Given
a triangle ABC, construct its orthocenter (call it H), then construct the orthocenters
of all the 'subtriangles' formed by the orthocenter
and two of the vertices of the triangle - HAB, HAC and HBC to be precise. Notice
that the orthocenter of the subtriangle
is the missing vertex. For triangle
HBC, the orthocenter is point A (also labeled H'). For
HAB, the orthocenter is point C (also labeled H''). For
triangle HAC, the orthocenter is point B (also H'''). Next,
construct the circumcircles of each of the
four triangles. If you connect
the centers of the circumcircles of the subtriangles
and construct the orthocenter of that triangle, the orthocenter is the circumcenter
of triangle ABC. Click
here for the sketch.

Can
we prove any of this? It looks
like there are two conjectures to prove. First
of all, are the orthocenters of the subtriangles
really the missing points, or is this a cruel joke that GSP is playing
on me?

Here's
a simplified picture to help with the proof. Show
that DHBC's
orthocenter is a vertex A, then extend that
to the other subtriangles. Take
as given triangle ABC and point H as the orthocenter of triangle ABC. The
altitudes from C to side AB, B to side A and A to side BC are shown as
well.

By
definition, the orthocenter of a triangle is the intersection of the lines
containing the altitudes of the triangle.
An altitude is a perpendicular segment from a vertex to the line defined
by the opposite side.

To
show that the orthocenter of triangle HBC is vertex A, consider the following. Triangles
ABC and HBC share side BC, and H falls on the altitude from A to BC by
definition. Therefore, the
altitude from H to BC is the same line defined by A to BC. Next,
again by definition, the altitude defined by CH is perpendicular to side
AB of triangle ABC. The altitude
from B to side CH of triangle HBC is then side AB, by definition of altitude,
and because there cannot be two lines perpendicular to a line through a
single point. Likewise, the
altitude from C to side BH is side AC. It
follows that lines defined by AB, AC and AH intersect at A, making A the
orthocenter of triangle HBC.

Repeating
the proof for triangle HAC proves that the orthocenter is B,
and C for triangle BAH.

The
second conjecture that's been made is that the triangle with vertices at
the circumcenters of the subtriangles
(call it DDEF)
has the circumcenter of triangle ABC as
its orthocenter. Here's a picture
with the circumcircles hidden and some construction
lines added. The circumcenter
of the black triangle (DABC)
and the orthocenter of DDEF is
labeled G. Take as given that
G is the circumcenter of DABC
and prove that G is also the orthocenter of DDEF. To
prove this conjecture, show that G is on the altitudes of DDEF. It
will be most helpful to show that DDEF
is congruent to DABC
and that corresponding sides of these triangles are parallel (for example,
show DE || to BC).

First,
here are some general observations.

1. AD
= BD as radii of the circumcircle of DHAB. Likewise,
AE = EC and CF = CB as radii of the circumcircles
of triangles HAC and HBC. Finally,
GA = GB = GC as radii of the circumcircle
of DABC. This
gives lots of isosceles triangles.

2. Look
at quadrilateral ADBG. Its diagonals
are AB and DG. Those diagonals
intersect in a right angle, since G and D fall on the perpendicular bisector
of AB by definition of the circumcenter. Diagonals
that intersect at right angles determine a rhombus, so ADBG is a rhombus:
a figure with four congruent sides whose opposite sides are parallel. This
same reasoning can be applied to give rhombi AGCE and GCFB.

3. By
definition of a rhombus, AD = BD = GA = GB and AD || BG and BD || GA for
rhombus ADBG. Similarly, GA
= GC = AE = EC and GA || EC and AE || GC for rhombus AGCE. For
rhombus GCFB, GB = BF = FC = GC and GB || FC and BF || GC.

4. Finally,
since each pair of rhombi share one side, all the sides listed above are
congruent. In addition, AD ||
EC since both are parallel to GA.

To
begin, prove that DE || BC. Here
is the same diagram with more lines hidden and some angles labeled.
For clarity, let a represent angle BEC; b represent angle EDB; c represent
angle BCE; d represent angle DBE; e represent angle BED; and f represent
angle EBC.

Quadrilateral
BDEC has one pair of parallel, congruent sides, EC and BD. Then
the measure of angle b + the measure of angle e + the measure of angle
a = 180 degrees using DE as a transversal, as same side interior angles
of EC and BD.Likewise,
the measure of angle c + the measure of angle d + the measure of angle
f = 180 degrees using BC as a transversal.

Draw
in diagonal EB. EB is also a
transversal of parallel sides EC and BD. As
alternate interior angles, angle a @
angle d. That gives DDEB @DCBE
by SAS (shared side BE; measure of angle a = measure of angle d and
BD = EC). Therefore,
ED @
BC, angle b @
anglec and angle e@
angle f as corresponding parts of congruent triangles. That
gives the measure of angle b + the measure of angle d + the measure of
angle f = 180 degrees after substitution.
This gives same side interior angles for parallel lines DE and BC with
BD as a transversal. Therefore,
DE || BC.

The
same logic can be applied to quadrilateral DACF to give AC congruent and
parallel to DF; and to quadrilateral AEFB to give AB congruent and parallel
to EF.So DDEF
is congruent to DABC
and pairs of sides of the triangles are parallel.

G
is the circumcenter of DABC. As
such, it lies at the intersection of the perpendicular bisectors of sides
AB, BC and AC, and is equidistant from vertices A, B and C. Vertices
D, F and E also lie on the perpendicular bisectors of AB, BC and AC, as
the circumcenters of DAHB, DBHC
and DAHC
respectively.

It
follows that DG^
AB. Since
AB || EF, DG ^
EF. Therefore, G lies
on the altitude from vertex D to side EF. Similarly,
EG ^
AC and EG ^
DF, so G lies on the altitude from vertex E to side EF. Finally,

Final
Notes (alternately titled 'Things I Noticed, But Have No Intention of Proving')

1. DDEF
is not only congruent to DABC,
it is a half turn rotation (180 degrees) from DABC. The
center of the rotation is at the midpoint of the segment from the orthocenter
of DABC
to its circumcenter (midpoint of HG in the
sketch above).

2. The
areas of all the circumcircles are the same. This
falls out pretty easily when you show that the radii are all the same.