Trig Ceva, for hexagons this time
by math_explorer, Sep 25, 2010, 6:33 AM
I've given up putting all the
signs in every single sine expression, at least for this post.
Lemma 1. If point
is inside a polygon $V_1V_2\ldotsV_n$, the product
![\[ \frac{\sin V_1V_2O \sin V_2V_3O \ldots \sin V_nV_1O}{\sin V_2V_1O \sin V_3V_2O \ldots \sin V_1V_nO} = 1 \]](//latex.artofproblemsolving.com/d/f/9/df9e2a2b1d56a5b7a19ed8d72354d41f2f832b32.png)
Proof:
. Take the product.
Click to reveal hidden text
Meh, I don't feel like finding a problem now. I've been putting off a lot of stuff.

Lemma 1. If point

![\[ \frac{\sin V_1V_2O \sin V_2V_3O \ldots \sin V_nV_1O}{\sin V_2V_1O \sin V_3V_2O \ldots \sin V_1V_nO} = 1 \]](http://latex.artofproblemsolving.com/d/f/9/df9e2a2b1d56a5b7a19ed8d72354d41f2f832b32.png)
Proof:

April wrote:
Let
and
be two points in interior of the given triangle
. Let
,
,
be the traces on sides
,
,
, respectively with the lines through
and parallel to
,
,
. Let
,
,
be the traces on sides
,
,
, respectively with the lines through
and parallel to
,
,
.
Prove that
,
,
are concurrent if and only if
,
,
are concurrent.























Prove that






Click to reveal hidden text
Reflect everything about the midpoint of
and
. Suppose
,
,
. Suppose
,
,
.
The problem becomes equivalent to proving that
,
,
are concurrent iff
,
,
are concurrent.
We also have that the lines through
parallel to
,
,
are just the lines
,
, and
, and similarly for the lines through
.
Therefore,
,
,
.
Then
(Law of Sines bash), and
. Therefore, since
and
, we have
.
Do the same thing for all three pairs of sides, and take the product, and you'll find that the left side is
, whose numerator is 1 iff
,
,
are concurrent, and whose denominator is 1 iff
,
,
are concurrent.
Meanwhile, the right side is equal to 1 because of Lemma 1 on hexagon
.
QED








The problem becomes equivalent to proving that






We also have that the lines through








Therefore,



Then





Do the same thing for all three pairs of sides, and take the product, and you'll find that the left side is







Meanwhile, the right side is equal to 1 because of Lemma 1 on hexagon

QED
Meh, I don't feel like finding a problem now. I've been putting off a lot of stuff.