[PDF] Chapter 1 The Medial Triangle - IrMO

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Chapter 1 The Medial Triangle - IrMO

Thus the orthocentre of A1B1C1 coincides with the circumcentre of ABC Figure 4: Let H be the orthocentre of the triangle ABC, that is the point of intersection of the altitudes of ABC Two of these altitudes AA2 and BB2 are shown (Figure 4) Since O is the orthocen-tre of A1B1C1 and H is the orthocentre of ABC then jAHj = 2jA1Oj The centroid

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Chapter 1. The Medial Triangle

2

Figure 1:

The triangle formed by joining the midpoints

of the sides of a given triangle is called the me- dial triangle. LetA1B1C1be the medial trian- gle of the triangleABCin Figure 1. The sides ofA1B1C1are parallel to the sides ofABCand half the lengths. SoA1B1C1is1 4 the area of ABC.

In fact

area(AC1B1) =area(A1B1C1) =area(C1BA1) =area(B1A1C) =1 4 area(ABC):

Figure 2:

The quadrilateralsAC1A1B1andC1BA1B1are parallelograms. Thus the line segmentsAA1andC1B1bisect one another, and the line segmentsBB1 andCA1bisect one another. (Figure 2)

Figure 3:

Thus the medians ofA1B1C1lie along the medians ofABC, so both tri- angles

A1B1C1andABChave the same centroidG.

3 Now draw the altitudes ofA1B1C1from verticesA1andC1. (Figure

3) These altitudes are perpendicular bisectors of the sidesBCandAB

of the triangleABCso they intersect atO, the circumcentre ofABC. Thus the orthocentre ofA1B1C1coincides with the circumcentre ofABC.

Figure 4:

LetHbe the orthocentre of the triangleABC,

that is the point of intersection of the altitudes ofABC. Two of these altitudesAA2andBB2 are shown. (Figure 4) SinceOis the orthocen- tre ofA1B1C1andHis the orthocentre ofABC then jAHj= 2jA1Oj

The centroidGofABClies onAA1and

jAGj= 2jGA1j . We also haveAA2kOA1, sinceOis the orthocentre ofA1B1C1. Thus H bAG=GcA1O; and so trianglesHAGandGA1Oare similar.

SinceHbAG=GbA1O,

jAHj= 2jA1Oj, jAGj= 2jGA1j. Thus A bGH=A1bGO, i.e.H;GandOare collinear. Furthermore,jHGj= 2jGOj: Thus Theorem 1The orthocentre, centroid and circumcentre of any trian- gle are collinear. The centroid divides the distance from the orthocentre to the circumcentre in the ratio

2 : 1.

4 The line on which these 3 points lie is called theEuler lineof the triangle. We now investigate the circumcircle of the medial triangleA1B1C1. First we adopt the notation

C(ABC)

to denote the circumcircle of the triangleABC.

Figure 5:

LetAA2be the altitude ofABCfrom the vertexA.

(Figure 5) Then A

1B1kABand

jA1B1j=1 2 jABj:

In the triangleAA2B;AcA2B= 90±andC1is the mid-

point ofAB. Thus jA2C1j=1 2 jABj. ThusC1B1A1A2is an isoceles trapezoid and thus a cyclic quadrilateral. It follows thatA2lies on the circumcircle ofA1B1C1. Similarly for the pointsB2andC2which are the feet of the altitudes from the verticesBandC.

Figure 6:

Thus we have

Theorem 2The feet of the altitudes of a trian-

gleABClie on the circumcircle of the medial triangle A

1B1C1.

LetA3be the midpoint of the line segmentAHjoining

the vertexAto the orthocentreH. (Figure 6) Then we claim thatA3belongs toC(A1B1C1), or equivalently A

1B1A3C1is a cyclic quadrilateral.

We haveC1A1kACandC1A3kBH, butBH?AC.

ThusC1A1?C1A3.

FurthermoreA3B1kHCandA1B1kAB.

But

HCkAB;thusA3B1?B1A1.

5 Thus, quadrilateralC1A1B1A3is cyclic, i.e.A32 C(A1B1C1). Similarly, ifB3C3are the midpoints of the line segmentsHBandHCre- spectively thenB3,C32 C(A1B1C1).

Thus we have

Theorem 3The 3 midpoints of the line segments joining the ortho- centre of a triangle to its vertices all lie on the circumcircle of the medial triangle. Thus we have the 9 pointsA1;B1;C1;A2;B2;C2andA3;B3;C3concyclic. This circle is theninepoint circleof the triangleABC. SinceC1A1B1A3is cyclic withA3C1?C1A1andA3B1?B1A1, thenA1A3 is a diameter of the ninepoint circle. Thus the centreNof the ninepoint circle is the midpoint of the diameterA1A3. We will show in a little while thatNalso lies on the Euler line and that it is the midpoint of the line segmentHOjoining the orthocentreHto the circumcentreO. De¯nition 1A pointA0is thesymmetric pointof a pointAthrough a third pointOifOis the midpoint of the line segmentAA0. (Figure 7)

Figure 7:

We now prove a result about points lying on the circumcircle of a triangle. LetA0be the symmetric point ofHthrough the pointA1which is the midpoint of the sideBCof a triangleABC. Then we claim thatA0belongs toC(ABC). To see this, proceed as follows.

Figure 8:

The pointA1is the midpoint of the segments

HA

0andBCsoHBA0Cis a parallelogram. Thus

A

0CkBH. ButBHextended is perpendicular toAC

6 and soA0C?AC. SimilarlyBA0kCHandCHis per- pendicular toABsoBA0?AB. Thus A bBA0=A0bCA= 90±:

ThusABA0Cis cyclic and furthermoreAA0is a di-

ameter. Thus A

02 C(ABC):

Similarly the other symmetric points ofHthroughB1andC1, which we denote byB0andC0respectively, also lie onC(ABC).

Figure 9:

Now consider the triangleAHA0. The points

A

3;A1andOare the midpoints of the sides

AH;HA

0andA0Arespectively. ThusHA1OA3is

a parallelogram soHObisectsA3A1. We saw earlier that the segmentA3A1is a diameter of

C(A1B1C1) so the midpoint ofHOis also then

the centre ofC(A1B1C1). Thus the centreNof the ninepoint circle, i.e.C(A1B1C1) lies on the

Euler line and is the midpoint of the segment

HO. Furthermore the radius of the ninepoint circle is one half of the radiusR of the circumcircleC(ABC). Now consider the symmetric points ofHthrough the pointsA2;B2and C

2where again these are the feet of the perpendiculars from the vertices. We

claim these are also onC(ABC).

Figure 10:

LetBB2andCC2be altitudes as shown

in Figure 10,His the point of intersec- tion.

ThenAC2HB2is cyclic soC2bAB2+C2bHB2=

180
7

ButBHC=C2bHB2so

BHC= 180±¡C2bAB2which we write as

= 180

±¡bA

By construction, the trianglesBHCandBA00Care congruent, soBbHC= BA

00C. Thus

BA

00C= 180±¡bA;

and soABA00Cis cyclic. Thus A

002 C(ABC)

Similarly forB00andC00.

8 Returning to the symmetric points throughA1;B1andC1of the ortho- centre, we can supply another proof of Theorem 1. Theorem 1The orthocentreH, centroidGand circumcentreOof a triangle are collinear points.

Figure 11:

ProofIn the triangleAHA0, the pointsO

andA1are midpoints of sidesAA0andHA0respec- tively. (Figure 11) Then the line segmentsAA1and

HOare medians, which intersect at the centroidG0

of4AHA0and furthermore jG0Hj jG0Oj= 2 =jG0Aj jG0A1j

ButAA1is also a median of the triangleABCso the

centroidGlies onAA1with jGAj jGA1j= 2 ThusG0coincides withGand soGlies on the lineOHwith jGHj jGOj= 2

Remark

On the Euler line the pointsH(orthocentre),N(centre of ninepoint circle),G(centroid) andO(circumcircle) are located as follows:

Figure 12:

with jHNj jNOj= 1 andjHGj jGOj= 2 9

Figure 13:

IfABCDis a cyclic quadrilateral, the four triangles formed by selecting 3 of the vertices are called the diag- onal triangles. The centre of the circumcircle ofABCD is also the circumcentre of each of the diagonal trian- gles. We adopt the notational convention of denoting points as- sociated with each of these triangles by using as subscript the vertex of the quadrilateral which is not a vertex of the diagonal triangle. ThusHA,GAandIAwill denote the orthocentre, the centroid and the incentre respectively of the triangleBCD. (Figure 13) Our ¯rst result is about the quadrilateral formed by the four orthocentres.

Figure 14:

Theorem 4LetABCDbe a cyclic quadrilateral and let

H A,HB,HCandHDdenote the orthocentres of the diagonal trianglesBCD;CDA;DABandABCrespectively. Then the quadrilateral H

AHBHCHD

is cyclic. It is also congruent to the quadrilateralABCD.

ProofLetMbe the midpoint ofCDand let

A

0andB0denote the symmetric points throughM

of the orthocentresHAandHBrespectively. (Figure 14) The linesAA0andBB0are diagonals of the circumcircleofABCDso ABB

0A0is a rectangle. Thus the sidesABandA0B0are parallel and of same

length. The linesHAA0andHBB0bisect one another soHAHBA0B0is a parallel- ogram so we get thatHAHBis parallel toA0B0and are of the same length. Thus H

AHBandABare parallel and are of the same length.

10 Similarly one shows that the remaining three sides ofHAHBHCHOare parallel and of the same length of the remaining 3 sides ofABCD. The result then follows. The next result is useful in showing that in a cyclic quadrilateral, various sets of 4 points associated with the diagonal triangles form cyclic quadrilat- erals.

Figure 15:

Proposition 1LetABCDbe a cyclic

quadrilateral and letC1;C2;C3andC4be circles through the pair of points

A;B;B;C;C;D;D;A

and which intersect at pointsA1;B1;C1and D

1. (Figure 15) ThenA1B1C1D1is cyclic.

ProofLetB1be the point where cir-

cles throughABandBCmeet. Similarly for the pointsA1,C1andD1. Join the pointsAA1,BB1,CC1andDD1, extend to points~A;~B;~Cand~D. (These exten- sions are for convenience of referring to angles later.)

We will apply to the previous diagram the

result that if we have a cyclic quadrilateral then an exterior angle is equal to the opposite angle of the quadrilateral. In Figure 16, ifCBis extended to~Bthen:

BbBA+AbBC= 180±

C bDA+AbBC= 180± Thus ~BbBC=C~DA In Figure 17 we apply the above result to 4 quadrilaterals. 11

Figure 16:

Figure 17:

InC1CDD1

~CbC1D1=CbDD1~DcD1C1=C1bCD

InBCC1B1

~CcC1B1=CbBB1~DcB1C1=C1bCB

InBB1A1A

BbB1A1=BbAA1~AA1B1=AbBB1

InAA1D1D

AA1D1=AbDD1~DD1A1=DbAA1

Adding up angles

C =C1bCD+A1bAD+C1bCB+BbAA1 =C1bCD+C1bCB+A1bAD+BbAA1 =BbCD+BbAD+ 180±

Similarly

D

1bC1B1+D1bA1B1= 180±

12

We now apply this to orthocentres and in-

centres of the diagonal triangles. Theorem 5LetABCDbe a cyclic quadrilateral and letHA;HB;HC andHDbe the orthocentres of the diagonal trianglesBCD;CDA;DABand ABC, respectively. ThenHAHBHCHDis a cyclic quadrilateral.

Figure 18:

ProofIn the triangleBCD, (Figure 18),

C bHAD= 180±¡CbBD:

In the triangleCDA,CbHBD= 180±¡DbBC

But D bAC=DbBCso C bHAD=CbHBD and so we conclude thatCDHBHAis cyclic. Similarly, show thatHBDAHC; HCABHD andHDBCHAare cyclic. Now apply the proposition to see thatHAHBHCHDis cyclic.

Figure 19:

Theorem 6IfABCDis a cyclic quadrilateral and

ifIA;IB;IC;IDare the incentres of the diagonal triangles

BCD;CDA;DABandABC, respectively, then the points

I

A;IB;ICandIDform a cyclic quadrilateral.

13

ProofRecall that ifABCis a triangle,Iis the in-

centre andP;Q;Rare feet of perpendiculars fromIto the sidesBC;CAandABthen B bIC=1 2 (RIP+PIQ) 1 2 (360±¡RIQ) 1 2 (360±¡180±+RbAQ) = 90

±+RbAQ

2

Now apply this to the trianglesBCDandACD. We get

(Figure 20) :

Figure 20:

C bIAD= 90±+1 2 (CbBD); C ^IBD= 90±+1 2 (CbAD): ButCbBD=CbADso it follows thatCbIAD=CbIBD. ThusIACDIBis a cyclic quadrilateral. The proof is now completed as in previous theorem. Theorem 7IfABCDis a cyclic quadrilateral andGA,GB,GCand G Dare the centroids of the diagonal trianglesBCD;CDA;DABandABC, respectively, then the quadrilateralGAGBGCGDis similar toABCD. Fur- thermore, the ratio of the lengths of their corresponding circles is1 3 14 ProofRecall that the quadrilateral formed by the orthocentresHAHBHCHD is congruent toABCD.

Figure 21:

We also have the fact that all four diagonal tri-

angles have a common circumcentre which is the cen- tre of the circleABCD. Let this be denoted by O.

Now joinOtoHA,HB,HCandHD. (Figure 21) The

centroidsGA,GB,GCandGDlie on these line segments and jOGAj jOHAj=jOGBj jOHBj=jOGCj jOHCj=jOGDj jOHDj=1 3 Then it follows thatGAGBGCGDis similar toHAHBHCHD.

The result now follows.

Finally we have

Theorem 8LetABCDbe a cyclic quadrilateral and letA1andC1 be the feet of the perpendiculars fromAandC, respectively, to the diagonal BDand letB1andD1be the feet of the projections fromBandDonto the diagonalAC. ThenA1B1C1D1is cyclic. ProofConsider the quadrilateralBCB1C1(Figure 22). Since

Figure 22:

15 B bC1C=BbB1C= 90±;thenBCB1C1is cyclic. Similarly,A;A1;B1;Bare cyclic,C;C1;D1;Dare cyclic andA;A1;D;D1 are cyclic. Then by results of Proposition 1,A1B1C1D1is cyclic. 16

Figure 23:

Four Circles Theorem

If 4 circles are pairwise externally tangent, then the points of tangency form a cyclic quadrilateral.

In Figure 23, the quadrilateralABCDis cyclic.

Remark

A similar theorem could not be true for 5 cir-

cles as 3 of the intersection points may lie on a line.

Figure 24:ABClie along a line

Figure 25:

Proof

Recall that ifTKis tangent to circle at

TandOis centre of circle, then the angle between chord TLand tangent lineTis one half of angle subtended at centreOby chordTL, i.e.KTL=1 2 (TbOL). This is becauseKbTL=TbRLwhereTRis the diameter at T. Draw tangent linesAA1;BB1;CC1andDD1at points of contact withA1;B1;C1;D1being points in region bounded by the circles. Thus B bAD+BbCD =BbAA1+A1bAD+BbCC1+C1bCD 1 2 (AbO1B+AbO2D+BbO3C+CbO4D) 1 2 (sum of angles of quadrilateralO1O2O3O4) = 180 17 Here we used the fact that the point of tangency of a pair of circles lies on the line joining their centres.

Remark

Ifr1;r2;r3andr4denote the radii of the four circles thenO

1O2+O3O4=r1+r2+r3+r4

andO1O3+O3O4=r1+r2+r3+r4 ThusO1O2O3O4is a quadrilateral with the sums of the oppo- site side lengths equal. Such a quadrilateral is calledinscribable, i.e. it has an incircle. In this situation, the circumcircle ofABCD is the incircle ofO1O2O3O4.quotesdbs_dbs29.pdfusesText_35