Difference between revisions of "2011 USAMO Problems/Problem 3"
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==Solutions== | ==Solutions== | ||
===Solution 1=== | ===Solution 1=== | ||
− | Let <math>\angle D = \alpha</math>, <math>\angle F = \gamma</math>, and <math>\angle B = \ | + | Let <math>\angle D = \alpha</math>, <math>\angle F = \gamma</math>, and <math>\angle B = \beta</math>, <math>AB=DE=p</math>, <math>BC=EF=q</math>, <math>CD=FA=r</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 D - \angle C - \angle B = 360^\circ - \alpha - 3\gamma - \beta = 180^\circ - 2\gamma</math>. | + | Let <math>AB</math> intersect <math>DE</math> at <math>X</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>. Define the points <math>M</math> and <math>N</math> on lines <math>AB</math> and <math>DE</math> respectively so that <math>\vec{MX} = \vec{AB}</math> and <math>\vec{XN} = \vec{DE}</math>. Then <math>\vec{u} = \vec{MN}</math>. As <math>XMN</math> is isosceles with <math>XM = XN = p</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{ | + | 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{EF}</math> and <math>\vec{v}</math> is <math>\alpha</math>. Since the angle between <math>\vec{NE}</math> and <math>\vec{EF}</math> is <math>\angle E = 3\beta</math>, the angle between <math>\vec{u}</math> and <math>\vec{v}</math> is <math>360^\circ - \gamma - 3\beta - \alpha = 180^\circ - 2\beta</math>. Similarly, the angle between <math>\vec{v}</math> and <math>\vec{w}</math> is <math>180^\circ - 2\gamma</math>, and the angle between <math>\vec{w}</math> and <math>\vec{u}</math> is <math>180^\circ - 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 \ | + | 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 corresponding angles of <math>\gamma</math>, <math>\alpha</math>, and <math>\beta</math>. It follows by SAS congruency that this triangle is congruent to <math>FAB</math>, <math>BCD</math>, and <math>DEF</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>. |
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===Solution 2=== | ===Solution 2=== |
Revision as of 11:57, 21 August 2019
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.
Solutions
Solution 1
Let ,
, and
,
,
,
. Define the vectors:
Clearly,
.
Let intersect
at
. Note that
. Define the points
and
on lines
and
respectively so that
and
. 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
. Since the angle between
and
is
, the angle between
and
is
. 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 corresponding angles of
,
, and
. It follows by SAS congruency that this triangle is congruent to
,
, and
, so
,
, and
, and
,
, and
are the reflections of the vertices of triangle
about the sides. So
,
, and
concur at the orthocenter of triangle
.
Solution 2
We work in the complex plane, where lowercase letters denote point affixes. Let denote hexagon
. Since
, the condition
is equivalent to
.
Construct a "phantom hexagon" as follows: let
be a triangle with
,
, and
(this is possible since
by the angle conditions), and reflect
over its sides to get points
, respectively. By rotation and reflection if necessary, we assume
and
have the same orientation (clockwise or counterclockwise), i.e.
. It's easy to verify that
for
and opposite sides of
have equal lengths. As the corresponding sides of
and
must then be parallel, there exist positive reals
such that
,
, and
. But then
, etc., so the non-parallel condition "transfers" directly from
to
and
If
, then
must be similar to
and the conclusion is obvious.
Otherwise, since and
, we must have
and
. Now let
,
,
be the feet of the altitudes in
; by the non-parallel condition in
,
are pairwise distinct. But
, whence
are three distinct collinear points, which is clearly impossible. (The points can only be collinear when
is a right triangle, but in this case two of
must coincide.)
Alternatively (for the previous paragraph), WLOG assume that is the unit circle, and use the fact that
, etc. to get simple expressions for
and
.
Solution 3
We work in the complex plane to give (essentially) a complete characterization when the parallel condition is relaxed.
WLOG assume are on the unit circle. It suffices to show that
uniquely determine
, since we know that if we let
be the reflection of
over
,
be the reflection of
over
, and
be the reflection of
over
, then
satisfies the problem conditions. (*)
It's easy to see with the given conditions that
Note that
so plugging into the third equation we have
Simplifying, this becomes
Of course, we can also "conjugate" this equation -- a nice way to do this is to note that if
then
whence
If
, then eliminating
, we get
The first case corresponds to (*) (since
uniquely determine
and
), the second corresponds to
(or equivalently, since
,
), and by symmetry, the third corresponds to
.
Otherwise, if , then we easily find
from the first of the two equations in
(we actually don't need this, but it tells us that the locus of working
is a line through the origin). It's easy to compute
and
, so
, and we're done.
Comment. It appears that taking the unit circle is nicer than, say
or
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.
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 |