# Difference between revisions of "Orthic triangle"

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In [[geometry]], given any <math>\triangle ABC</math>, let <math>D</math>, <math>E</math>, and <math>F</math> denote the feet of the altitudes from <math>A</math>, <math>B</math>, and <math>C</math>, respectively. Then, <math>\triangle DEF</math> is called the '''orthic triangle''' of <math>\triangle ABC</math>. | In [[geometry]], given any <math>\triangle ABC</math>, let <math>D</math>, <math>E</math>, and <math>F</math> denote the feet of the altitudes from <math>A</math>, <math>B</math>, and <math>C</math>, respectively. Then, <math>\triangle DEF</math> is called the '''orthic triangle''' of <math>\triangle ABC</math>. | ||

− | + | The orthic triangle does not exist if <math>\triangle ABC</math> is right. The two cases of when <math>\triangle ABC</math> is either acute or obtuse each carry different characteristics and must be handled separately. | |

− | Orthic triangles are not unique to their mother | + | 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. |

== Cyclic quadrilaterals == | == Cyclic quadrilaterals == | ||

− | In both the acute and obtuse case, quadrilaterals <math>ADEB</math>, <math>BEFC</math>, <math> | + | In both the acute and obtuse case, quadrilaterals <math>ADEB</math>, <math>BEFC</math>, <math>CFAD</math>, <math>AEHF</math>, <math>BFHD</math>, and <math>CDHE</math> are [[Cyclic quadrilateral | 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 <math>\measuredangle</math> instead of the conventional <math>\angle</math>. We know that <cmath>\measuredangle ADB = 90^{\circ} = \measuredangle AEB,</cmath> and thus <math>ADEB</math> is cyclic. In addition, <cmath>\measuredangle AEH = 90^{\circ} = \measuredangle AFH,</cmath> so <math>AEHF</math> is also cyclic. It follows that the other cyclic quadrilaterals are also cyclic. | ''Proof'': we will be using [[directed angles]], denoted by <math>\measuredangle</math> instead of the conventional <math>\angle</math>. We know that <cmath>\measuredangle ADB = 90^{\circ} = \measuredangle AEB,</cmath> and thus <math>ADEB</math> is cyclic. In addition, <cmath>\measuredangle AEH = 90^{\circ} = \measuredangle AFH,</cmath> so <math>AEHF</math> is also cyclic. It follows that the other cyclic quadrilaterals are also cyclic. | ||

<math>\square</math> | <math>\square</math> | ||

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== Connection with incenters and excenters == | == Connection with incenters and excenters == | ||

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''Proof'': First, we show that the orthic triangle leads to the excenters. Let <math>D</math>, <math>E</math>, and <math>F</math> be on <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math>, respectively. Because <math>ADEB</math> is cyclic, <math>\angle EDC = \angle A</math>. Likewise, <math>\angle BDF = \angle A</math> as well. Then because <math>\angle BDF + \angle D + \angle EDC = 180^{\circ}</math>, <math>\angle A + \angle D + \angle A = 180^{\circ}</math> and so <math>\angle D = 180^{\circ} - 2\angle A</math>. | ''Proof'': First, we show that the orthic triangle leads to the excenters. Let <math>D</math>, <math>E</math>, and <math>F</math> be on <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math>, respectively. Because <math>ADEB</math> is cyclic, <math>\angle EDC = \angle A</math>. Likewise, <math>\angle BDF = \angle A</math> as well. Then because <math>\angle BDF + \angle D + \angle EDC = 180^{\circ}</math>, <math>\angle A + \angle D + \angle A = 180^{\circ}</math> and so <math>\angle D = 180^{\circ} - 2\angle A</math>. | ||

− | Thus, the exterior angle of <math>\angle D</math> is <math>2\angle A</math>. But <math>\angle BDF = \angle EDC = \angle A</math>, so <math>\overline{BC}</math> bisects the exterior angle of <math>\angle D</math>. Similarly, <math>\overline{CA}</math> and <math>\overline{AB}</math> bisect the exterior angles of <math>\angle E</math> and <math>\angle F</math> respectively. Thus, the intersections of <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math> (namely <math>A</math>, <math>B</math>, and <math>C</math>) are the excenters of <math>\triangle DEF</math>, and we | + | Thus, the exterior angle of <math>\angle D</math> is <math>2\angle A</math>. But <math>\angle BDF = \angle EDC = \angle A</math>, so <math>\overline{BC}</math> bisects the exterior angle of <math>\angle D</math>. Similarly, <math>\overline{CA}</math> and <math>\overline{AB}</math> bisect the exterior angles of <math>\angle E</math> and <math>\angle F</math> respectively. Thus, the intersections of <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math> (namely <math>A</math>, <math>B</math>, and <math>C</math>) are the excenters of <math>\triangle DEF</math>, and we are halfway finished. |

− | Next, we show that the excenters lead to the orthic triangle. Let <math>A</math>, <math>B</math>, and <math>C</math> be the <math>D</math>-excenter, <math>E</math>-excenter, and <math>F</math>-excenter of <math>\triangle DEF</math>, and let <math>H</math> be the incenter of <math>\triangle DEF</math>. <math>A</math> is equidistant from <math>\overline{DE}</math> and <math>\overline{FE}</math>, so <math>A</math> is on <math>\overline{DH}</math>, | + | Next, we show that the excenters lead to the orthic triangle. Let <math>A</math>, <math>B</math>, and <math>C</math> be the <math>D</math>-excenter, <math>E</math>-excenter, and <math>F</math>-excenter of <math>\triangle DEF</math>, and let <math>H</math> be the incenter of <math>\triangle DEF</math>. <math>A</math> is equidistant from <math>\overline{DE}</math> and <math>\overline{FE}</math>, so <math>A</math> is on <math>\overline{DH}</math>; as a result, <math>\overline{AH}</math> is an internal angle bisector of <math>\angle D</math>. |

We know that <math>\angle ADE = \frac{1}{2} \angle D</math> and because <math>C</math> is an <math>F</math>-excenter of <math>\triangle DEF</math>, <math>\angle EDC = 90^{\circ} - \frac{1}{2} \angle D</math>. Thus, <math>\angle ADC = 90^{\circ}</math>, and because <math>D</math> is on <math>\overline{BC}</math>, <math>D</math> is the foot of the altitude from <math>A</math> of <math>\triangle ABC</math>. Similarly, <math>E</math> and <math>F</math> are feet of the altitudes from <math>B</math> and <math>C</math>, respectively. Then <math>\triangle DEF</math> is the orthic triangle of <math>\triangle ABC</math>, and we are done. | We know that <math>\angle ADE = \frac{1}{2} \angle D</math> and because <math>C</math> is an <math>F</math>-excenter of <math>\triangle DEF</math>, <math>\angle EDC = 90^{\circ} - \frac{1}{2} \angle D</math>. Thus, <math>\angle ADC = 90^{\circ}</math>, and because <math>D</math> is on <math>\overline{BC}</math>, <math>D</math> is the foot of the altitude from <math>A</math> of <math>\triangle ABC</math>. Similarly, <math>E</math> and <math>F</math> are feet of the altitudes from <math>B</math> and <math>C</math>, respectively. Then <math>\triangle DEF</math> is the orthic triangle of <math>\triangle ABC</math>, and we are done. | ||

<math>\square</math> | <math>\square</math> | ||

− | 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. | + | 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 <math>\triangle ABC</math> are the excenters. | In the obtuse case, the two vertices with acute angles and the [[orthocenter]] of <math>\triangle ABC</math> are the excenters. | ||

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

− | * | + | * [[Incenter/excenter lemma]] |

+ | * [[Geometry/Olympiad | Olympiad geometry]] | ||

[[Category:Geometry]] | [[Category:Geometry]] | ||

− | |||

[[Category:Definition]] | [[Category:Definition]] |

## Latest revision as of 16:29, 18 May 2021

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*:

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

### 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)