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: added solution tag)
(Added a solution)
Line 1: Line 1:
 
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.
 
==Solution==
 
==Solution==
−
{{solution}}
+
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>.
  
 +
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>.
 +
 +
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>.
 
==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 15:02, 30 April 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.

Solution

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 $BDF$.

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=46735"