Difference between revisions of "2011 USAMO Problems/Problem 3"
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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== | ||
− | {{ | + | 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 14:02, 30 April 2012
In hexagon , which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy
,
, and
. Furthermore
,
, and
. Prove that diagonals
,
, and
are concurrent.
Solution
Let ,
, and
,
,
,
,
intersect
at
,
intersect
at
, and
intersect
at
. Define the vectors:
Clearly,
.
Note that . By sliding the vectors
and
to the vectors
and
respectively, then
. As
is isosceles with
, the base angles are both
. Thus,
. Similarly,
and
.
Next we will find the angles between ,
, and
. As
, the angle between the vectors
and
is
. Similarly, the angle between
and
is
, and the angle between
and
is
. Thus, the angle between
and
is
, or just
in the other direction if we take it modulo
. Similarly, the angle between
and
is
, and the angle between
and
is
.
And since , we can arrange the three vectors to form a triangle, so the triangle with sides of lengths
,
, and
has opposite angles of
,
, and
, respectively. So by the law of sines:
and the triangle with sides of length
,
, and
has corrosponding angles of
,
, and
. But then triangles
,
, and
. So
,
, and
, and
,
, and
are the reflections of the vertices of triangle
about the sides. So
,
, and
concur at the orthocenter of triangle
.
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 |