\vec{u}</math> is <math>2 \alpha</math>.
 
Next we will find the angles between <math>\vec{u}</math>, <math>\vec{v}</math>, and <math>\vec{w}</math>.  As <math>\angle MNX = \gamma</math>, the angle between the vectors <math>\vec{u}</math> and <math>\vec{NE}</math> is <math>\gamma</math>.  Similarly, the angle between <math>\vec{NE}</math> and <math>\vec{EF}</math> is <math>180^\circ-\beta</math>, and the angle between <math>\vec{EF}</math> and <math>\vec{v}</math> is <math>\alpha</math>.  Thus, the angle between <math>\vec{u}</math> and <math>\vec{v}</math> is <math>\gamma + 180^\circ-\beta+\alpha = 360^\circ - 2\beta</math>, or just <math>2\beta</math> in the other direction if we take it modulo <math>360^\circ</math>.  Similarly, the angle between <math>\vec{v}</math> and <math>\vec{w}</math> is <math>2 \gamma</math>, and the angle between <math>\vec{w}</math> and <math>\vec{u}</math> is <math>2 \alpha</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 \epsilon</math> has opposite angles of <math>180^\circ - 2\gamma</math>, <math>180^\circ - 2\alpha</math>, and <math>180^\circ - 2\epsilon</math>, respectively.  So by the law of sines (double sine formula): <cmath> \frac{2p \cos \gamma}{\sin 2\gamma} = \frac{2q \cos \alpha}{\sin 2\alpha} = \frac{2r \cos \epsilon}{\sin 2\epsilon} </cmath> <cmath> \frac{p}{\sin \gamma} = \frac{q}{\sin \alpha} = \frac{r}{\sin \epsilon}, </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>D</math>, <math>F</math>, and <math>B</math> are the reflections of the vertices of triangle <math>ACE</math> about the sides.  So <math>AD</math>, <math>BE</math>, and <math>CF</math> concur at the orthocenter of triangle <math>ACE</math>, with <math>ACE</math> being the smaller triangle:
+
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 \epsilon</math> has opposite angles of <math>180^\circ - 2\gamma</math>, <math>180^\circ - 2\alpha</math>, and <math>180^\circ - 2\epsilon</math>, respectively.  So by the law of sines (double sine formula): <cmath> \frac{2p \cos \gamma}{\sin 2\gamma} = \frac{2q \cos \alpha}{\sin 2\alpha} = \frac{2r \cos \epsilon}{\sin 2\epsilon} </cmath> <cmath> \frac{p}{\sin \gamma} = \frac{q}{\sin \alpha} = \frac{r}{\sin \epsilon}, </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>\epsilon</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>D</math>, <math>F</math>, and <math>B</math> are the reflections of the vertices of triangle <math>ACE</math> about the sides.  So <math>AD</math>, <math>BE</math>, and <math>CF</math> concur at the orthocenter of triangle <math>ACE</math>, with <math>ACE</math> being the smaller triangle:
 
<asy>
 
<asy>
 
pair A,B,C,D,E,F;
 
pair A,B,C,D,E,F;

Revision as of 22:36, 15 April 2016

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 $\angle B = \epsilon$, $AB=DE=p$, $BC=EF=q$, $CD=FA=r$, $AB$ 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, $\vec{u}+\vec{v}+\vec{w}=\textbf{0}$.

Note that $\angle X = 360^\circ - \angle D - \angle C - \angle B = 360^\circ - \alpha - 3\gamma - \beta = 180^\circ - 2\gamma$. 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 $|\vec{w}| = 2r \cos \epsilon$.

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 $2r \cos \epsilon$ has opposite angles of $180^\circ - 2\gamma$, $180^\circ - 2\alpha$, and $180^\circ - 2\epsilon$, respectively. So by the law of sines (double sine formula): \[\frac{2p \cos \gamma}{\sin 2\gamma} = \frac{2q \cos \alpha}{\sin 2\alpha} = \frac{2r \cos \epsilon}{\sin 2\epsilon}\] \[\frac{p}{\sin \gamma} = \frac{q}{\sin \alpha} = \frac{r}{\sin \epsilon},\] and the triangle with sides of length $p$, $q$, and $r$ has corrosponding angles of $\gamma$, $\alpha$, and $\epsilon$ (but then triangles $FAB$, $CDB$, and $FDE$?). 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 $ACE$, with $ACE$ being the smaller triangle: [asy] pair A,B,C,D,E,F; B=(0,0); D=(200,0); F=(100,173.2); A=(B+F)/2; C=(B+D)/2; E=(D+F)/2; draw(B--D--F--B); draw(A--C--E--A,red); label("$\textcolor{red}{A}$",A,dir(150)); label("$B$",B,dir(210)); label("$\textcolor{red}{C}$",C,S); label("$D$",D,dir(330)); label("$\textcolor{red}{E}$",E,dir(30)); label("$F$",F,N); [/asy]

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 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).

These problems are copyrighted © by the Mathematical Association of America, as part of the 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
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