Difference between revisions of "Orthic triangle"
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If <math>\triangle ABC</math> is acute, then the incenter of the orthic triangle of <math>\triangle ABC</math> is the orthocenter <math>H</math>. | If <math>\triangle ABC</math> is acute, then the incenter of the orthic triangle of <math>\triangle ABC</math> is the orthocenter <math>H</math>. | ||
− | ''Proof'': Let <math>\theta = \angle EAH</math>. Since <math>\angle ADC = 90^\circ</math>, we have that <math>\theta = 90^\circ - C</math>. The quadrilateral <math>EAFH</math> is cyclic and, in fact <math>E</math> and <math>F</math> lie on the circle with diameter <math>AH</math>. Since <math>EH</math> subtends <math>\theta</math> as well as <math>\angle EFH</math> on this circle, so <math>\angle EFH = \theta = 90^\circ - C</math>. The same argument (with <math> | + | ''Proof'': Let <math>\theta = \angle EAH</math>. Since <math>\angle ADC = 90^\circ</math>, we have that <math>\theta = 90^\circ - \angle C</math>. The quadrilateral <math>EAFH</math> is cyclic and, in fact <math>E</math> and <math>F</math> lie on the circle with diameter <math>AH</math>. Since <math>EH</math> subtends <math>\theta</math> as well as <math>\angle EFH</math> on this circle, so <math>\angle EFH = \theta = 90^\circ - \angle C</math>. The same argument (with <math>FBDH</math> instead of <math>EAFH</math>) shows that <math>\angle DFH = 90^\circ - \angle C</math>. Hence <math>\angle EFH = \angle DFH</math>, i.e. the line <math>HF</math> bisects <math>\angle EFD</math>. By the same reasoning <math>HD</math> bisects <math>\angle EDF</math> and <math>HE</math> bisects <math>\angle FED</math>. |
<math>\square</math> | <math>\square</math> | ||
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== Problems == | == Problems == | ||
+ | * In acute triangle <math>ABC</math> points <math>P</math> and <math>Q</math> are the feet of the perpendiculars from <math>C</math> to <math>\overline{AB}</math> and from <math>B</math> to <math>\overline{AC}</math>, respectively. Line <math>PQ</math> intersects the circumcircle of <math>\triangle ABC</math> in two distinct points, <math>X</math> and <math>Y</math>. Suppose <math>XP=10</math>, <math>PQ=25</math>, and <math>QY=15</math>. Compute the product <math>AB\cdot AC</math>. (AIME II, 2019, 15) | ||
===Olympiad=== | ===Olympiad=== | ||
* Let <math>\triangle ABC</math> be an acute triangle with <math>D</math>, <math>E</math>, <math>F</math> the feet of the altitudes lying on <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{AB}</math> respectively. One of the intersection points of the line <math>\overline{EF}</math> and the circumcircle is <math>P</math>. The lines <math>\overline{BP}</math> and <math>\overline{DF}</math> meet at point <math>Q</math>. Prove that <math>|AP| = |AQ|</math>. (IMO Shortlist 2010 G1) | * Let <math>\triangle ABC</math> be an acute triangle with <math>D</math>, <math>E</math>, <math>F</math> the feet of the altitudes lying on <math>\overline{BC}</math>, <math>\overline{CA}</math>, and <math>\overline{AB}</math> respectively. One of the intersection points of the line <math>\overline{EF}</math> and the circumcircle is <math>P</math>. The lines <math>\overline{BP}</math> and <math>\overline{DF}</math> meet at point <math>Q</math>. Prove that <math>|AP| = |AQ|</math>. (IMO Shortlist 2010 G1) |
Latest revision as of 09:39, 10 July 2024
In geometry, given any , let
,
, and
denote the feet of the altitudes from
,
, and
, respectively. Then,
is called the orthic triangle of
.
The orthic triangle does not exist if is right. The two cases of when
is either acute or obtuse each carry different characteristics and must be handled separately.
Orthic triangles are not unique to their mother triangle; one acute and one to three obtuse triangles are guaranteed to have the same orthic triangle. To see this, take an acute triangle and swap its orthocenter and any vertex to get an obtuse triangle. It is easy to verify that this placement of the orthocenter is correct and that the orthic triangle will remain the same as before the swapping, as seen in the diagrams to the right.
Contents
Cyclic quadrilaterals
In both the acute and obtuse case, quadrilaterals ,
,
,
,
, and
are cyclic. These cyclic quadrilaterals make frequent appearances in olympiad geometry and are the most crucial section of this article.
Proof: we will be using directed angles, denoted by instead of the conventional
. We know that
and thus
is cyclic. In addition,
so
is also cyclic. It follows that the other cyclic quadrilaterals are also cyclic.
Connection with incenters and excenters
Incenter of the orthic triangle
If is acute, then the incenter of the orthic triangle of
is the orthocenter
.
Proof: Let . Since
, we have that
. The quadrilateral
is cyclic and, in fact
and
lie on the circle with diameter
. Since
subtends
as well as
on this circle, so
. The same argument (with
instead of
) shows that
. Hence
, i.e. the line
bisects
. By the same reasoning
bisects
and
bisects
.
If is obtuse, then the incenter of the orthic triangle of
is the obtuse vertex.
Excenters of the orthic triangle
For any acute and any
,
is the orthic triangle of
if and only if
,
, and
are the excenters of
.
Proof: First, we show that the orthic triangle leads to the excenters. Let ,
, and
be on
,
, and
, respectively. Because
is cyclic,
. Likewise,
as well. Then because
,
and so
.
Thus, the exterior angle of is
. But
, so
bisects the exterior angle of
. Similarly,
and
bisect the exterior angles of
and
respectively. Thus, the intersections of
,
, and
(namely
,
, and
) are the excenters of
, and we are halfway finished.
Next, we show that the excenters lead to the orthic triangle. Let ,
, and
be the
-excenter,
-excenter, and
-excenter of
, and let
be the incenter of
.
is equidistant from
and
, so
is on
; as a result,
is an internal angle bisector of
.
We know that and because
is an
-excenter of
,
. Thus,
, and because
is on
,
is the foot of the altitude from
of
. Similarly,
and
are feet of the altitudes from
and
, respectively. Then
is the orthic triangle of
, and we are done.
This lemma makes frequent appearances in olympiad geometry. Problems written in either excenters or the orthic triangle can often be solved by shifting perspective to the other, via the medium of this lemma. Also, note the converse works as well.
In the obtuse case, the two vertices with acute angles and the orthocenter of are the excenters.
Relationship with the incenter/excenter lemma
With this knowledge in mind, we can transfer results about the incenter and excenters to the orthic triangle. In particular, the incenter/excenter lemma can be translated into the language of the orthic triangle. It tells that all six cyclic quadrilaterals of the orthic triangle have a circumcenter on the nine-point circle of .
In the case where is acute, quadrilaterals
,
, and
follow immediately from the lemma. Actually, because
, the circumcenter of
is the midpoint of
and
, called an Euler point. It follows that the circumcenters of
and
are the other two Euler points.
As for ,
, and
, via the inscribed angle theorem, their circumcenters are the midpoints of the side lengths of
, which we know to be on the nine-point circle.
Identical reasoning follows that in the obtuse case, the six cyclic quadrilaterals still have circumcenters on the nine-point circle.
Problems
- In acute triangle
points
and
are the feet of the perpendiculars from
to
and from
to
, respectively. Line
intersects the circumcircle of
in two distinct points,
and
. Suppose
,
, and
. Compute the product
. (AIME II, 2019, 15)
Olympiad
- Let
be an acute triangle with
,
,
the feet of the altitudes lying on
,
, and
respectively. One of the intersection points of the line
and the circumcircle is
. The lines
and
meet at point
. Prove that
. (IMO Shortlist 2010 G1)