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The Orthocentre and the Pedal Triangle of a

Triangle

Axel Hagerud, Rilind Hoti, Neo Dahlfors & Isak Fleig

Norra real gymnasieskola

May 2021

Contents

1 Introduction 2

2 The location of the orthocentre 2

3 The pedal and orthic triangles 4

4 Inscribed quadrilaterals associated with the orthocentre 6

5 The re

ections of the orthocentre 8

6 Conclusion 12

7 Exercises for the reader 12?

Minervagymnasium

1

1 IntroductionIn geometry one of the most common objects is the triangle. Being able to make

further constructions from the information given is the key to solving all but the most basic problems you may encounter in geometry. This document is about the orthocentre and the pedal triangle of a triangle, which will be introduced shortly. The orthocentre is one of the most important points of a triangle. The pedal triangle, especially the orthic pedal triangle, is also frequently key to solving problems.

2 The location of the orthocentre

To nd the orthocentre of a triangleABC, letR,S,Tbe the orthogonal projections of the verticesA,B,Conto the lines generated byBC;CA;AB respectively. The orthocentreHis then the point where the linesAR,BS,CT intersect.A HT BCRS AH T BCRS

Figure 1:His the orthocentre of triangleABC.

Exercise 2.1.

Investigare when the orthocentre lies inside, and when it lies outside the triangle. Does the orthocentre ever lie on the perimeter of the triangle? The very existence of an orthocentre relies on the fact thatAR,BS,CT always intersect at a single point. Theorem 2.1.The altitudes of a triangle meet at a point, the orthocentre.

Proof.

To prove this, the fact that the perpendicular bisectors of a triangle meet at a point will be used. This can be shown as follows: 2 A D BE COA BE C ODFigure 2: The perpendicular bisectors of a triangle meet at one point, the centre of the circumscribed circle. Here an acute and an obtuse triangle are shown. Lemma (The perpendicular bisectors of a triangle intersect). Consider the triangleABC. The perpendicular bisectors of the sidesABand ACmeet at a pointOand intersectABandACatDandErespectively. Since jADj=jBDjand\ADO= 90=\BDO, the trianglesADOandBDOare congruent by side-angle-side. This implies thatjAOj=jBOj. Similarly, it can be shown thatjAOj=jCOj, implying thatjBOj=jCOj. Using simple trigonometry, this implies thatOlies on the perpendicular bisector of sideBC, meaning that the perpendicular bisectors of a triangle meet at one point. Given a triangleABC, three additional triangles, congruent withABC, can be constructed as in Figure 3:XZ BCYA Figure 3: Triangles4ABZ;4ACYand4BCXcongruent with triangle4ABC are constructed on the sides of4ABC. From the way the triangles are constructed it immediately follows thatBC is parallel withY Z, which can be noticed by the angles in Figure 3. Since the 3 altitude of triangleABCthroughAis perpendicular toBC, it must also be perpendicular toY Z. This, in addition to the fact thatjAYj=jAZjmeans that said altitude also is the perpendicular bisector of segmentY Z. Using identical arguments, it can be shown that all the altitudes of triangleABCare the perpendicular bisectors of triangleXY Z. Since the perpendicular bisectors of a triangle are concurrent, it follows that the altitudes of any given triangle

ABCalways meet at one point.

The proof above works without need for modication for acute and obtuse triangles. The same result can also be achieved using Ceva's Theorem or the

Euler Line.

3 The pedal and orthic trianglesA

PT BCRSA HT BCRS Figure 4: There are many pedal trianglesSTRof the triangleABC. The picture to the right shows a pedal triangle that is also the orthic triangle. A pedal triangle can be obtained by projecting a point onto the sides of a triangle and then connecting the projections, as in Figure 4 where4SRTis a pedal triangle of4ABC. Depending on the chosen point for projection, a pedal triangle will have dierent properties. Both the medial triangle and the intouch triangle are pedal triangles, generated by projecting the centre of the circumscribed or inscribed circle of a triangle onto its sides. If the point being projected is the orthocentre, the pedal triangle becomes the orthic triangle. The orthic triangle can also be obtained by connecting the three points where the altitudes intersect with the lines generated by the sides of the triangle. The orthic triangle has some very interesting properties.

Theorem 3.1.

The orthocentre of the acute angled triangle4ABCis also the centre of the inscribed circle of the triangle4STR.

Proof.

The centre of the inscribed circle is the point where the internal angle bisectors meet. Thus, proving the theorem is equivalent to showing that the 4 angles\RST,\RTSand\SRThave the respective angular bisectorsSH,TH andRH. Due to rotational symmetrical reasons this only needs to be proven for one of the three vertices mentioned above. To show that\TRH=\SRHrst construct a circle with the sideABas its diameter. The inverse of the inscribed angle theorem implies that the points RandSalso lie on the arc of this circle. The inscribed angle theorem is now used to show that\ABS=\ARS. By constructing an additional circle with diameterACthe same reasoning can be used to show that\ART=\ACT:A HT BCRS Figure 5: The inscribed angle theorem is used to show that\ABS=\ARS and\ART=\ACT: The angles\BTHand\CSHare equal, both are 90. Since\BHTand \SHCare opposite angles this means that the triangles4BHTand4CHS share two angles. This implies that the triangles are similar, which means that

4TBH=\SCH.BCTHS

Figure 6: The triangles are similar. That implies that\TBH=\SCH. Returning to the conguration in Figure 5 the above shown statement now 5 means that\TRH=\SRH. This proves thatRHis the bisector of\TRS. This implies thatHis the centre of the inscribed circle of the orthic triangle, since the bisectors of the three anglesSH,THandRHall meet inH, the orthocentre of4ABC. The proof is thus complete.Exercise 3.1. In Figure 5, show that the trianglesABCandASTare similar. Exercise 3.2.It is well known that the area of a triangle isbaseheight2. Show that this formula gives the same area irregardless of the choice of base of the triangle.

4 Inscribed quadrilaterals associated with the

orthocentreA HT BCRS Figure 7:His the orthocentre and4RSTis the orthic triangle of4ABC. When studying the conguration of the orthocentreHof a triangleABCalong with the verticesS;RandTof its orthic triangle as in Figure 7, one can observe a strong correlation between these points and inscribed quadrilaterals, which can be very useful for problem solving. In Figure 7, the quadrilateralATHSis inscribed in a circle. This is because a quadrilateral is inscribed in a circle if and only if two opposite angles add up to 180, and \ATH+HSA= 90+ 90= 180: The quadrilateralBCSTis also inscribed in a circle by the inverse of the inscribed angle theorem, since \BTC=\BSC= 90: Similarly, it can be shown that all of the quadrilaterals

ATHS;BRHT;CSHR;BCST;CATRandABRS

6 are inscribed in circles, and since both 90= 90and 90+ 90= 180, the circles will remain even if dierent congurations permute the order of the points. An example will illustrate how these inscribed quadrilaterals may be used to solve problems.

Example 4.1

(IMO Shortlist 2010 G1).LetABCbe an acute triangle with D;E;Fthe feet of the altitudes lying onBC;CA;ABrespectively. One of the

intersection points of the lineEFand the circumcircle isP:The linesBPandDFmeet at pointQ:Prove thatjAPj=jAQj:CH

BP DQ A E F Figure 8: The construction given in the problem, with the added orthocentreH of4ABC. The following solution holds only for the conguration given in Figure 8. If the other intersection ofEFand the circumcirle of4ABCis chosen asP, the proof is similar but minor changes are needed in the angle calculations.

Solution.Note that

\QPA= 180\BPA(Supplementary angles) =\BCA(Inscribed quadrilateralAPBC) =\DCA(Don ray!CB) = 180 \DFA(Inscribed quadrilateralAFDC) =\QFA(Supplementary angles) which implies that the quadrilateralAQPFis inscribed in a circle because of the inverse of the inscribed angle theorem (P;Flie on the same side ofAQ). 7

But then

\PQA= 180\PFA(Inscribed quadrilateralAQPF) = 180 \BFE(Vertical angles) =\BCE(Inscribed quadrilateralBFEC) =\BCA(Aon ray!CE) = 180 \BPA(Inscribed quadrilateralAPBC) =\QPA(Supplementary angles)gives that\PQA=\QPA. This means that4APQis isosceles and it follows that segmentsAPandAQare of equal length.5 The re ections of the orthocentre One of the more unexpected, but surprisingly useful properties of the orthocentre, is it's symmetry with respect to the sides of the triangle. As a quick reminder, the re ection of a pointPover a line`is the pointP0such thatPP0is perpendicular to`and the distances from both points to`are equal.P P 0`

Figure 9:P0is the re

ection ofPover the line`.

Theorem 5.1.

In the triangleABCwith orthocentreH, the re

ections ofH over the linesAB;BCandCAlie on the circumscribed circle of the triangle ABC. 8 HA BC HA BC

Figure 10: The re

ections of the orthocentre over the sides of the triangle.The following proof assumes that the triangle is acute, as in the leftmost

picture of Figure 10. The proof of the obtuse case can be obtained by minor changes in the angle calculations.

Proof.

Consider only the re

ection ofHoverBCsince the two other calculations will be identical in approach. Also, letDandEbe the feet of the perpendiculars fromAandBtoBCandCArespectively, and letH0be the re ection ofH overBC.HA BC H 0DE

Figure 11:H0is the re

ection ofHoverBC. By denitions,\HDB= 90=\BDH0and segmentsHDandDH0are 9 equal in length. TrianglesHBDandH0BDare then congruent by side-angle-side, allowing for the following calculation: \H0BC=\H0BD(Don ray!BC) =\DBH(4H0BD=4HBD) = 90 \BHD(Sum of angles in4HBD) = 90 \EHA(Vertical angles) = 90 (90\HAE) =\HAE(Sum of angles in4EHA) =\H0AC(H0;Con rays!AH;!AE): Thus\H0BC=\H0AC, and sinceBandAlie on the same side ofH

0C, the

inverse of the inscribed angle theorem gives thatA;B;H0andCall lie on a circle. Since this circle passes through bothA;BandC, it is the circumscribed circle of triangleABCwhich concludes the proof.Exercise 5.1.

It is possible to re

ect objects over points as well as lines. A pointP0is said to be the re ection ofPover the pointQifQis the midpoint of segmentPP0. Show that the re ections of the orthocentre over the midpoints of a triangle also lie on the circumscribed circle of the triangle.

Example 5.1.

(P olishMO Finals 2019)

LetABCbe an acute triangle. The

pointsXandYlie on the segmentsABandAC, respectively, and are such that jAXj=jAYjand the segmentXYpasses through the orthocentre of the triangle ABC. The lines tangent to the circumcircle of the triangleAXYat the points XandYintersect at the pointP. Prove that the pointsA;B;C;Pare concyclic.A H X BCY P Figure 12: The construction given in the problem statement. Before contemplating the solution, the reader should be aware of the following result. 10 Theorem 5.2(Tangent-chord theorem).LetA;BandCbe points on a circle, and letPbe a point on the tangent to the circle atAsuch thatBandPare on opposite sides ofAC. Then\ABC=\PACas in Figure 13.ACB P

Figure 13: The tangent-chord theorem.

It is a very useful tool in handling tangency conditions, but sadly often goes overlooked in the Swedish mathematics curriculum.

Solution.

LetRandQbe the intersections closest toAofPXandPYwith the circumscribed circle of triangleABC.A H XR BC PYQ

Figure 14:RandQare dened in the section above.

Close inspection of Figure 14 suggests thatQandRare re ections ofHover the linesACandAB. This is indeed the case since the calculation \HY A=XY A(Xon ray!Y X) =\Y XA(4AXYisosceles) =\QY A(Tangent-chord theorem) 11 gives that\HY A=\QY A, implying that the re ection ofHoverCAlies on ray!Y Qdue to symmetry. The fact that the re ection ofHoverCAlies on the circumscribed circle of triangleABCnow implies that this re ection isQ, since it is the only point on ray!Y Qthat lies on this circumscribed circle. In a similar way it can be established thatRis the re ection ofHoverAB.

Now note that

\QPR=\Y PX(Y;Xon rays!PQ;!PR) = 180 \PXY\XY P(Sum of angles in4PXY) = 180

2\XAY(Tangent-chord theorem)

= 180

2\BAC(B;Con rays!AX;!AY)

implies that\QPR= 1802\BAC. From here, showing that\QCR=

1802\BACwould be enough to prove thatA;B;PandClie on a circle

because of the inverse of the inscribed angle theorem. This is just straightforward angle chasing and the solution is thus complete.Exercise 5.2. Complete the angle chasing in the solution above by showing that \QCR= 1802\BAC.

6 Conclusion

The results and examples discussed above have hopefully been both insightful and entertaining. The reader is highly encouraged to investigate the concepts of theEuler lineand theNine point circle, both of which are related to the orthocentre and the orthic triangle, but too extensive too do them justice in this short text. To conclude, a few practice problems are given for the reader to apply the ideas presented and build new problem solving strategies.

7 Exercises for the reader

1. Show that ifHis the orthocentre of triangleABC, thenAis the orthocentre of triangleHBC. This is also called the orthocentric system, where four points on a plane always can form a triangle and its corresponding orthocentre. 2. Find four similar triangles with vertices among the pointsA;B;C;D;E;F, where4DEFis the orthic triangle of4ABC. 3. Given a triangleABCand its orthocentreH, prove that the re ections ofH over the midpoints of segmentsAB;BCandCAall lie on the circumscribed circle of triangleABC. 4. In triangleABCwith orthocentreHand midpointsK;LandMof sides AB;BCandCArespectively, the pointPis chosen on the circumscribed 12 circle of4ABCsuch that\HAP= 90. Show that the midpoint of segmentHPlies on the circumscribed circle of4KLM.

Extra challenging problem:

Prove that the orthic triangle has the smallest perimeter among the triangles that can be inscribed in an acute triangle. This problem is also called Fagnano's problem and is very dicult. The reader is encouraged to rst try to prove the problem geometrically and then look up some famous solutions to the problem and attempt to complete those solutions. Some interesting geometric solutions have been carried out by Hermann Schwarz and Lipot Fejer. 13quotesdbs_dbs29.pdfusesText_35

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