ELMO 2014 SL G4
by Wolstenholme, Aug 1, 2014, 9:39 PM
Let 
be a quadrilateral inscribed in circle 
. Define 
and 
. Prove that 
, and the 
-symmedian of triangle 
are concurrent.
Proof:
Lemma 1: Let
= 
with respect to . Then 
is the -symmedian of triangle 
.
Proof: Let the reflection of about the bisector of 
meet 
at 
. Then:
![\[ \frac{AM}{CM} = \frac{BM * \frac{\sin{ABM}}{\sin{BAM}}}{BM * \frac{\sin{CBM}}{\sin{BCM}}} = \frac{\sin{ABM} * \sin{BAK}}{\sin{CBM} * \sin{BCK}} = \frac{\sin{CBK} * \sin{BAK}}{\sin{BCK} * \sin{ABK}} = \frac{CK * BK}{BK * AK} = 1 \]](//latex.artofproblemsolving.com/6/5/e/65ef50065006c297f29a8f357efee140cedaa46f.png)
which implies the desired result.
Lemma 2:
is a harmonic quadrilateral
Proof: Taking perspective at and projecting the quadrilateral onto line 
it suffices to show that 
where 
. Now let 
. It is well-known that if and only if 
are collinear (this can be proven by subsequent applications of Ceva and Menelaus). But by Pascal's Theorem on 
we have that are collinear as desired.
Now, by a well-known result, this implies that
. And now by Pascal's theorem on 
we have that 
as well so the three lines in the problem all concur at as desired.
be a quadrilateral inscribed in circle 
. Define 
and 
. Prove that 
, and the 
-symmedian of triangle 
are concurrent.Proof:
Lemma 1: Let
= 
with respect to . Then 
is the -symmedian of triangle 
.Proof: Let the reflection of
about the bisector of 
meet 
at 
. Then:![\[ \frac{AM}{CM} = \frac{BM * \frac{\sin{ABM}}{\sin{BAM}}}{BM * \frac{\sin{CBM}}{\sin{BCM}}} = \frac{\sin{ABM} * \sin{BAK}}{\sin{CBM} * \sin{BCK}} = \frac{\sin{CBK} * \sin{BAK}}{\sin{BCK} * \sin{ABK}} = \frac{CK * BK}{BK * AK} = 1 \]](http://latex.artofproblemsolving.com/6/5/e/65ef50065006c297f29a8f357efee140cedaa46f.png)
which implies the desired result.
Lemma 2:
is a harmonic quadrilateralProof: Taking perspective at
and projecting the quadrilateral onto line 
it suffices to show that 
where 
. Now let 
. It is well-known that if and only if 
are collinear (this can be proven by subsequent applications of Ceva and Menelaus). But by Pascal's Theorem on 
we have that are collinear as desired.Now, by a well-known result, this implies that
. And now by Pascal's theorem on 
we have that 
as well so the three lines in the problem all concur at as desired.
August 2016
June 2016