Difference between revisions of "2011 USAMO Problems/Problem 3"
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&=(r-t)(a'-b'+d'-e')+(s-t)(b'-c'+e'-f'). | &=(r-t)(a'-b'+d'-e')+(s-t)(b'-c'+e'-f'). | ||
\end{align*}</cmath> | \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 | + | 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== | ==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 15:35, 29 October 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.
Solutions
Solution 1
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
.
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
If
, then
must be similar to
and the conclusion is obvious.
Otherwise, since and
, we must have
and
. Construct parallelograms
and
; if
is the reflection of
over
and
is the reflection of
over
, then by simple angle chasing we can show that
and
. But
means
and
must be linearly dependent (note that
and
), so we must have
. But then
, which is impossible, so we're done.
Alternatively (for the previous paragraph), WLOG assume the is the unit circle, and compute
, 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 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).
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 |