Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
(→‎Solution 1)
Line 2: Line 2:
 
==Solutions==
 
==Solutions==
 
===Solution 1===
 
===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 D = \alpha</math>, <math>\angle F = \gamma</math>, and <math>\angle B = \epsilon</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}=\textbf{0}</math>.
  
−
Note that <math>\angle X = 360^\circ - \angle A - \angle F - \angle E = 360^\circ - \alpha - 3\gamma - \beta = 180^\circ - 2\gamma</math>. By sliding the vectors <math>\vec{AB}</math> and <math>\vec{DE}</math> to the vectors <math>\vec{MX}</math> and <math>\vec{XN}</math> respectively, then <math>\vec{u} = \vec{MN}</math>.  As <math>XMN</math> is isosceles with <math>XM = XN</math>, the base angles are both <math>\gamma</math>.  Thus, <math>|\vec{u}|=2p \cos \gamma</math>.  Similarly, <math>|\vec{v}|=2q \cos \alpha</math> and <math>|\vec{w}| = 2r \cos \beta</math>.
+
Note that <math>\angle X = 360^\circ - \angle D - \angle C - \angle B = 360^\circ - \alpha - 3\gamma - \beta = 180^\circ - 2\gamma</math>. By sliding the vectors <math>\vec{AB}</math> and <math>\vec{DE}</math> to the vectors <math>\vec{MX}</math> and <math>\vec{XN}</math> respectively, then <math>\vec{u} = \vec{MN}</math>.  As <math>XMN</math> is isosceles with <math>XM = XN</math>, the base angles are both <math>\gamma</math>.  Thus, <math>|\vec{u}|=2p \cos \gamma</math>.  Similarly, <math>|\vec{v}|=2q \cos \alpha</math> and <math>|\vec{w}| = 2r \cos \epsilon</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>.
 
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 \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 \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:
 +
[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("<math>\textcolor{red}{A}</math>",A,dir(150));
 +
label("<math>B</math>",B,dir(210));
 +
label("<math>\textcolor{red}{C}</math>",C,S);
 +
label("<math>D</math>",D,dir(330));
 +
label("<math>\textcolor{red}{E}</math>",E,dir(30));
 +
label("<math>F</math>",F,N);
 +
[/asy]
  
 
===Solution 2===
 
===Solution 2===

Revision as of 21:33, 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 $\beta$ (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).

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_USAMO_Problems/Problem_3&oldid=78085"