Abstract When we meet the simple problem, it does not usually leave us the impression. But if you think deeply then you can get many interesting results from it. This problem is an example. Although it is a simples problem, we constructed many results about collinearity and concurrency from it. Therefore we should not skip the simple problem. Think deeply and you will see the pleasure when you solve math problem!

To make the problems succinct, in this note we let (I) be the incircle of triangle ABC, (Ia), (Ib), (Ic) be the excircles opposite points A,B,C, respectively, a,b,c be the lenghs of three sides BC, CA, AB and p be the semiperimeter of triangle ABC. Problem 1(Main problem). Let ABC be a triangle. (I) touches AC, AB at E, F, respectively; (Ia) touches BC, CA, AB at M, N, P, o respectively. BI cuts EF at H, BIa cuts PN at K, AI cuts PM at J. Prove that ∠BHC = ∠BKC = ∠AJC = 90 .

Proof.

The idea in this proof is to show that the quadrilateral IEHC,KCNIa and JMCIa be cyclic. We have: ACB ABC HEC = FEA = 90o − BAC = ∠ + ∠ = IBC + ICB = HIC ∠ ∠ ∠ 2 2 ∠ ∠ ∠ o Then IEHC is a cyclic quadrilateral. So ∠BHC = ∠IHC = ∠IEC = 90 BAC On the other hand, NKI = PKB = 180o − BPK − PBK = 180o −(90o − IBC)−(90o − ∠ ) = ∠ a ∠ ∠ ∠ ∠ 2 ACB ABC ∠ + ∠ = NCI then KCNI is a cyclic quadrilateral. 2 2 ∠ a a o o Thus ∠CKIa = ∠CNIa = 90 or ∠BKC = 90 We have ∠BMP = ∠IBC, but the quadrilateral IBIaC is cyclic hence ∠IBC = ∠IIaC Therefore ∠BMP = ∠JIaC so JMCIa is a cyclic quadrilateral. o We get ∠IaJC = ∠IaMC = 90 o Finally, ∠BHC = ∠BKC = ∠AJC = 90 2 Comment 1 It is easy to show that KH be the median agree to the side AB of triangle ABC. o Because ∠HBK = 90 and from problem 1, we obtain BHCK is a rectangle. Then ∠BHK = ∠HBC = ∠ABH so HK//AB, but HK passes through the midpoint of BC therefore HK is the median agree to the side AB of triangle ABC.

1 Problem 2. Let ABC be the triangle. (Ia) touches AB, AC at M,N, (Ib) touches AB, BC at F,E, (Ic) touches AC, BC at D,J, respectively. BIa,CIa cut MN at K, H, AIb,CIb cut EF at Q, P, AIc,BIc cut DJ at R, S, respectively. Prove that six points K, H, P, Q, R, S are concyclic.

Proof.

We will use the concept of radical center to prove this problem. o Applying problem 1 we get ∠AQC = ∠AP C = 90 so the quadrilateral AQPC is cyclic. Arcording to comment 1, R, H lie on the median of triangle ABC agree to the side AC hence RH//AC. Since ∠IbQP = ∠PCA = ∠RHP ,we claim the quadrilateral RQPH is cyclic. Similarly, the quadrilateral KSRQ, SKHP are cyclic. If six points K, H, P, Q, R, S are not concyclic then using the theorem about radical center, we obtain three radical axises RQ, HP, SK of the circumcircles of RQPH, KSRQ, SKHP must be concurrent, but RQ, HP, SK make the triangle IaIbIc so the supposition can not happen. Therefore six points K, H, P, Q, R, S are concyclic 2 Comment 2. Since IaA, IbB,IcC are three altitudes of triangle IaIbIc, we can express problem 2 as the Taylor’s circle for the acute triangle: 0 0 0 Given the triangle ABC and three altitude AA ,BB ,CC . Let A1,A2 be the projection of A’ on AB, AC; B1,B2 be the projection of B’ on BA, BC, C1,C2 be the projection of C’ on CA, CB, respectively. Show that six points A1,A2,B1,B2,C1,C2 are concyclic. Problem 3. Let ABC be a triangle. (Ib) touches BC, AB at P, Q, (Ic) touches BC, AC at M,N, respectively. MN inter- sects PQ at J. Prove that AJ ⊥ BC

Proof.

Let PQ cut IbA, IbC at K, L, MN cut IcA, IcB at F,E, respectively, and let AH be the altitude of triangle ABC. o We have ∠AEB = ∠AF B = ∠AHB = 90 so H lies on the circumcircle of quadrilateral AFEB, similarly H lies on the circumcircle of quadrilateral AKLC. Hence AH is the radical axis of the circumcircles of quadrilaterals AF EB, AKLC. On the other hand, by problem 2 we get EFKL is a cyclic quadrilateral. Applying the theorem about radical center of the circumcircles of quadrilaterals AF EB, AKLC, EF KL we claim EF, KL, AH are concurrent (because EF, KL, AH can not be parallel)

2 Therefore AJ ⊥ BC 2 Comment 3. Using problem 3, we can prove the following problem easily: Let ABC be a triangle, (Ia) touches AB, AC at R,S;(Ib) touches BC, AB at P,Q;(Ic) touches BC, AC at M,N, respectively. MN ∩PQ = A0,MN ∩RS = B0,PQ∩RS = C0 . Prove that AA0,BB0,CC0 are concurrent.

Proof. AA0,BB0,CC0 are 3 altitude of triangle ABC so they are concurrent. Problem 4. Let AH be altitude of given triangle ABC. (I) touches 3 side BC, CA, AB at D,E,F, (Ia) touches BC, CA, AB at M, S, N, respectively. Show that AH, ED, MN are concurrent, AH, DF, MS are concurrent.

Proof.

Let AI, BI cut ED at G, R, respectively, J be the intersection of AI and MN, (Ib) touches BC, AB at P,Q, respectively. PQ ∩ IbC = L. o Using problem 1 we get ∠AJC = 90 then the quadrilateral AJCL is cyclic the circle which has a diameter AC. So ∠JLC = ∠JAC o On the other hand, ∠CLP = ∠IbAC thus ∠JLP = ∠JLC + ∠CLP = ∠JAC + ∠IbAC = 90 Moreover, MN//BIb and BIb ⊥ PQ we obtain MN ⊥ PQ. But J ∈ MN,JL ⊥ PQ therefore M,N,L are collinear. o Using problem 1 again we have ∠AGB = ∠ARB = 90 = ∠AHB so the quadrilateral AHGR is cyclic.

3 On the other hand, ∠AGR = ∠ABR = ∠RBC = ∠BMN = ∠CML By comment 1, R, L lie on the median of triangle ABC agree to BC hence RL//BC We claim ∠CML = ∠MLR So ∠AGR = ∠MLR Therefore GRLJ is a cyclic quadrilateral. Applying the theorem about the radical center of the circumcircles of AHGR, GRLJ, AHJL we get AH, GR, JL are concurrent. Or AH, ED, MN are concurrent. Similarly, AH, DF, MS are concurrent.2

Comment 4. M, N, L are collinear is the interesting result. It was appeared in Germany TST 2004: The A-excircle of a triangle ABC touches the side BC at the point K and the extended side AB at the point L. The B-excircle touches the lines BA and BC at the points M and N, respectively. The lines KL and MN meet at the point X. Show that the line CX bisects the angle CAN. Problem 5. Let ABC be a triangle. (Ib) touches BC, AC at P,R;(Ic) touches BC, AB at E, D, respectively. Prove that PR, ED and the altitude AH of triangle ABC are concurrent.

Proof.

Let Q, F be the tangencies of (Ib) and AB, (Ic) and AC, respectively. EF ∩ PQ = J, ED ∩ PR = K We have ED ⊥ P Q, P R ⊥ EF so ED,PR are two altitudes of triangle JEP Then JK ⊥ BC By problem 3, JA ⊥ BC so K lies on AH. 2 Problem 6. Let ABC be a triangle. (Ib) touches BC, AC at P,R;(Ic) touches BC, AB at E, D, respectively. PR∩ED = K. Prove that AK = r (r is the radius of (I) ) Proof. To prove this problem we will use two lemmas: Lemma 1. (Ib) touches BC, AC at P,R;(Ic) touches BC, AB at E, D, respectively. Let M be the midpoint of BC then MI,ED,PR are concurrent.

Proof.

Let K be the intersection of ED and PR Since EB = CP = p − a then M is the midpoint of EP. We have EK//BI, P K//IC so ∆EKP v ∆BIC BM BC BI Thus = = EM EP EK

4 Therefore M,I,K are collinear. (QED) Lemma 2. Let AH be altitude of given triangle ABC, let M be the midpoint of BC. MI intersects AH at D then AD = r (r is the radius of (I) )

Proof.

Let (O) be the circumcircle of triangle ABC. AI intersects (O) at E. (I) touches AC at K. Because E is the midpoint of BC and AH ⊥ BC, we have AH//EM AD AI AI Then = = ME EI EB AI IK On the other hand,∆AKI ∆BME so = v BE ME AD IK Thus = ME ME Therefore AD = IK = r (QED) Return to problem 6. Applying problem 5 and lemma 1, we get AH, ED, P R, IM concur at K. Applying lemma 2, we obtain AK = r 2 Problem 7. Let AH be altitude of given triangle ABC, (Ia) touches AB, AC at G, H;(Ib), (Ic) touches BC at P, E, respectively. Prove that EG, PH and AH are concurrent. Proof. Lemma (Desargues’s theorem). Given two triangles ABC and A0B0C0. Show that AA0,BB0,CC0 are concurrent if and only if the intersections of BC and B0C0,CA and C0A0, AB and A’B’ are collinear (*)

Proof.

5 Let X = BC ∩ B0C0,Y = AC ∩ A0C0,Z = AB ∩ A0B0. +If AA0,BB0,CC0 concur at O. Using Menelaus’s theorem for triangles OBC, OCA, OAB with the lines (X,B’,C’),(Y,A’,C’), (Z,A’,B’) we have:  XB C0C B0O  . . = 1  0 0  XC C O B B  YC A0A C0O . . = 1 0 0  YA A O C C  ZA B0B A0O  . . = 1 ZB B0O A0A Multiplying three above equalities with side to side we obtain:

XB YC ZA . . = 1 XC YA ZB Using Menelaus’s theorem again for triangle ABC we get X,Y,Z are collinear. +To the contrary, if X,Y,Z are collinear. Let O = AA0 ∩ CC0 Applying the above theorem for the triangles AZA’ and CXC’ we claim O,B,B’ are collinear. Or AA0,BB0,CC0 are concurrent. (QED)

Return to this problem.

6 Assume that (Ib) touches AB at Q, (Ic) touches AC at F. PQ ∩ EF = J.GH ∩ BC = X,PQ ∩ AC = Y,EF ∩ AB = Z,EG ∩ PH = L Applying Menelaus’s theorem for triangle ABC with the lines (X,H,G), (Q, Y, P ), (F,E,Z) we have:   XB HC GA  . . = 1  XC HA GB  YC QA PB . . = 1 YA QB PC   ZA EB FC  . . = 1  ZB EC FA Moreover HC = FA = p−b, GB = QA = p−c, EB = PC = p−a, HA = GA = QB = PB = EC = FC = p then multiplying three above equalities with side to side we get:

XB YC ZA . . = 1 XC YA ZB Applying Menelaus’s theorem again for triangle ABC we obtain X,Y,Z are collinear. Because X = GH ∩ EP,Y = JP ∩ AH, Z = JE ∩ AG then using lemma (*) for the triangles JEP and AGH we claim J,A,L are collinear. By problem 3, AJ is the altitude of triangle ABC therefore L lies on AH. 2 Problem 8. Let ABC be a triangle. (Ib), (Ic) touches BC at Q, G, respectively. (I) touches AB,AC at R, S, respectively. Show that GS, QR and the altitude AA’ of triangle ABC are concurrent.

Proof. First, we will prove this lemma: Lemma. Given the triangle ABC. D, E, F lie on BC, CA, AB, respectively. EF intersects BC at G then AD, BE, CF are concurrent if and only if (BCDG) = −1 (*) Proof.

Applying Menelaus’s theorem for triangle ABC with the line GEF we have:

GB EC FA . . = 1(1) GC EA FB Applying Cva’s theorem for triangle ABC then AD,BE,CF are concurrent if and only if:

DB EC FA . . = −1(2) DC EA FB DB GB From (1) and (2) we obtain AD, BE, CF are concurrent if and only if = − or (BCDG) = −1 (QED) DC GC

7 Return to this problem.

Assume that (Ib) touches AB at P, (Ic) touches AC at H, GH ∩ PQ = J (Ia) touches AB, AC at E, F, respectively.GE ∩ QF = T, GE ∩ IbIc = M, QF ∩ IbIc = N,IbIc intersects P Q, GH at L, K, respectively. By comment 4, GL, QK are two altitudes of triangle JGQ then GL, QK, AJ are concurrent. Let D = KL ∩ BC. Using the lemma (*) we get (DA0GQ) = −1 Then T (DA0GQ) = −1 if and only if (TD,TA0, T G, T Q) = −1 or (T D, T A, T M, T N) = −1 Thus (DAMN) = −1 Applying lemma (*) again we claim TA, MQ, NG are concurrent. The idea is to show that G, S, N are collinear, Q, R, M are collinear. We can assume that b > c without losing the general. rb p − c We have SABC = rb(p − b) = rc(p − c) so = rc p − b DG IcG p − b DG p − b p − b DQ Since IbQ//IcG then = = we can infer that = = then = DQ IbQ p − c GQ p − c − p + b b − c GQ p − b + b − c p − c = b − c b − c CQ p − a CQ p − a p − a On the other side, = therefore = = (1) CG p GQ p + p − a b + c DQ (p − c)(b + c) Thus = CQ (p − a)(b − c) GC b + c − p + a p DG (p − b)(b + c) From (1) we get = = then = GQ b + c b + c GC (b − c)p ND FA QC Applying Menelaus’s theorem for triangle ADC with the line FQN we obtain: . . = 1 or NA FC QD ND p (p − a)(b − c) . . = 1(2) NA p − b (p − c)(b + c) Let GS intersect IbIc at N’. Applying Menelaus’s theorem again for triangle ADC with the line GSN’ we N 0D SA GC N 0D p − a p(b − c) obtain . . = 1 or . . = 1(3) N 0A SC GD N 0A p − c (p − b)(b + c) ND N 0D From (2) and (3) we claim = hence N 0 ≡ N or G, S, N are collinear. NA N 0A Similarly, Q, R, M are collinear. 2

8 Therefore, from problems 3,4,5,7,8 we get this result: Combine result. Let ABC be the triangle. (Ia) touches BC, CA, AB at S, H, G;(Ib) touches BC, CA, AB at P,R,Q;(Ic) touches BC, CA, AB at E,F,D.(I) touches BC, CA, AB at A’,B’,C’.EF ∩PQ = J, ED ∩PR = K,A0C0 ∩SH = X,A0B0 ∩ GS = Y,EG ∩ PH = L . Prove that six points J,K,X,Y,L,W are collinear.

Problem 1 has other interesting and nice applications for us to discover . Because the scope is limit, we will finish at here and see you again to introduce other applications of problem 1. Finally, thanks the teacher Do Thanh Son and Tran Quang Hung very much for your profound comments for this note.

Group of students in class 10A2 Math Tran Minh Chau Tran Thi Mai Dung Nguyen Van Linh Nguyen Vu Da My

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