2011 USAMO Problems/Problem 3: Difference between revisions

Revision as of 15:35, 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 (for the previous paragraph), WLOG assume the $(A'C'E')$ is the unit circle, and compute $b'=a'+c'-\frac{a'c'}{e'}$ , etc.

Solution 3

We work in the complex plane to give (essentially) a complete characterization when we remove the condition that opposite sides are not parallel.

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).

@@ Line 20: / Line 20: @@
 &=(r-t)(a'-b'+d'-e')+(s-t)(b'-c'+e'-f').
 \end{align*}</cmath>
-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 <math>C'D'\parallel A'F'</math>, which is impossible, so we're done.
+If <math>r-t=s-t=0</math>, then <math>P</math> must be similar to <math>P'</math> and the conclusion is obvious.
-Alternatively, WLOG assume the <math>(A'C'E')</math> is the unit circle, and compute <math>b'=a'+c'-\frac{a'c'}{e'}</math>, etc.
+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 <math>C'D'\parallel A'F'</math>, which is impossible, so we're done.
+Alternatively (for the previous paragraph), WLOG assume the <math>(A'C'E')</math> is the unit circle, and compute <math>b'=a'+c'-\frac{a'c'}{e'}</math>, etc.
+===Solution 3===
+We work in the complex plane to give (essentially) a complete characterization when we remove the condition that opposite sides are not parallel.
+WLOG assume <math>a,b,c</math> are on the unit circle. It suffices to show that <math>a,b,c</math> uniquely determine <math>d,e,f</math>, since we know that if we let <math>E</math> be the reflection of <math>B</math> over <math>AC</math>, <math>D</math> be the reflection of <math>A</math> over <math>CE</math>, and <math>F</math> be the reflection of <math>C</math> over <math>AE</math>, then <math>ABCDEF</math> satisfies the problem conditions. (*)
+It's easy to see with the given conditions that
+<cmath>\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*}</cmath>
+Note that
+<cmath>\frac{e-d}{c-d}=\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)},</cmath>
+so plugging into the third equation we have
+<cmath>\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*}</cmath>
+Simplifying, this becomes
+<cmath>\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*}</cmath>
+Of course, we can also "conjugate" this equation -- a nice way to do this is to note that if
+<cmath>x=\frac{(a-b)[c(e-c)+a(e-a)]}{c(c-e)(c-b)-a(a-e)(a-b)},</cmath>
+then
+<cmath>\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)},</cmath>
+whence
+<cmath>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)].</cmath>
+If <math>a+c\ne 0</math>, then eliminating <math>\overline{e}</math>, we get
+<cmath>e\in\left\{a+c-\frac{ac}{b},a+\frac{2c(c-b)}{a+c},c+\frac{2a(a-b)}{a+c}\right\}.</cmath>
+The first case corresponds to (*) (since <math>a,b,c,e</math> uniquely determine <math>d</math> and <math>f</math>), the second corresponds to <math>AB\parallel DE</math> (or equivalently, since <math>AB=DE</math>, <math>a-b=e-d</math>), and by symmetry, the third corresponds to <math>CB\parallel FE</math>.
+Otherwise, if <math>c=-a</math>, then we easily find <math>b^2e=a^4\overline{e}</math> from the first of the two equations in <math>e,\overline{e}</math> (we actually don't need this, but it tells us that the locus of working <math>e</math> is a line through the origin). It's easy to compute <math>d=e+\frac{a(a-b)}{b}</math> and <math>f=e+\frac{a(a+b)}{b}</math>, so <math>a-c=2a=f-d\implies c-d=a-f\implies CD\parallel AF</math>, and we're done.
+'''Comment.''' It appears that taking <math>(ABC)</math> the unit circle is nicer than, say <math>e=0</math> or <math>(ACE)</math> the unit circle (which may not even be reasonably tractable).
 ==See Also==
 {{USAMO newbox|year=2011|num-b=2|num-a=4}}

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