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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 find 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,CPmeets (O) again at A1, B1, C1. Denote A2, B2, C2 the reflections of A1, B1, C1 with respect toBC,CA,AB, respectively. Then H,A1, B1, C1 are concyclic. The circle (H,A1, B1, C1) 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, B1, C1. Denote A2, B2, C2 the reflections of A1, B1, C1 across the midpoints of BC,CA,AB,respectively. Then H,A1, B1, C1 are concyclic.

Proof.We introduce a lemma:Lemma 1. (B2C2, B2A2) ≡ (PC,PA)(modπ)Proof.

1

B'

C'

A'

M

N

B2

C2

A2

X

ZY

C1

B1

A1

O

P

A

BC

Denote M,N the reflections of A2, C2 across the midpoint of AC then ∆MB1N is the reflectionof ∆A2B2C2 across the midpoint of AC.

Since AM ‖ CA2, AM = CA2 and BA1 ‖ CA2, BA1 = CA2 we get AMA1B is a parallelogram.Similarly, BC1NC is a parallelogram too.

Let A′, B′, C ′ be the midpoints of AA1, BB1, CC1 then A′, C ′ are the midpoints of BM,BN ,

respectively. The homothety H12

B : M 7→ A′, B1 7→ B′, N 7→ C ′ therefore ∆MB1N 7→ ∆A′B′C ′.This means (B1M,B1N) ≡ (B′A′, B′C ′)(modπ).

On the other side, OA′, OB′, OC ′ are perpendicular to AA1, BB1, CC1 hence O,A′, B′, C ′, Pare concyclic, which follows that:

(B2A2, B2C2) ≡ (B1M,B1N) ≡ (B′A′, B′C ′) ≡ (PA′, PC ′) ≡ (PA,PC)(modπ). We are done.Back to problem 2.

C3

B3

A3

B4

C4

A4

H

A2

B2

C2

ZY

X

C1

B1

A1

OP

A

BC

2

Construct three lines through A1, B1, C1 and perpendicular to AA1, BB1, CC1, respectively.They intersect each other and form triangle A3B3C3, intersect (O) again at A4, B4, C4, respectively.

Note that AA4, BB4, CC4 are diameters of (O) therefore A4, C4 are the reflections of H acrossthe midpoints of BC,AB. We get HA2A4A1, HC2C4C1 are parallelograms, which follows thatHA2 ‖ A1A4, HC2 ‖ C1C4.

But PC1B3A1 is a cyclic quadrilateral and applying lemma 1 we have (HC2, HA2) ≡ (B3C1, B3A1)≡ (PC1, PA1) ≡ (PC,PA) ≡ (B2C2, B2A2)(modπ) or H,A2, B2, C2 are concyclic. Our proof iscompleted.

Remark 1: Now come back to problem 1. Let A3, B3, C3 be the reflections of A2, B2, C2 acrossthe midpoints of BC,CA,AB, respectively. Since (BA2C) is the reflections of (ABC) wrt BC thenA3 ∈ (ABC). It is easy to see that BC is the midline of the triangle A1A2A3 hence A1A3 ‖ BC. Weclaim that AA3, BB3, CC3 concur at the isogonal conjugate Q of P wrt ∆ABC. Then accordingto problem 2 we obtain H,A2, B2, C2 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 = ∠EFA. Therefore AG ‖ EF .Back to problem 1.

3

C4

B4

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 reflections of A,B,C wrt Q;A3B3C3 be the circumcevian triangle of Q;A4B4C4 be the median triangle of ∆ABC.

The symmetry with center Q takes A to X,B to Y,C to Z then ∆ABC 7→ XY Z. But A4B4C4

is the median triangle of ∆ABC hence there exist a point K which is the center of the homothety

with ratio1

2. This transformation maps ∆XY Z to ∆A4B4C4.

On the other hand, let L be the second intersection of A1A2 and (O). Applying lemma 2,AL ‖ HA2.

Moreover, LA1 is perpendicular to BC and A3A1 is parallel to BC therefore A3A1 ⊥ LA1,which follows that AA3 ⊥ AL or AA3 ⊥ HA2.

From remark 1, A4 is the midpoint of A2A3 hence XA2KA3 is a parallelogram. This meansKA2 ‖ AA3.

So KA2 ⊥ HA2 or A2 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, B1, C1.Denote A2, B2, C2 the reflections of A1, B1, C1 with respect to BC,CA,AB, respectively. Then∆A2B2C2 ∼ ∆A1B1C1.

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.

4

L

K

G

HI

J

O

A

BC

F

E

See on the figure, 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 ‖ FG and FHGC is a cyclic quadrilateralthus (KG,KL) ≡ (GK,GF ) ≡ (CA,CF )(modπ)(1)

Similarly, (KL,KJ) ≡ (BE,BA)(modπ)(2)From (1) and (2) we are done.Back to problem 3.

C2

B2

A2C

1

B1

A1

H

O

P

A

BC

According to Hagge circle, we have H,A2, B2, C2 are concyclic.Since A2 is the reflections of A1 wrt BC we get HA2 is the Steiner line of A1, which is parallel

to it’s Simson line.Analogously, HB2 is the Steiner line ofB1. Then applying lemma 3, (C2A2, C2B2) ≡ (HA2, HB2)

≡ (C1A1, C1B1)(modπ). Do similarly with other direct angles we refer to the similarity of two tri-angles A1B1C1 and A2B2C2.

Problem 4. Using the same notations as problem 1, let A3, B3, C3 be the second intersectionsof AH,BH,CH and (A2B2C2), respectively. Then A2A3, B2B3, C2C3 concur at P .

Proof.

5

A3

C3

B3

H

B2

B1

A2

A1

C2

C1

P

A

BC

(A2C2, A2A3) ≡ (HC2, HA3) ≡ (HC2, BC)+(BC,HA) ≡ π/2+(HC2, BC)(modπ). Let L,Z bethe projections of C1 on BC,AB, respectively then LZ is the Simson line of C1. We get HC2 ‖ LZ

Therefore (HC2, BC) ≡ (LZ,LC) ≡ (C1Z,C1B)(modπ)Hence (A2C2, A2A3) ≡ π/2 + (C1Z,C1B) ≡ (BA,BC1) ≡ (CA,CC1)(modπ)Since H,A3, B3, C3 are concyclic then (C3B3, C3A3) ≡ (HB3, HA3) ≡ (CA,CB)(modπ) . Sim-

ilarly we get ∆A3B3C3 ∼ ∆ABC. But from problem 3, ∆A2B2C2 ∼ ∆A1B1C1.Thus there exist a similarity transformation which takes A3C2B3A2C3 to AC1BA1CB1 and it

follows that A2A3, B2B3, C2C3 concur at P ′. Moreover,AP

A1P=A3P

′

A2P ′ hence PP ′ ‖ AH. Similarly,

PP ′ ‖ BH,CH. This means P ≡ P ′. Our proof is completed then.

Problem 5. Let I be the circumcenter of triangle A2B2C2. K is the reflection of H across I.AK,BK,CK cut (I) again at A4, B4, C4. Then A3A4, B3B4, C3C4 are concurrent.

Proof.

A4

B4

C4

KA3

B3

C3

I

H

O

P

A

BC

6

Since KA3 ⊥ AH we have (C3C4, C3A3) ≡ (KC4,KA3) ≡ (CK,CB)(modπ).Similarly, (C3C4, C3B3) ≡ (CK,CA)(modπ). Do the same with other angles then applying

Ceva-sine theorem for triangle A3B3C3 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. ThenPQ ‖ O1O2 and PQ = O1O2.

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, B1, C1, respectively. DenoteA2, B2, C2 the reflections of A1, B1, C1 across the midpoints of BC,CA,AB, respectively, O2 thecenter of (A2B2C2). Then OO2HP is a parallelogram.

Proof.

H'

U Z

Y

X

O'2

A'

C'

H

O2

B2

C2

A2

C1

B1

A1

O

P

A

BC

Denote A′, C ′ the reflections of A2, C2 across the midpoint of AC, respectively. Then ∆A′B1C′

is the reflections of ∆A2B2C2 across the midpoint of AC.Since AA′ ‖ A2C,AA

′ = A2C,A2C ‖ BA1, A2C = BA1 we get AA′A1B is a parallelogram. (1)Similarly, BCC ′C1 is a parallelogram. (2)Let X,Y, Z be the midpoints of AA1, BB1, CC1, respectively then X,Y, Z lies on (OP ).But from (1) and (2), X, Z are the midpoints of BA′, BC ′.

A homothety H12

B : A′ 7→ X,B1 7→ Y,C ′ 7→ Z therefore ∆A′B1C′ 7→ ∆XY Z

A symmetry with center is the midpoint of BC:S : B 7→ B′, H 7→ H ′ the orthocenter of triangle AB′C, O2 → O′

2

⇒ O2H → O′2H

′

It is easy to show that B,O,H ′ are collinear. Hence the homothety H12

B : H ′ 7→ O,O′2 7→ U the

circumcenter of triangle XY Z.We conclude that OP ‖= O′

2H ‖= O2H. Therefore HPOO2 is a parallelogram. We are done.Back to our problem.From the lemma above, OO1 ‖= QH,OO2 ‖= PH thus PQ ‖= O1O2

7

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 reflection of P across point E.

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

always the reflection 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, B1, B2, C1, C2, respectively. Let E,F be the Miquelpoints of triangle ABC wrt (A1, B1, C1) and (A2, B2, C2) respectively. Then E-Hagge circle andF-Hagge circle are congruent.

Proof.We will introduce a lemma.Lemma 5. Given triangleABC. A circle (O) intersectsBC,CA,AB at 6 pointsA1, A2, B1, B2, C1,

C2, respectively. Let P,Q be the Miquel points of ∆ABC wrt (A1, B1, C1) and (A2, B2, C2). ThenP and Q are isogonal conjugate and OP = OQ.

Proof.

Y

Z

X

Q

P

C2

C1

B2

B1

A2

A

B C

O

A1

Let X,Y, Z be the second intersections of A1P,B1P,C1P and (O), respectively.We have (ZY,ZX) ≡ (A1Y,A1X) ≡ (A1Y, PY ) + (PY,A1X) ≡ (A2C,A2B1) + (CA2, CB1) ≡

(B1B2, B1A2)(modπ)This means XY = A2B2. Similarly, Y Z = B2C2, XZ = A2C2. Therefore ∆XY Z = ∆A2B2C2.On the other side, (PX,PY ) ≡ (QA2, QB2)(modπ), (PY, PZ) ≡ (QB2, QC2)(modπ) then there

exist a rotation with center O which takes ∆XY Z to ∆A2B2C2 and P to Q. We conclude thatOP = OQ.

Moreover, (AP,AB) ≡ (B1P,B1C1) ≡ (ZY,ZP ) ≡ (C2B2, C2Q) ≡ (AB2, AQ)(modπ).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.

8

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 the

triangle ABC. Then Q,G,K are collinear and QG =1

2GK.

Proof.

X

Q

G

Ma

K

H

C2

C1

B2

B1

A2

A1

P

A

B C

Let X be the reflection of A wrt Q,Ma be the midpoint of BC then from proof 2 of problem 1,we obtain Ma is the midpoint of XK.

Then applying Menelaus theorem for the triangle AXMa with the line (Q,G,K) we have Q,G,Kare collinear. Applying Menelaus theorem again for the triangle QXK with the line (A,G,Ma) then

QG =1

2.

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

C1

B3

C3

A3

C2

A2

B2

B1

A1

G

A

B C

Let A1B1C1, A2B2C2 be the circumcevian triangles of G and L, respectively; A3, B3, C3 be thereflections of A2, B2, C2 wrt BC,CA,AB, respectively.

It is easy to see that A3, B3, C3 lie on AG,BG,CG, respectively.From problem 3, ∆A3B3C3 ∼ ∆A2B2C2. Then (A2B2, A2C2) ≡ (A3C3, A3B3)(modπ)

On the other side, (GC3, GB3) ≡ 1

2(

_

CB1 +_

C1B) ≡ 1

2(

_

B2A +_

AC2)

≡ (A2B2, A2C2) ≡ (A3C3, A3B3)(modπ).

9

This means G,A3, B3, C3 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 oftriangle ABC then HN = 2OI.

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 ′ thereflections of P across the midpoints of BC,CA,AB. Then H,A′, B′, C ′ are concyclic.

4. When P lies at the infinity, problem 1 can be re-written as below:Given triangle ABC with its circumcircle (O), orthocenter H. Denote A1, B1, C1 points on (O)

such that AA1 ‖ BB1 ‖ CC1, A2, B2, C2 the reflections of A1, B1, C1 across the lines BC,CA,AB,respectively. Then H,A2, B2, C2 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 reflections” in the magazine Forum Geometricorum. In that note,he constructed two triads of congruent circles through the vertices, one associated with reflectionsin the altitudes, and the other reflections 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 reflections 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, Ca bethe reflections of B and C across the line AH. Similarly, consider the reflections Cb, Ab of C,A,respectively across the line BH, and Ac, Bc of A,B across the line CH. Then the circumcircles ofthree triangles ACbBc, BAcCa, CAbBa and O-Hagge circle are congruent.

10

C2

C1 B

1

A1

Ab

Cb

Bc

Ca

Ac

Ba

HO

A

B C

Note. In [2], Quang Tuan Bui proved that three circles (ACbBc), (BAcCa), (CAbBa) are con-gruent with the circle (H,HO). By remark 2, HO is the radius of O-Hagge circle. We are done.

Problem 13. Let I be the incenter of triangle ABC. Consider the reflections of the verticesin the angles bisectors: Ba, Ca of B,C in AI,Ab, Cb of A,C in BI,Ac, Bc of A,B in CI. Then thecircumcircles of three triangles ACbBc, BAcCa, CAbBa and Fuhrmann circle are congruent.

H

S

B2

C2 A

2

C1

B1

A1

T

Cb

Ca

Ab

Ac

Bc

Ba

I

O

A

BC

Note. In his article, Quang Tuan Bui also proved that three circles (ACbBc), (BAcCa), (CAbBa)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 Ba, Ca be the reflectionsof B and C across the line AE. Similarly, consider the reflections Cb, Ab of C,A, respectively acrossthe line BE, and Ac, Bc of A,B across the line CE. Then the circumcircles of three trianglesACbBc, BAcCa, CAbBa and E-Hagge circle are congruent.

Proof.

11

Cb

B2

A2

A1

B1

Q

Oc

C2

JC1

Bc

E

H

O

A

BC

Denote A1B1C1 the circumcevian triangle of E. A2, B2, C2 the reflections of A1, B1, C1 acrossBC,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 AHB.

Since (Oc) and (O) are symmetric wrt AB,C2 is the reflection of C1 wrt AB we obtain C2 ∈ (Oc).It is not hard to see that Oc lies on CE. However, B and Bc are symmetric wrt CE we concludethat Bc ∈ (Oc).

We have (BA,BBc) ≡ (CQ,CC1)(modπ), which follows from the fact that CQ ⊥ AB,BBc ⊥CC1. This means the direct angle (BA,BBc) is half the measure of the arc QC1 or HC2. ThereforeABc = HC2 or AHC2Bc is a isosceles trapezoid.

Analogously, AHB2Cb is also a isosceles trapezoid. We claim (ABcCb) is the image of (HC2A2)across the symmetry whose the axis is the perpendicular bisector of AH. This means two circles(ABcCb) and E-Hagge circle are congruent. Similarly we are done.

Remark 3. Other synthetic proofs of the congruent of three circles (ACbBc), (BAcCa), (CAbBa)can be found in [6] or [7]. We have some properties:

1. The radius of three circles (ACbBc), (BAcCa), (CAbBa) is equal to the distance of O andKosnita point K (the isogonal conjugate of E).

2. The orthocenter of triangle IaIbIc is the reflection of H wrt the center of N-Hagge circle(Ia, Ib, Ic are the circumcenters of three triangles ACbBc, BAcCa, CAbBa, 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). Denote

A1, B1, C1 the reflections of A,B,C wrt O, respectively, P an arbitrary point in the plane. LetA2B2C2 be the pedal triangle of P wrt ∆ABC,A3, B3, C3 be the reflections of A1, B1, C1 wrtA2, B2, C2, respectively. Then 4 points H,A3, B3, C3 are concyclic.

Proof.

12

A5

B5

C5

H

A3

B3

C3

C4

B4

A4

C2

B2

A2

C1

B1

A1

O

P

A

B C

Let A4, B4, C4 be the midpoints of BC,CA,AB, respectively then A4 is the midpoint of A3H.Similar with B4, C4.

We get HA3 ‖= 2A2A4, HB3 ‖= 2B2B4, HC3 ‖= 2C2C4.(1)Construct three rectangles OA4A2A5, OB4B2B5, OC4C2C5 hence A5, B5, C5 lie on the circle

with diameter OP.(2)But OA5 ‖= A4A2, OB5 ‖= B4B2, OC5 ‖= C4C2(3)From (1), (2) and (3) we claim H,A3, B3, C3 are concyclic. Moreover, we can show that∆A3B3C3 ∼ ∆ABC.

Problem 16. Given triangle ABC with its circumcircle (O) and its orthocenter H. Let HaHbHc

be the orthic triangle of triangle ABC. P is an arbitrary point on Euler line. AP,BP,CP intersect(O) again at A1, B1, C1. Let A2, B2, C2 be the reflections of A1, B1, C1 wrt Ha, Hb, Hc, respectively.Then H,A2, B2, C2 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, B1, C1, AY,BY,CY inter-sect (O) again at A2, B2, C2. Let A3, B3, C3 be the reflections of A2, B2, C2 wrt the midpoints ofA1X,B1Y,C1Z. Then A3, B3, C3, X lie on a circle (O′) and O′ lies on XY .

Proof.

13

U

T

S

Z

W

V

B3

C3

A3

P

Q

C1

C2

B2

B1

A2A

1

O

X

A

BC

Y

Let S be the intersection of A1C2 and A2C1. Applying Pascal’s theorem for 6 points C1, C2, A1,A2, A,C we get S ∈ XY .

Denote T,U the second intersections of the perpendicular lines through A1, C1 to A1A2, C1C2

with (O), respectively.Applying Pascal’s theorem again for 6 points A1, A2, C1, C2, T, U We conclude that P ∈ XY .

Similarly we obtain the perpendicular line through B1 to B1B2 also passes through P , three lineswhich pass through A2, B2, C2 and perpendicular to A1A2, B1B2, C1C2, respectively concur at Q ∈XY .

Construct three rectangles PA1A2V, PB1B2W,PZC2C1 then XA3 ‖ A1A2 ‖ PV and XA3 =A1A2 = PV . Let L be the midpoint of XP then LA3 = LV = 1

2A3V.Similarly we get (A3B3C3) is the reflection of (VWZ) wrt L. But P, V,W,Z lie on a circle with

diameter PQ so X lies on (A3B3C3). Moreover, O′ is the reflection 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, B1, C1, respectively. LetA2B2C2 be the pedal triangle of P wrt ∆ABC,A3, B3, C3 be the reflections of A1, B1, C1 wrtA2, B2, C2, respectively. Then A3, B3, C3, H are concyclic.

Proof.We introduce two lemmas.Lemma 6. Three triangles A1B1C1, A2B2C2, A3B3C3 are similar.Proof.

14

j

B4

C4

B3

A3

C3

C1

B1

A1

C2

B2

A2

O

P

A

BC

The result ∆A1B1C1 ∼ ∆A2B2C2 is trivival. So we only show that ∆A3B3C3 ∼ ∆A1B1C1.Let B4, C4 be the reflections of B3, C3 wrt A2, respectively. We get ∆A1B4C4 ∼ ∆A3B3C3.(1)Since A2C2 is the midline of triangle C1C3C4 then C1C4 ‖ A2C2. Analogously, B1B4 ‖ A2B2.

Denote R the intersection of B1B4 and C1C4. We conclude that (RB1, RC1) ≡ (A2B2, A2C2) ≡(A1B1, A1C1)(modπ), which follows that R ∈ (O).

Hence (C1A1, C1C4) ≡ (B1A1, B1B4)(modπ).

On the other side,A1C1

A1B1=A2C2

A2B2=C1C4

B1B4. Therefore ∆A1C1C4 ∼ ∆A1B1B4.

This means (A1B1, A1B4) ≡ (A1C1, A1C4)(modπ), then (A1B4, A1C4) ≡ (A1B1, A1C1)(modπ).Similarly we get ∆A1B1C1 ∼ ∆A1B4C4.(2)

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

Lemma 7. Given triangle ABC. Let A1 be the projection of A on BC,A2, B2, C2 be themidpoints 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 (A2B2C2). 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'

C2

B2

A2

A

B C

P

Let V be the intersection of B′C ′ and B2C2. According to the first Fontene’s 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 AA”A′P is a parallelogram then I, L, I ′ are collinear and II ′ ⊥ BC. But L lies on

B2C2 then I ′ is the reflection of I wrt B2C2. Thus (I ′, I ′A) is the reflection of (I, IA′′) wrt B2C2,which follows that two circles intersect at two points M and N which lie on B2C2.

Since B′, C ′ ∈ (I ′, I ′A) we get V F ′.V A′ = V B′.V C ′ = VM.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

C2

B2

A2

O

P

A

B C

Let L be the intersection of (A2B2C2) 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 Mbe the midpoint of AH,K be the projection of A on BC,R be the midpoint of AA1.

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 (A2T ) passes through L.Then (A2T,A2L) ≡ (KT,KL) ≡ (NM,NL)(modπ), which follows that MN ‖ A2T ‖ AP .

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

Then OR ‖= XN , we obtain ORNX is a parallelogram. This means RN ‖= OX ‖= 1

2AH.

Therefore N is the midpoint of HA1 or NA1 is the midline of triangle HA3A1. We haveHA3 ‖ A2L. Similarly, HB3 ‖ B2L.

Then (HA3, HB3) ≡ (LA2, LB2) ≡ (C2A2, C2B2)(modπ).Applying lemma 6 we get (HA3, HB3) ≡ (C3A3, C3B3)(modπ) or H,A3, B3, C3 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 reflections, 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.


[6] Nguyen Van Linh, 3 equal circumcircles.


[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 (NewMathematical Library), The Mathematical Association of America.

[9] Shobber, Artofproblemsolving topic 97484 (Circumcircle passes through orthocentre [Fuhrmannextended]).

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

[10] Livetolove212, Artofproblemsolving topic 319520 (Concyclic).


[11] Livetolove212, Artofproblemsolving topic 321333 (Concyclic again).


[12] vittasko, Artofproblemsolving topic 345850 (Parallelism relative to the Euler line of a triangle.)


[13] Nguyen Van Linh, Hagge circle.


[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]

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