hagge circles revisited

8/11/2019 Hagge Circles Revisited

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Hagge circles revisited

Nguyen Van Linh

24/12/2011

Abstract

In 1907, Karl Hagge wrote an article on the construction of circles that always pass through theorthocenter of a given triangle. The purpose of his work is to nd an extension of the Wallace-Simsontheorem when the generating point is not on the circumcircle. Then they was named Hagge circles.In this paper, we present the new insight into Hagge circles with many corollaries. Furthermore, wealso introduce three problems which are similar to Hagge circles.

1 Hagge circles and their corollaries

Problem 1 (The construction of Hagge circles). Given triangle ABC with its circumcircle(O) and orthocenter H . Choose P an arbitrary point in the plane. The cevian lines AP,BP,CP meets ( O) again at A1 , B 1 , C 1 . Denote A2 , B 2 , C 2 the reections of A1 , B 1 , C 1 with respect toBC,CA,AB , respectively. Then H, A 1 , B 1 , C 1 are concyclic. The circle ( H, A 1 , B 1 , C 1 ) is calledP-Hagge circle.

Here we give two proofs for problem 1. Both of them start from the problem which appeared inChina Team Selection Test 2006.

Problem 2 (China TST 2006). Given triangle ABC with its circumcircle ( O) and orthocenterH . Let P be an arbitrary point in the plane. The cevian lines AP,BP,CP meets ( O) again atA1 , B 1 , C 1 . Denote A2 , B 2 , C 2 the reections of A1 , B 1 , C 1 across the midpoints of BC,CA, AB ,respectively. Then H, A 1 , B 1 , C 1 are concyclic.

Proof.We introduce a lemma:Lemma 1. (B 2 C 2 , B 2 A2 ) (PC ,PA )(mod )Proof.

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B'

C'

A'

M

N

B 2

C 2

A2

X

Z Y

C 1

B 1

A1

O

P

A

BC

Denote M, N the reections of A2 , C 2 across the midpoint of AC then MB 1 N is the reectionof A2 B 2 C 2 across the midpoint of AC .

Since AM CA2 , AM = CA2 and BA1 CA2 , BA 1 = CA2 we get AM A 1 B is a parallelogram.Similarly, BC 1 NC is a parallelogram too.

Let A , B , C be the midpoints of AA1 , BB 1 , CC 1 then A , C are the midpoints of BM,BN ,respectively. The homothety H

12

B : M A , B 1 B , N C therefore MB 1 N A B C .This means ( B 1 M, B 1 N ) (B A , B C )(mod ).

On the other side, OA , OB , OC are perpendicular to AA1 , BB 1 , CC 1 hence O, A , B , C , P are concyclic, which follows that:

(B 2 A2 , B 2 C 2 ) (B 1 M, B 1 N ) (B A , B C ) (P A , P C ) (PA,PC )(mod ). We are done.Back to problem 2.

C 3

B 3

A3

B 4

C 4

A4

H

A2

B 2

C 2

Z Y

X

C 1

B 1

A1

OP

A

BC

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Construct three lines through A1 , B 1 , C 1 and perpendicular to AA1 , BB 1 , CC 1 , respectively.They intersect each other and form triangle A3 B 3 C 3 , intersect ( O) again at A4 , B 4 , C 4 , respectively.

Note that AA 4 , BB 4 , CC 4 are diameters of ( O) therefore A 4 , C 4 are the reections of H acrossthe midpoints of BC,AB . We get HA 2 A4 A1 , HC 2 C 4 C 1 are parallelograms, which follows thatHA 2 A1 A4 , H C 2 C 1 C 4 .

But P C 1 B 3 A1 is a cyclic quadrilateral and applying lemma 1 we have ( HC 2 , HA 2 ) (B 3 C 1 , B 3 A1 ) (P C 1 , P A 1 ) (PC,PA ) (B 2 C 2 , B 2 A2 )(mod ) or H, A 2 , B 2 , C 2 are concyclic. Our proof iscompleted.

Remark 1: Now come back to problem 1. Let A3 , B 3 , C 3 be the reections of A2 , B 2 , C 2 acrossthe midpoints of BC,CA, AB , respectively. Since ( BA 2 C ) is the reections of ( ABC ) wrt BC thenA3 (ABC ). It is easy to see that BC is the midline of the triangle A1 A2 A3 hence A1 A3 BC . Weclaim that AA3 , BB 3 , CC 3 concur at the isogonal conjugate Q of P wrt ABC . Then accordingto problem 2 we obtain H, A 2 , B 2 , C 2 are concyclic.

Another proof (Nsato) [9]We introduce a lemma.Lemma 2. Given triangle ABC and its circumcircle ( O). Let D be an arbitrary point on ( O).

A line d through D and d BC.d (O) = {G} then AG is parallel to the Simson line of point D .Proof.

F

E

G A

B C

D

Denote d BC = {E }, F the projection of D on AC then EF is the Simson line of D .Since DEFC is a cyclic quadrilateral we get GAC = GDC = EF A . Therefore AG EF .Back to problem 1.

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C 4

B 4

Y

A1

LZ

K

A4

X

Q

A3

H

A2

P

A

B C

Let Q be the isogonal conjugate of P wrt ABC ; X,Y,Z be the reections of A,B, C wrt Q;A3 B 3 C 3 be the circumcevian triangle of Q; A4 B 4 C 4 be the median triangle of ABC .

The symmetry with center Q takes A to X, B to Y, C to Z then ABC XY Z . But A 4 B 4 C 4is the median triangle of ABC hence there exist a point K which is the center of the homothetywith ratio

12

. This transformation maps XY Z to A4 B 4 C 4 .On the other hand, let L be the second intersection of A1 A2 and ( O). Applying lemma 2,

AL HA 2 .Moreover, LA1 is perpendicular to BC and A3 A1 is parallel to BC therefore A3 A1 LA1 ,

which follows that AA 3 AL or AA 3 HA 2 .

From remark 1, A4 is the midpoint of A2 A3 hence XA 2 KA 3 is a parallelogram. This meansKA 2 AA 3 .So KA 2 HA 2 or A 2 lies on the circle with diamenter KH . Similarly we are done.

Problem 3. Given triangle ABC with its circumcircle ( O) and orthocenter H . Let P bean arbitrary point in the plane. The cevian lines AP,BP,CP meets ( O) again at A1 , B 1 , C 1 .Denote A2 , B 2 , C 2 the reections of A1 , B 1 , C 1 with respect to BC,CA, AB , respectively. Then A2 B 2 C 2 A1 B 1 C 1 .

Proof.Firstly we introduce a lemma:Lemma 3. Given triangle ABC and its circumcircle ( O). Let E , F be two arbitrary points on

(O). Then the angle between the Simson lines of two points E and F is half the measure of the arcEF .

Proof.

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L

K

G

HI

J

O

A

B C

F

E

See on the gure, IJ, HG are the Simson lines of two points E and F , respectively. IJ HG ={K }

Denote L the projection of K on BC . We have KL F G and FH GC is a cyclic quadrilateral

thus ( KG,KL ) (GK, GF ) (CA,CF )(mod )(1)Similarly, ( KL,KJ ) (BE,BA )(mod )(2)From (1) and (2) we are done.Back to problem 3.

C 2

B 2

A2C 1

B 1

A1

H

O

P

A

BC

According to Hagge circle, we have H, A 2 , B 2 , C 2 are concyclic.Since A 2 is the reections of A 1 wrt BC we get H A 2 is the Steiner line of A 1 , which is parallel

to its Simson line.Analogously, HB 2 is the Steiner line of B 1 . Then applying lemma 3, ( C 2 A2 , C 2 B 2 ) (HA 2 , HB 2 )

(C 1 A1 , C 1 B 1 )(mod ). Do similarly with other direct angles we refer to the similarity of two tri-angles A 1 B 1 C 1 and A 2 B 2 C 2 .

Problem 4. Using the same notations as problem 1, let A 3 , B 3 , C 3 be the second intersectionsof AH,BH,CH and ( A2 B 2 C 2 ), respectively. Then A 2 A3 , B 2 B 3 , C 2 C 3 concur at P .

Proof.

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A3

C 3

B 3

H

B 2

B1

A2

A1

C 2

C 1

P

A

B C

(A2 C 2 , A2 A3 ) (HC 2 , HA 3 ) (HC 2 , BC )+( BC,HA ) / 2+( HC 2 , BC )(mod ). Let L, Z bethe projections of C 1 on BC,AB , respectively then LZ is the Simson line of C 1 . We get H C 2 LZ

Therefore ( HC 2 , BC ) (LZ,LC ) (C 1 Z, C 1 B )(mod )Hence (A2 C 2 , A2 A3 ) / 2 + ( C 1 Z, C 1 B ) (BA,BC 1 ) (CA,CC 1 )(mod )Since H, A 3 , B 3 , C 3 are concyclic then ( C 3 B 3 , C 3 A3 ) (HB 3 , HA 3 ) (CA,CB )(mod ) . Sim-

ilarly we get A3 B 3 C 3 ABC . But from problem 3, A2 B 2 C 2 A1 B 1 C 1 .Thus there exist a similarity transformation which takes A 3 C 2 B 3 A2 C 3 to AC 1 BA 1 CB 1 and it

follows that A 2 A3 , B 2 B 3 , C 2 C 3 concur at P . Moreover, AP A1 P

= A3 P A2 P

hence P P AH . Similarly,

P P BH, CH . This means P P . Our proof is completed then.

Problem 5. Let I be the circumcenter of triangle A 2 B 2 C 2 . K is the reection of H across I .AK,BK,CK cut ( I ) again at A 4 , B 4 , C 4 . Then A 3 A4 , B 3 B 4 , C 3 C 4 are concurrent.Proof.

A4

B 4

C 4K A3

B 3 C 3

I

HO

P

A

BC

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Since K A 3 AH we have (C 3 C 4 , C 3 A3 ) (KC 4 , KA 3 ) (CK,CB )(mod ).Similarly, ( C 3 C 4 , C 3 B 3 ) (CK,CA )(mod ). Do the same with other angles then applying

Ceva-sine theorem for triangle A 3 B 3 C 3 we are done.

Problem 6. Given triangle ABC with its circumcircle ( O). Let P and Q be two isogonalconjugate points, O1 , O2 be the circumcenters of P-Hagge and Q-Hagge circles, respectively. ThenP Q O1 O2 and P Q = O1 O2 .

Proof.Lemma 4. Given triangle ABC with its orthocenter H , circumcenter O and let P be an

arbitrary point in the plane. AP,BP,CP intersect ( O) again at A1 , B 1 , C 1 , respectively. DenoteA2 , B 2 , C 2 the reections of A1 , B 1 , C 1 across the midpoints of BC,CA, AB , respectively, O2 thecenter of ( A2 B 2 C 2 ). Then OO2 HP is a parallelogram.

Proof.

H'

U ZY

X

O' 2

A'

C'

H

O 2 B 2

C 2

A2

C 1

B 1

A1

O

P

A

B

C

Denote A , C the reections of A 2 , C 2 across the midpoint of AC , respectively. Then A B 1 C is the reections of A2 B 2 C 2 across the midpoint of AC .

Since AA A2 C,AA = A2 C, A 2 C BA1 , A2 C = BA1 we get AA A1 B is a parallelogram. (1)Similarly, BCC C 1 is a parallelogram. (2)Let X,Y,Z be the midpoints of AA 1 , BB 1 , CC 1 , respectively then X,Y,Z lies on (OP ).But from (1) and (2) ,X ,Z are the midpoints of BA , BC .

A homothety H 12

B : A X, B 1 Y, C Z therefore A B 1 C XY Z A symmetry with center is the midpoint of BC :S : B B , H H the orthocenter of triangle AB C , O2 O2 O2 H O2 H It is easy to show that B,O,H are collinear. Hence the homothety H

12

B : H O, O2 U thecircumcenter of triangle X Y Z .

We conclude that OP = O2 H = O2 H . Therefore HPOO 2 is a parallelogram. We are done.Back to our problem.From the lemma above, OO1 = QH, OO 2 = P H thus P Q = O1 O2

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Our proof is completed.

Remark 2. The radius of P-Hagge circle is equal to the distance of Q and O.

Problem 7. Given triangle ABC , its Nine-point circle ( E ) and an arbitrary point P on (E ).Let Q be the isogonal conjugate point of P wrt ABC . Then the center I of Q-Hagge circle lieson (E ) too. Moreover, I is the reection of P across point E .

Proof.The conclusion follows immediately from remark 2. Note that the center of Q-Hagge circle is

always the reection of P wrt E .

Problem 8. Given triangle ABC and its circumcenter O. P and Q are isogonal conjugate wrttriangle ABC such that O,P, Q are collinear. Then P-Hagge circle and Q-Hagge circle are tangent.

Proof.This problem is also a corollary of remark 2. We will leave the proof for the readers.

Problem 9. Given triangle ABC and its circumcircle ( O, R ). A circle which has center O andradius r < R meets BC,CA, AB at A1 , A2 , B 1 , B 2 , C 1 , C 2 , respectively. Let E, F be the Miquelpoints of triangle ABC wrt ( A1 , B 1 , C 1 ) and ( A2 , B 2 , C 2 ) respectively. Then E-Hagge circle andF-Hagge circle are congruent.

Proof.

We will introduce a lemma.Lemma 5. Given triangle ABC . A circle (O) intersects BC,CA,AB at 6 points A1 , A2 , B 1 , B 2 , C 1 ,C 2 , respectively. Let P, Q be the Miquel points of ABC wrt ( A1 , B 1 , C 1 ) and ( A2 , B 2 , C 2 ). ThenP and Q are isogonal conjugate and OP = OQ.

Proof.

Y

Z

X

Q

PC 2

C 1

B 2

B 1

A2

A

B C

O

A1

Let X,Y,Z be the second intersections of A 1 P, B 1 P, C 1 P and ( O), respectively.We have ( ZY,ZX ) (A1 Y, A1 X ) (A1 Y,PY ) + ( PY,A 1 X ) (A2 C, A 2 B 1 ) + ( CA 2 , CB 1 )

(B 1 B 2 , B 1 A2 )(mod )This means X Y = A2 B 2 . Similarly, Y Z = B2 C 2 , XZ = A2 C 2 . Therefore XY Z = A2 B 2 C 2 .On the other side, ( PX,PY ) (QA 2 , QB 2 )(mod ), (PY,PZ ) (QB 2 , QC 2 )(mod ) then there

exist a rotation with center O which takes XY Z to A2 B 2 C 2 and P to Q. We conclude thatOP = OQ.

Moreover, ( AP,AB ) (B 1 P, B 1 C 1 ) (ZY,ZP ) (C 2 B 2 , C 2 Q) (AB 2 , AQ)(mod ).Similarly we get P and Q are isogonal conjugate wrt ABC.Back to our problem.According to lemma 5, P and Q are isogonal conjugate wrt ABC and OP = OQ. Then

applying remark 2 the conclusion follows.

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Problem 10. Given triangle ABC with its orthocenter H . Let P, Q be two isogonal conjugatepoints wrt ABC . Construct the diameter HK of P-Hagge circle. Let G be the centroid of thetriangle ABC . Then Q, G, K are collinear and QG =

12

GK .Proof.

X

Q

G

Ma

K

H

C 2

C 1

B 2

B 1

A2

A1

P

A

B C

Let X be the reection of A wrt Q, M a be the midpoint of BC then from proof 2 of problem 1,we obtain M a is the midpoint of X K .

Then applying Menelaus theorem for the triangle AXM a with the line ( Q,G,K ) we have Q,G,K are collinear. Applying Menelaus theorem again for the triangle QXK with the line ( A,G,M a ) then

QG = 12

.

Problem 11. The centroid G lies on L-Hagge circle where L is the Symmedian point.Proof.

C 1

B 3

C 3 A3

C 2

A2

B 2

B 1

A1

G

A

B C

Let A 1 B 1 C 1 , A2 B 2 C 2 be the circumcevian triangles of G and L, respectively; A 3 , B 3 , C 3 be thereections of A 2 , B 2 , C 2 wrt BC,CA, AB , respectively.

It is easy to see that A 3 , B 3 , C 3 lie on AG, BG, CG , respectively.From problem 3, A3 B 3 C 3 A2 B 2 C 2 . Then ( A2 B 2 , A2 C 2 ) (A3 C 3 , A3 B 3 )(mod )

On the other side, ( GC 3 , GB 3 ) 12

(CB 1 + C 1 B ) 12

(B 2 A + AC 2 ) (A2 B 2 , A2 C 2 ) (A3 C 3 , A3 B 3 )(mod ).

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This means G, A3 , B 3 , C 3 are concyclic. We are done.

More results.1. When P I the incenter of triangle ABC , P-Hagge circle is called Fuhrmann circle of

ABC , after the 19th-century German geometer Wilhelm Fuhrmann . The line segment con-necting the orthocenter H and the Nagel point N is the diameter of Fuhrmann circle. The solutionfor this result can be found in [8]. Moreover, denote O, I the circumcenter and the incenter of triangle ABC then H N = 2 OI .

2. When P lies on the circumcircle of triangle ABC , P-Hagge circle becomes the Steiner line of P .

3. When Q lies on the circumcircle of triangle ABC , we have a problem:

Given triangle ABC and an arbitrary point P on its circumcircle ( O). Denote A , B , C thereections of P across the midpoints of BC,CA, AB . Then H, A , B , C are concyclic.

4. When P lies at the innity, problem 1 can be re-written as below:Given triangle ABC with its circumcircle ( O), orthocenter H . Denote A 1 , B 1 , C 1 points on ( O)

such that AA1 BB 1 CC 1 , A2 , B 2 , C 2 the reections of A1 , B 1 , C 1 across the lines BC,CA, AB ,respectively. Then H, A 2 , B 2 , C 2 are concylic.

5. When P H the orthocenter of ABC , P-Hagge circle becomes point H .6. When P O the circumcenter of ABC , the center of P-Hagge circle coincides with O.

2 Results from the triads of congruent circles

Three year ago, Quang Tuan Bui , the Vietnamese geometer, wrote an article about twotriads of congruent circles from reections in the magazine Forum Geometricorum. In that note,he constructed two triads of congruent circles through the vertices, one associated with reectionsin the altitudes, and the other reections in the angle bisectors. In 2011, Tran Quang Hung , theteacher at High School for Gifted Student, Hanoi University of Science, also constructed anothertriad which associated with reections in the lines joining the vertices and the Nine-point center.His work was published in a abook called Exploration and Creativity. Here we introduce theinvolvement of Hagge circles in these problems.

Problem 12. Given triangle ABC with its orthocenter H , circumcenter O. Let Ba , C a bethe reections of B and C across the line AH . Similarly, consider the reections C b , Ab of C, A,respectively across the line BH , and A c , B c of A, B across the line C H . Then the circumcircles of three triangles AC b B c , BA c C a , CA b B a and O-Hagge circle are congruent.

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C 2

C 1 B 1

A1 Ab

C b

B c

C a

Ac

B a

H O

A

B C

Note. In [2], Quang Tuan Bui proved that three circles ( AC b B c ), (BA c C a ), (CA b B a ) are con-

gruent with the circle ( H,HO ). By remark 2, H O is the radius of O-Hagge circle. We are done.

Problem 13. Let I be the incenter of triangle ABC . Consider the reections of the verticesin the angles bisectors: Ba , C a of B, C in AI, A b , C b of A, C in BI, A c , B c of A, B in CI . Then thecircumcircles of three triangles AC b B c , BA c C a , CA b B a and Fuhrmann circle are congruent.

H

S

B 2

C 2 A2

C 1

B 1

A1

T

C b

C a

Ab

Ac

B c

B aI

O

A

BC

Note. In his article, Quang Tuan Bui also proved that three circles ( AC b B c ), (BA c C a ), (CA b B a )are congruent with the circle ( I , IO ). By remark 2, IO is the radius of Fuhrmann circle and it com-pletes the proof.

Problem 14. Given triangle ABC with its Nine-point center E . Let B a , C a be the reectionsof B and C across the line AE . Similarly, consider the reections C b , Ab of C, A, respectively acrossthe line BE , and Ac , B c of A, B across the line CE . Then the circumcircles of three trianglesAC b B c , BA c C a , CA b B a and E-Hagge circle are congruent.

Proof.

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C b

B 2

A2

A1

B 1

Q

O c

C 2J

C 1

B c

E

H

O

A

B C

Denote A 1 B 1 C 1 the circumcevian triangle of E . A2 , B 2 , C 2 the reections of A 1 , B 1 , C 1 across

BC,CA,AB , respectively. Let H be the orthocenter of triangle ABC . CH meets the circumcircle(O) of triangle ABC again at Q, (Oc ) be the circumcircle of triangle AH B .Since (Oc ) and ( O) are symmetric wrt AB,C 2 is the reection of C 1 wrt AB we obtain C 2 (Oc ).

It is not hard to see that Oc lies on C E . However, B and Bc are symmetric wrt C E we concludethat B c (Oc ).

We have ( BA,BB c ) (CQ,CC 1 )(mod ), which follows from the fact that CQ AB, BB c CC 1 . This means the direct angle ( BA,BB c ) is half the measure of the arc QC 1 or HC 2 . ThereforeAB c = HC 2 or AH C 2 B c is a isosceles trapezoid.

Analogously, AHB 2 C b is also a isosceles trapezoid. We claim ( AB c C b ) is the image of ( HC 2 A2 )across the symmetry whose the axis is the perpendicular bisector of AH . This means two circles(AB c C b ) and E-Hagge circle are congruent. Similarly we are done.

Remark 3. Other synthetic proofs of the congruent of three circles ( AC b B c ), (BA c C a ), (CA b B a )can be found in [6] or [7]. We have some properties:

1. The radius of three circles ( AC b B c ), (BA c C a ), (CA b B a ) is equal to the distance of O andKosnita point K (the isogonal conjugate of E ).

2. The orthocenter of triangle I a I b I c is the reection of H wrt the center of N-Hagge circle(I a , I b , I c are the circumcenters of three triangles AC b B c , BA c C a , CA b B a , respectively).

3. N is the center of K-Hagge circle.

3 Near Hagge circles

Finally, we give 3 interesting problems which involve in Hagge circles.

Problem 15. Given triangle ABC , its orthocenter H and its circumcircle ( O). DenoteA1 , B 1 , C 1 the reections of A,B,C wrt O, respectively, P an arbitrary point in the plane. LetA2 B 2 C 2 be the pedal triangle of P wrt ABC,A 3 , B 3 , C 3 be the reections of A1 , B 1 , C 1 wrtA2 , B 2 , C 2 , respectively. Then 4 points H, A 3 , B 3 , C 3 are concyclic.

Proof.

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A5

B 5

C 5H

A3

B 3

C 3

C 4 B 4

A4

C 2

B2

A2

C 1 B 1

A1

O

P

A

B C

Let A 4 , B 4 , C 4 be the midpoints of BC,CA, AB , respectively then A 4 is the midpoint of A 3 H .

Similar with B 4 , C 4 .We get H A 3 = 2 A2 A4 , HB 3 = 2 B 2 B 4 , HC 3 = 2 C 2 C 4 .(1)Construct three rectangles OA4 A2 A5 , OB 4 B 2 B 5 , OC 4 C 2 C 5 hence A5 , B 5 , C 5 lie on the circle

with diameter OP. (2)But OA5 = A4 A2 , OB 5 = B4 B 2 , OC 5 = C 4 C 2 (3)From (1) , (2) and (3) we claim H, A 3 , B 3 , C 3 are concyclic. Moreover, we can show that A3 B 3 C 3 ABC .

Problem 16. Given triangle ABC with its circumcircle ( O) and its orthocenter H . Let H a H b H cbe the orthic triangle of triangle ABC . P is an arbitrary point on Euler line. AP,BP,CP intersect(O) again at A1 , B 1 , C 1 . Let A2 , B 2 , C 2 be the reections of A1 , B 1 , C 1 wrt H a , H b , H c , respectively.Then H, A 2 , B 2 , C 2 are concyclic ( See [12] or [14]).

Proof.

We will prove this problem in the general case as below.Generalization.Given triangle ABC and its circumcircle ( O). Let X, Y be two points inside ( O) such that O

lies on the line segment XY . AX,BX,CX intersect ( O) again at A1 , B 1 , C 1 , AY,BY,CY inter-sect (O) again at A2 , B 2 , C 2 . Let A3 , B 3 , C 3 be the reections of A2 , B 2 , C 2 wrt the midpoints of A1 X, B 1 Y, C 1 Z . Then A 3 , B 3 , C 3 , X lie on a circle (O ) and O lies on X Y .

Proof.

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U

T

S

Z

W

V

B3

C 3

A3

P

Q

C 1

C 2

B 2

B 1

A2 A1

O

X

A

B C

Y

Let S be the intersection of A 1 C 2 and A2 C 1 . Applying Pascals theorem for 6 points C 1 , C 2 , A1 ,A2 ,A,C we get S XY .

Denote T, U the second intersections of the perpendicular lines through A1 , C 1 to A1 A2 , C 1 C 2with ( O), respectively.

Applying Pascals theorem again for 6 points A1 , A2 , C 1 , C 2 , T, U We conclude that P XY .Similarly we obtain the perpendicular line through B 1 to B1 B 2 also passes through P , three lineswhich pass through A 2 , B 2 , C 2 and perpendicular to A 1 A2 , B 1 B 2 , C 1 C 2 , respectively concur at Q XY .

Construct three rectangles P A1 A2 V,PB 1 B 2 W,PZC 2 C 1 then XA 3 A1 A2 P V and XA 3 =A1 A2 = P V . Let L be the midpoint of X P then LA 3 = LV = 12 A3 V.

Similarly we get ( A3 B 3 C 3 ) is the reection of ( V W Z ) wrt L. But P,V,W, Z lie on a circle withdiameter P Q so X lies on (A3 B 3 C 3 ). Moreover, O is the reection of O wrt L, which follows thatO XY . We are done.

Problem 17. Given triangle ABC and its circumcircle ( O), its orthocenter H . Let P bean arbitrary point in the plane. AP,BP,CP intersect ( O) again at A1 , B 1 , C 1 , respectively. LetA2 B 2 C 2 be the pedal triangle of P wrt ABC,A 3 , B 3 , C 3 be the reections of A1 , B 1 , C 1 wrtA2 , B 2 , C 2 , respectively. Then A 3 , B 3 , C 3 , H are concyclic.

Proof.We introduce two lemmas.Lemma 6. Three triangles A 1 B 1 C 1 , A2 B 2 C 2 , A3 B 3 C 3 are similar.Proof.

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B4

C 4

B3

A3

C3

C 1

B1

A1

C 2

B2

A2

O

P

A

B C

The result A1 B 1 C 1 A2 B 2 C 2 is trivival. So we only show that A3 B 3 C 3 A1 B 1 C 1 .Let B 4 , C 4 be the reections of B 3 , C 3 wrt A 2 , respectively. We get A1 B 4 C 4 A3 B 3 C 3 .(1)Since A 2 C 2 is the midline of triangle C 1 C 3 C 4 then C 1 C 4 A 2 C 2 . Analogously, B1 B 4 A 2 B 2 .

Denote R the intersection of B1 B 4 and C 1 C 4 . We conclude that ( RB 1 , RC 1 ) (A2 B 2 , A2 C 2 ) (A1 B 1 , A1 C 1 )(mod ), which follows that R (O).

Hence (C 1 A1 , C 1 C 4 ) (B 1 A1 , B 1 B 4 )(mod ).

On the other side, A1 C 1A1 B 1

= A2 C 2A2 B 2

= C 1 C 4B 1 B 4

. Therefore A1 C 1 C 4 A1 B 1 B 4 .

This means ( A1 B 1 , A1 B 4 ) (A1 C 1 , A1 C 4 )(mod ), then ( A1 B 4 , A1 C 4 ) (A1 B 1 , A1 C 1 )(mod ).Similarly we get A1 B 1 C 1 A1 B 4 C 4 .(2)

From (1) and (2) we are done.

Lemma 7. Given triangle ABC . Let A1 be the projection of A on BC,A 2 , B 2 , C 2 be the

midpoints of BC,CA, AB , respectively. Let P be an arbitrary point in the plane, A B C be thepedal triangle of P wrt ABC,F and F be the intersections of ( A B C ) and ( A2 B 2 C 2 ). A linethrough A and parallel to AP intersects AA1 at A . Then the circle with diameter A A passesthrough one of two point F, F .

Proof.

MN

I'

L

I

A''

A1

F'

V

C'

A'

B'

C 2B 2

A2

A

B C

P

Let V be the intersection of B C and B2 C 2 . According to the rst Fontenes theorem (see [4]or [5]), we obtain A , V, F are collinear.

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On the other side, denote I , L the midpoints of AP,AA , respectively, I the center of ( A A ).Because AAA P is a parallelogram then I ,L , I are collinear and II BC . But L lies on

B 2 C 2 then I is the reection of I wrt B 2 C 2 . Thus ( I , I A) is the reection of ( I , IA ) wrt B 2 C 2 ,which follows that two circles intersect at two points M and N which lie on B 2 C 2 .

Since B , C (I , I A) we get V F .V A = V B .V C = V M.V N hence F , A ,M,N are concyclicor (I , IA ) passes through F .

Back to our problem.

M

O'

T

L

R

S

N

H

Z Y

X

A3

A1

C 2

B 2

A2

O

P

A

B C

Let L be the intersection of ( A2 B 2 C 2 ) and the Nine-point circle of triangle ABC,X,Y, Z be themidpoints of BC,CA, AB , respectively, N be the second intersection of LA2 and ( XY Z ). Let M be the midpoint of AH, K be the projection of A on BC, R be the midpoint of AA 1 .

Since OX = AM then AO = XM .Denote T the intersection of the lines through A2 and parallel to AA1 and AK . Applying lemma

7 we obtain ( A2 T ) passes through L.Then ( A2 T, A2 L) (KT,KL ) (NM,NL )(mod ), which follows that MN A2 T AP .

Hence (MX,MN ) (AO,AR )(mod ).But ( NM,NX ) (OR,OA ) / 2(mod ) thus XMN = OAR .

Then OR = XN , we obtain ORNX is a parallelogram. This means RN = OX = 12

AH.Therefore N is the midpoint of HA 1 or NA1 is the midline of triangle HA 3 A1 . We have

HA 3 A2 L. Similarly, H B 3 B2 L.Then ( HA 3 , HB 3 ) (LA 2 , LB 2 ) (C 2 A2 , C 2 B 2 )(mod ).Applying lemma 6 we get ( HA 3 , HB 3 ) (C 3 A3 , C 3 B 3 )(mod ) or H, A 3 , B 3 , C 3 are concyclic.

Our proof is completed then.

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References

[1] Christopher J. Bradley and Geoff C.Smith, On a Construction of Hagge, Forum Geom., 7 (2007) 231-247.

[2] Quang Tuan Bui, Two triads of congruent circles from reections, Forum Geom., 8 (2008)7-12.

[3] Nguyen Van Linh, Two similar geometry problems.http://nguyenvanlinh.wordpress.com

[4] Wolfram Mathworld, Fontene theorems.http://mathworld.wolfram.com/FonteneTheorems.html

[5] Nguyen Van Linh, Fontene theorems and some corollaries.http://nguyenvanlinh.wordpress.com

[6] Nguyen Van Linh, 3 equal circumcircles.http://nguyenvanlinh.wordpress.com

[7] Exploration and Creativity 2011, High School for Gifted Student, Hanoi University of Science.[8] Ross Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry (New

Mathematical Library), The Mathematical Association of America.[9] Shobber, Artofproblemsolving topic 97484 (Circumcircle passes through orthocentre [Fuhrmann

extended]).http://www.artofproblemsolving.com/Forum/viewtopic.php?t=97484

[10] Livetolove212, Artofproblemsolving topic 319520 (Concyclic).http://www.artofproblemsolving.com/Forum/viewtopic.php?t=319520

[11] Livetolove212, Artofproblemsolving topic 321333 (Concyclic again).http://www.artofproblemsolving.com/Forum/viewtopic.php?t=321333

[12] vittasko, Artofproblemsolving topic 345850 (Parallelism relative to the Euler line of a triangle.)http://www.artofproblemsolving.com/Forum/viewtopic.php?t=345850

[13] Nguyen Van Linh, Hagge circle.http://nguyenvanlinh.wordpress.com

[14] Nguyen Van Linh, A concyclic problem, Hyacinthos message 18801.http://tech.groups.yahoo.com/group/Hyacinthos/message/18801

[15] Jean-Louis Ayme, Le P-Cercle De Hagge, Une preuve purement synthetique.http://perso.orange.fr/jl.ayme

Nguyen Van Linh, Foreign Trade University of Vietnam, 91 Chua Lang, Hanoi.E-mail address: [email protected]

hagge circles revisited

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