Harmonic Divisions (Lemmas)
by Generic_Username, Aug 12, 2016, 8:07 PM
Let
be four points on a line or circle. The cross-ratio
where the segments are directed and we take the
if
is not in the interior of the circle in the case of a circle and
otherwise.
Let
be four points on a line or circle in this order. Then if
the points
form a harmonic bundle. In the case of a circle,
is called a harmonic quadrilateral.
Lemma 1:
is harmonic iff
where
is the midpoint of segment 
Proof: Let
on the number line. Then
done.
Lemma 2:
Let
be the midpoint of segment
Then
is harmonic iff
is the point at infinity on line 
Proof: By definition,
iff
done.
Lemma 3:
In
let
be concurrent cevians. If
then
is harmonic. Converse also holds.
Proof: We only prove one direction, as the proofs are similar.
Assuming the concurrency of cevians, by Ceva we have
By Menelaus WRT
we have
and dividing yields the desired result.
Lemma 4: Let
be three points lying on circle
Let the tangents at
and
intersect at a point
and let the line
intersect
again at
. Then
is harmonic. Converse also holds.
Proof: Again, we only prove one direction. Since
is a symmedian in
we have by Steiner's
so 
Lemma 5: Let
be points lying in this order on line
and let
be a point not lying on this line. Take another line
and consider the intersections
of the lines
with another line
Then 
Proof: Let
By Ratio Lemma,
which is a ratio solely dependent on angles, so we are done.
Lemma 6: Let
be four points lying on a line in this order. If
does not lie on this line, two of these statements imply the last;
i)
ii)
is the internal bisector of 
iii)
Proof:
:
are the internal and external bisectors of
so we are immediately done.
:
By the angle bisector theorem we may immediately conclude.
:
Using notation from the previous lemma,
iff
1. Let
be a triangle with orthocenter
and let
be the feet of the altitudes lying on the sides
Let
Prove
is perpendicular to the
-median of 
Solution
2. In
let
be the midpoint of
let
be the second intersection of
with the circumcircle of
and let
be the circle which is tangent to
at
and passes through
Furthermore, let
Prove that
are concurrent.
Solution (incomplete)
3. (IMO 2012) Given triangle ABC the point
is the centre of the excircle opposite the vertex
. This excircle is tangent to the side
at
, and to the lines
and
at
and
, respectively. The lines
and
meet at
, and the lines
and
meet at
. Let
be the point of intersection of the lines
and
, and let
be the point of intersection of the lines
and
. Prove that
is the midpoint of
. (The excircle of
opposite the vertex
is the circle that is tangent to the line segment
, to the ray
beyond
, and to the ray
beyond
.)
Solution
4. (IMO 2003) Let
be a cyclic quadrilateral. Let
be the feet of the perpendiculars from
to the lines
respectively. Show that
if and only if
is harmonic.
Solution
5. (Sharygin 2013) Let
be the foot of the
internal angle bisector of
Points
are the incenters of
respectively. The line
meets
at point
Prove that 
Solution
6. (USA TST 2011) In an acute scalene triangle
points
lie on sides
respectively, such that
Altitudes
meet at orthocenter
Points
and
lie on segment
such that
and
Lines
and
intersect at point
Compute 
Solution (projective)
Solution (Yufei)
7. Let
be a circle with center
and
a point outside it. Denote by
the points where the tangents from
to
meet the circle,
the point on
for which
lies on line
the foot of the perpendicular from
to
the midpoint of segment
and
the second intersection of
with
Prove that 
Solution
8. (APMO 2013) Let
be a quadrilateral inscribed in a circle
, and let
be a point on the extension of
such that
and
are tangent to
. The tangent at
intersects
at
and the line
at
. Let
be the second point of intersection between
and
. Prove that
,
,
are collinear.
Solution
9. (Sharygin 2013) Let
be a bisector of triangle
Points
and
are projections of
and
respectively on
The circle with diameter
intersects
at points
and
Prove that 
Solution
10. (ELMO SL 2014) Let
in
and let
be a point on segment
The tangent at
to the circumcircle
of triangle
hits
at
The other tangent from
to
touches it at
and
Prove that 
Solution





Let




Lemma 1:




Proof: Let



Lemma 2:
Let





Proof: By definition,


Lemma 3:
In




Proof: We only prove one direction, as the proofs are similar.
Assuming the concurrency of cevians, by Ceva we have



Lemma 4: Let









Proof: Again, we only prove one direction. Since




Lemma 5: Let








Proof: Let


Lemma 6: Let


i)

ii)


iii)

Proof:




By the angle bisector theorem we may immediately conclude.

Using notation from the previous lemma,


1. Let








Solution
It suffices to show that
is the orthocenter of
where
is the midpoint of
But this is equivalent to
which is true since
in
and
by Lemma 3.








2. In












Solution (incomplete)
From the tangency condition,
so
Hence
so
is harmonic.




3. (IMO 2012) Given triangle ABC the point





























Solution
Let
Then
so
is cyclic. Thus
so
and thus
done.







4. (IMO 2003) Let






Solution
By Ratio Lemma,
and if either of these quantities is
we are done.
These equalities hold since a spiral similarity at
sends
to
and an angle chase in cyclic quadrilaterals 


These equalities hold since a spiral similarity at




5. (Sharygin 2013) Let









Solution
Since
bisects
it suffices to show that
is harmonic by Lemma 6. Now by the converse of Lemma 3, it suffices to show that
concur where
is the incenter of
Now by Ceva it remains to show
but by the angle bisector theorem in
we get
done.









6. (USA TST 2011) In an acute scalene triangle















Solution (projective)
Let
By Lemma 5 and 7,
where
is the point at infinity on
and
Hence by Lemma 2 we may conclude.





Solution (Yufei)
Diameter of incircle Lemma WRT 

7. Let






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Solution
By Lemma 2,
is harmonic. Projecting from
we have that
is harmonic where
is the intersection of
and the parallel to
through
Now by Lemma 4 we have
collinear so
as desired.
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8. (APMO 2013) Let
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Solution
Note that from the fact that common tangents intersect at the extension of the diagonal, quadrilaterals
and
are trivially harmonic.
Let
Then
form a harmonic bundle by projecting
from
onto line 
Now projecting
from
yields
harmonic so thus
and we are done.
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Let
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Now projecting
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9. (Sharygin 2013) Let
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Solution
From
and
we have
so
is harmonic. Thus by Lemma 6
bisects
and
bisects
so
is the incenter of
done.
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10. (ELMO SL 2014) Let
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Solution
send help
This post has been edited 4 times. Last edited by Generic_Username, Aug 23, 2016, 5:55 PM