# Difference between revisions of "Orthic triangle"

Etmetalakret (talk | contribs) |
Etmetalakret (talk | contribs) |
||

Line 35: | Line 35: | ||

== Problems == | == Problems == | ||

===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 | + | * 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) |

== See also == | == See also == |

## Revision as of 13:14, 10 May 2021

In geometry, given any , let , , and denote the feet of the altitudes from , , and , respectively. Then, is called the **orthic triangle** of .

It's easy to see that there is no orthic triangle if is right. The only two cases are when is either acute or obtuse.

Orthic triangles are not unique to their mother triangles, as 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's easy to verify that this placement of the orthocenter is correct, and that the orthic triangle will remain the same as before.

## Contents

## Cyclic quadrilaterals

In both the acute and obtuse case, quadrilaterals , , , , , and are cyclic.

*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.

These cyclic quadrilaterals make frequent appearances in olympiad geometry and are the most crucial section of this article.

## Connection with incenters and excenters

### Incenter of the orthic triangle

If is acute, then the incenter of the orthic triangle is the orthocenter .

If is obtuse, then the incenter of the orthic triangle is the obtuse vertex.

### Excenters of the orthic triangle

For any and , is the orthic triangle of if and only if , , and are the excenters of .

*Proof*: (for excenters -> orthic triangle, note that AD perp BC. Thus, AD is an altitude of , with foot . Similarly, the others go and we have and are the feet of the altitudes, so is the orthic trianlge) (for orthic triangle -> excenters, note that and . then BC is the exterior bisector of FDE. similarly, AB and CA are exterior bisectors, so they intersect at the excenters, ie A, B, and C)

This lemma also applies in reverse. In the acute case, , , and are the excenters of the orthic triangle, while in the obtuse case, the two vertexes 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. 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 the six cyclic quadrilaterals also have circumcenters on the nine-point circle.

## Problems

### 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)