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SOME NATIONAL OLYMPIAD PROBLEMS IN PLANE GEOMETRY Nguyen Dang Phat Khoa Toán - Trường ĐHSP Hà Nội Problem Let A1 be the center of the square inscribed in acute triangle ABC with two vertices of the square on side BC Thus one of the two remaining vertices of the square is on side AB and the other is on AC Points B1 , C1 are defined in a similar way for inscribed squares with two vertices on sides AC and AB, respective1y Prove that lines AA1 , BB1 , CC1 are concurrent Solution Let α = ∠CAB, β = ∠ABC, and γ = ∠BCA be the angles of triangle ABC Let the line through A and A1 meet side BC at X Similarly, let the line through B and B1 meet side CA at Y , and the line through C and C1 meet side AB at Z (Fig 72) By the converse of Ceva’s Theorem, it suffices to prove that BX CY AZ = XC Y A ZB BX Let the square with center A1 have side s, vertices P and Q on sides AB and AC, respectively, Consider first XC and vertices S and T on BC with S between B and T Since AX passes through the center of the square QP ST , if it cuts side P Q of the square into segments of length u and v, then it cuts side ST into segments of length v and u as shown We then have BX u BX + u BT BS + s scotβ + cotβ + = = = = = = XC v XC + v SC T C + s scotγ + cotγ + Similarly, CY cotγ + AZ cotα + = and = Y A cotα + ZB cotβ + Hence, BX CY AZ =1 XC Y A ZB completing the proof Problem In acute triangle ABC with circumcenter O and altitude AP, ∠C ∠B + 30◦ Prove that ∠A + ∠COP < 90◦ Solution Let α = ∠CAB, β = ∠ABC, γ = ∠BCA and δ = ∠COP (Fig 73) Let K and Q be the reflections of A and P , respectively, across the perpendicular bisector of BC Let R denote the circumradius of ABC, then OA = OB = OC = OK = R Furthermore, we have QP = KA because KQP A is a rectangle Now note that ∠AOK = ∠AOB − ∠KOB = ∠AOB − ∠AOC = 2γ − 2β ≥ 60◦ 26 It follows from this and from OA = OK = R that KA ≥ R and QP ≥ R Therefore, using the Triangle Inequality, we have OP + R = OQ + OC > QC = QP + P C ≥ R + P C It follows that OP > P C, and hence in COP, P CO > δ Now since 1 α = ∠BOC = (180◦ − 2∠P CO) = 90◦ − ∠P CO, 2 it indeed follows that α + β < 90◦ Problem Let ABC be a triangle with centroid G Determine, with proof, the position of the point P in the plane of ABC such that AP.AG + BP.BG + CP.CG is a minimum, and express this minimum value in terms of the side lengths of ABC Solution As usual, let a, b, c denote the sides of the triangle facing the vertices A, B, C, respectively We will show that the desired minimum value of AP.AG + BP.BG + CP.CG is attained when P is the centroid G, and that the minimum value is AG2 + BG2 + CG2 = {(2b2 + 2c2 − a2 ) + (2c2 + 2a2 − b2 ) + (2a2 + 2b2 − c2 )} a2 + b2 + c2 = The latter follows by using Stewart’s Theorem to compute the lengths of the medians AL, BM, CN , along with 2 the relations AG = AL, BG = BM, 3 CG = CN Let S be the circle passing through B, G and C The median AL meets S at G and K Let θ, φ, χ be the angle measures as shown By the Law of Sines, we find (Fig 74) BG sin ϕ AG sin χ = and = CG sin θ BG sin ϕ 27 Also BK = 2R sin θ, CK = 2R sin ϕ, BC = 2R sin χ where R is the radius of S Hence CG BG AG = = BK CK BC (*) Let P be any point in the plane of ABC By Ptolemy’s Theorem, P K.BC BP.CK + BK.CP, with equality if and only if P lies on S In view of (*), we have P K.AG BP.BG + CG.CP Addition of AP.AG to both sides gives (AP + P K).AG Since AK AP.AG + BP.BG + CP.CG AP + AK (by the Triangle Inequa1ity, we have AK.AG AP.AG + BP.BG + CP.CG Equality holds if and only if P lies on the segment AK and P lies on S as well Hence equality holds if and only if P = G Problem Let M be a point in the interior of triangle ABC Let A lie on BC with M A perpendicular to BC Define B on CA and C on AB similarly Define p(M ) = M A M B M C M A.M B.M C Determine, with proof, the location of M such that p(M ) is maximal Let µ(ABC) denote this maximum value For which triangles ABC is the value of µ(ABC) maximal? Solution Let α, β, γ denote the angles A, B, C respectively Also let α1 = ∠M AB α2 = ∠M AC β1 = ∠M BC β2 = ∠M BA γ1 = ∠M CA γ2 = ∠M CB We have M B M C M B M A M A M C = sin α1 sin α2 , = sin γ1 sin γ2 , = sin β1 sin β2 , 2 (M A) ) (M C) ) (M B)2 ) so that p(M )2 = sin α1 sin α2 sin β1 sin β2 sin γ1 sin γ2 Observe that sin α1 sin α2 = [cos(α1 − α2 ) − cos(α1 + α2 )] α (1 − cos α) = sin2 2 (1) Likewise, sin β1 sin β2 sin2 β and sin γ1 sin γ2 γ sin2 (2) Therefore α β γ sin sin 2 Clearly, equality is achieved in (1) and (2) if and only if α1 = α2 , β1 = β2 , γ1 = γ2 ; in other words, p(M ) achieves its maximum value when M is the center of the inscribed circle of triangle ABC and this maximum value is α β γ p(ABC) = sin sin sin 2 It is well known that this quantity is maximal when the triangle is equilateral This can be proven in many ways; for example, using Jensen’s inequality A more elementary proof uses the first equality of (1) to deduce that if π x, y and x + y is fixed, the value of sin x sin y will increase as the difference |x − y| decreases Thus, if π π x + y + z = , the value of sin x sin y sin z can be increased if any of the x, y, z are not equal to (For example, if π π π π π x < and z > and x is closer to than z is, replace x by x = and z by z = z − + x The sum x + y + z 6 6 remains unchanged, but the product sin x sin y sin z increases.) p(M ) sin 28 Problem Let ABC be an acute triangle Let DAC, EAB, and F BC be isosceles triangles exterior to ABC, with DA = DC, EA = EB and F B = F C, such that ∠ADC = 2∠BAC, ∠BEA = 2∠ABC, ∠CF B = 2∠ACB Let D be the intersection of lines DB and EF , let E be the intersection of EC and DF , and let F be the intersection of F A and DE Find, with proof, the value of the sum EC FA DB + + DD EE FF Solution Note that ∠ADC, ∠BEA, ∠CF B < π since ABC is an acute triangle Also, (Fig.75) ∠DAC = π π − ∠ADC = − ∠BAC 2 ∠BAE = π π − ∠BEA = − ∠ABC 2 and Hence ∠DAE = ∠DAC + ∠BAC + ∠BAE = π − ∠ABC < π Likewise, ∠EBF < π and ∠F CD < π Thus the polygon DAEBF C is convex and ∠ADC + ∠BEA + ∠CF B = 2(∠BAC + ∠ABC + ∠ACB) = 2π Let ω1 , ω2 , ω3 be circles with centers at D, E, F , respectively, and radii DA, EB, F C, respectively Using ∠ADC + ∠BEA + ∠CF B = 2π, it is easy to see by the Inscribed Angle theorem that these three circles are concurrent; let O be the common point The O is the reflection of C with respect to DF Likewise, O is also the reflestion of A with respect to DE and the reflection of B with respect to EF Let [XY Z] denote the area of triangle XY Z We have DB [OEF ] DD + D B DB [EBF ] = =1+ =1+ =1+ DD DD DD [DEF ] [DEF ] Likewise, EC [ODF ] FA [ODE] =1+ and =1+ EE [DEF ] FF [DEF ] Thus DB EC FA [OEF ] + [ODF ] + [ODE] + + =3+ = DD EE FF [DEF ] 29 Problem Let ABC be a triangle and P an exterior point in the plane of the triangle Suppose AP, BP, CP meet the sides BC, CA, AB (or extensions thereof) in D, E, F , respectively.Suppose further that the areas of triangles P BD, P CE, P AF are all equal Prove that each of these areas is equal to the area of triangle ABC itself z x y , , and , respectively y z x Since AD, BE, and CF are concurrent (at P ), by Ceva’s thoerem we may choose the ratios in this manner Let us assume that [ABC] = 1, where [U V W ] denotes the signed area of U V W Note that for P to lie outside the triangle at least one of x, y, z must be positive and at least one must be negative Also, Solution Let D, E, and F divide the sides BC, CA, and AB in the signed ratios [ABC] [P BC] [P CA] [P AB] = = = x y z x+y+z Now [P BD] = x [P BD][P BC] z zx [ABC] = = [P BC][ABC] y + z x + y + z (y + z)(x + y + z) Similarly, [P CE] = yz xy and [P AF ] = (z + x)(x + y + z) (x + y)(x + y + z) Since these three areas are equal, we have y(y + z) = z(z + x) = x(x + y) We may assume at this stage that z = This yields y(y + 1) = + x = x(x + y) So x = y + y − from the first equation, and hence we have (y + y − 1)2 + (y + y − 1)y = y + y Simplification gives y + 3y − y − 4y + = 0, which can be factored as (y − 1)(y + 4y + 3y − 1) = If y = then we have x = 1, implying that P coincides with the centroid of that P lies outside the triangle Therefore, y + 4y + 3y − = ABC, contradicting the hypothesis Using this fact, it follows that [P BD] = = [P CE] = = [P AF ] = = x zx = (y + z)(x + y + z) (1 + y)(x + y + 1) y2 + y − y2 + y − y2 + y − = = = −1, (y + 1)(y + 2y) y + 3y + 2y −y − y + xy xy = (z + x)(x + y + z) (x + 1)(x + y + 1) (y + y − 1)y y2 + y − = = −1, (y + y)(y + 2y) y + 3y + 2y yz y = (x + y)(x + y + z) (x + y)(x + y + 1) 1 y = = = −1 (y + 2y − 1)(y + 2y) y + 4y + 3y − −1 These calculations also imply that not both x and y are positive Hence P lies outside ABC Moreover, [P BD] = [P CE] = [P AF ] = −1 = −[ABC] Hence, the desired result The negative sign only indicates that triangles P BD, P CE, and P AF are oriented opposite to ABC Problem Let O be an interior point of acute triangle ABC Let A1 lie on BC with OA1 perpendicular to BC Define B1 on CA and C1 on AB similarly Prove that O is the circumcenter of ABC if and only if the perimeter of A1 B1 C1 is not less than any one of the perimeters of AB1 C1 , BC1 A1 , and CA1 B1 Solution If O is the circumcenter of ABC, then A1 , B1 and C1 are the midpoints of BC, CA, and AB, respectively, and hence PA1 B1 C1 = PAB1 C1 = PBC1 A1 = PCA1 B1 , where PXY Z denotes the perimeter of XY Z Conversely, suppose that PA1 B1 C1 PAB1 C1 , PBC1 A1 , PCA1 B1 Let (Fig 76) ∠CAB = α ∠CA1 B1 = α1 ∠BA1 C1 = α2 , ∠ABC = β ∠AB1 C1 = β1 ∠CB1 A1 = β2 , ∠BCA = γ ∠BC1 A1 = γ1 ∠AC1 B1 = γ2 30 Let A2 be the point of intersection of the lines through B1 and C1 , which are parallel to AB and AC, respectively, as shown in the figure above Assume that γ1 α and β2 α If one of these inequalities is strict, then A1 is an interior point of B1 C1 A2 Hence PA1 B1 C1 < PA2 B1 C1 = PAB1 C1 , which is a contradiction If γ1 = α and β2 = α, then A1 = A2 and therefore B1 O⊥A1 C1 and C1 O⊥A1 B1 Hence O is the orthocenter (intersection of the altitudes) of A1 B1 C1 , and thus OA1 ⊥B1 C1 Hence B1 C1 //BC This implies that A1 , B1 and C1 are the midpoints of BC, CA and AB, respectively : i.e., triangles AB1 C1 , A1 B1 C1 , A1 B1 C, and A1 BC1 are congruent Hence, O is the circumcenter of ABC Analogously, the same conclusion holds if α1 ≥ β and γ2 ≥ β, or β1 ≥ γ and α2 ≥ γ Suppose now that none of these cases are satisfied; i.e., it is not true that γ1 ≥ α and β2 ≥ α or α1 ≥ β and γ2 ≥ β, or β1 ≥ γ and α2 ≥ γ Suppose without loss of generality that γ1 < α Then α2 > γ, since γ1 + α2 = π − β = α + γ Hence β1 < γ, which implies that γ2 > β Hence α1 < β, implying that β2 > α In conclusion, γ1 < α < β2 , α1 < β < γ2 , and β1 < γ < α2 Since AC1 OB1 and A1 CB1 O are cyclic, we have ∠AOB1 = γ2 and OB1 OB1 ∠COB1 = α1 Hence, AO = > = CO In the same way, the inequalities γ1 < β2 and β1 < α2 imply cos γ2 cos α1 that CO > BO and BO > AO, a contradiction Problem Let ABC be a triangle with ∠BAC = 60◦ Let AP bisect ∠BAC and let BQ bisect ∠ABC, with P on BC and Q on AC If AB + BP = AQ + QB, what are the angles of the triangle ? Solution Denote the angles of ABC by α = 60◦ , β, and γ Extend AB to P so that BP = BP , and construct β P ” on AQ so that AP ” = AP Then BP P is an isosceles triangle with base angle Since AQ + QP ” = AB + BP = AB + BP = AQ + QB, it follows that QP ” = QB Since AP P ” is equilateral and AP bisects the angle at A, we have P P = P P ” (Fig 77a) Claim Points B, P, P ” are collinear, so P ” coincides with C Proof Suppose to the contrary that BP P ” is a nondegenerate triangle (Fig 77b) We have that ∠P BQ = β ∠P P B = ∠P P ”Q = Thus the diagram appears as below, or else with P is on the other side of BP ” In either case, the assumption that BP P ” is nondegenerate leads BP = P P ” = P P , thus to the conclusion that BP P is β equilateral, and finally to the absurdity = 60◦ so α + β = 60◦ + 120◦ = 180◦ 31 Thus points B, P, P ” are collinear, and P ” = C as claimed β Since triangle BCQ is isosceles, we have 120◦ − β = γ = so β = 80◦ and γ = 40◦ Thus ABC is a 60 − 80 − 40 degree triangle Problem In the plane we are given two circles intersecting at X and Y Prove that there exist four points with the following property : For every circle touching the two given circles at A and B, and meeting the line XY at C and D, each of the lines AC, AD, BC, BD passes through one of those four points Solution Let Ω be a circle as specified (Fig 78) Since it meets the line XY , it touches the two given circles either both externally or both internally; in the latter case the two given circles can be situated inside Ω or outside Ω The last subcase is illustrated in the diagram; however, the reasoning that follows is valid for any other situation, without any modifications It is enough to consider the lines CA and CB Let CA meet the circle (AXY ) again at P and let CB meet the circle (BXY ) again at Q Since C lies on XY , we have by the power of a point theorem CA.CP = CX.CY = CB.CQ Thus the triangles CAB and CQP are similar so that ∠CAB = ∠CQP Draw the line CR tangent to Ω, with R lying on the same side of line XY as B Then ∠BCR = ∠CAB = ∠CQP , implying CR//P Q Consider the two homotheties, centred respectively at A and B, and mapping Ω onto the two given circles One of these homotheties transforms line CR to the line tangent at P to one of these circles, and the other homothety takes CR to the line tangent at Q to the other circle Both these image lines are parallel to CR; hence they coincide with the line P Q, which is therefore a common tangent to those circles Conclusion the four points of contact of the two given circles with their common tangents have the required property Comment In each one of the two remaining cases (out of the three mentioned in the beginning) one can also use 32 the inversion ccntred at C which sends X to Y , A to P , and B to Q It maps Ω onto line P Q and each one of the two given circles onto itself, and it follows immediately that P Q is a common tangent Comment The problem was received in the following formulation Suppose m is the radical axis of the given circles Γ1 and Γ2 in the plane and ν is the set consisting of all circles Λ touching Γ1 and Γ2 both internally or both externally and intersecting m ν = ∅ : Suppose Λ ∈ ν, touching Γi in Ai and intersecting m in Bi , i ∈ {1, 2} Prove, there exists a set W of four points in the plane, such that for every Λ ∈ ν every line Ai Bj , i, j ∈ {1, 2} is incident with at least one point in W Problem 10 Two circles Γ1 and Γ2 intersect at M and N Let AB be the line tangent to these circles at A and B, respectively, so that M lies closer to AB than N Let CD be the line parallel to AB and passing through M , with C on circle Γ1 and D on Γ2 Lines AC and BD meet at E; lines AN and CD meet at P ; lines BN and CD meet at Q Show that EP = EQ Solution Let K be the point of intersection of M N and AB By the power of a point theorem, AK = KN.KM = BK , i.e., K is the midpoint of AB Since P Q//AB M is the midpoint of P Q So it suffices to prove that EM ⊥P Q (Fig 79) Since CD//AB, the points A and B are the midpoints of the are CM and DM of Γ1 and Γ2 Thus the triangles ACM and BDM are isosceles Hence, by parallel lines ∠BAM = ∠AM C = ∠ACM = ∠EAB, ∠ABM = ∠BM D = ∠BDM = ∠EBA, showing that the points E and M are symmetric across AB, and so EM ⊥AB And since P Q//AB, we obtain EM ⊥P Q, as needed Comment The proposer has also suggested an alternative version of the problem: Under the same assumptions, prove that N E is the bisector of the angle CN D Problem 11 Let O be the circumcentre and H the orthocentre of an acute triangle ABC Show that there exist points D, E, F on sides BC, CA, AB respectively, such that OD + DH = OE + EH = OF + F H and the lines AD, BE, CF are concurrent Solution Let AH extended meet the circumcircle of triangle ABC in L and BC in K (Fig 80) Let OL meet BC in D Join the points H and D It is known that HK = KL So HD = LD as well Hence OD + DH = OD + DL = OL = R, the circumradius of triangle ABC Similarly, points E and F may be chosen on CA and AB such that OE + EH = R = OF + F H We claim that AD, BE, CF concur Draw OB, OC and BL Now ∠OBC = 90◦ − A and ∠CBL = ∠CAL = 90◦ −C Therefore ∠OBL = (90◦ −A)+(90◦ −C) = B As OB = OL, we have ∠OLB = B, too Hence ∠BOL = 180◦ −2B That is, ∠BOD = 180◦ −2B in triangle BOD Analogously, we have ∠COD = 180◦ −2C in triangie COD By the sine rule BD OD = sin ∠BOD sin ∠OBD and CD OD = sin ∠COD sin ∠OCD 33 The right-hand sides of these two equations are equal Division yields BD sin(180◦ − 2B) sin 2B = = ◦ CD sin(180 − 2C) sin 2C Similarly, sin 2C AF sin 2A CE = , = EA sin 2A F B sin 2B Thus BD CE AF = (*) DC EA F B By Ceva’s theorem, AD, BE, CF are concurrent, as desired Note (by the proposer) If P is a variable point on BC, then the position of P for which OP + P H is a minimum is given by P = D Comment The equality (*) can be obtained without any calculation Let AO and LO extended cut the circumcircle again at A and L and let the circumdiameter AA cut BC at D Then ALA L is a rectangle symmetric with respect to the perpendicular bisector of BC (because A L//BC) Consequently the points D and D are situated symmetrically with respect to the midpoint of BC Hence BD = D C and DC = BD Defining analogously points E and F on sides CA and AB we get analogous equalities Thus the product in DC EA F B (*) is equal to the product and the latter has value because the lines AD , BE and CF evidently BD CE AF are concurrent (at O) Problem 12 Let A1 A2 An be a convex polygon, n ≥ Prove that A1 A2 An is cyclic if and only if to each vertex Aj one can assign a pair (bj ; cj ) of real numbers, j = 1, 2, , n, so that Ai Aj = bj ci − bi cj for all i, j with ≤ i < j ≤ n (*) Solution Suppose that there are pairs (bj , cj ) of real numbers satisfying (*) To prove that A1 A2 An is cyclic, it suffices to show that, for each j = 4, , n, the points A1 , A2 , A3 , Aj lie on a circle Note that they are the consecutive vertices of a convex quadrilateral A1 A2 A3 Aj Then, by the converse of Ptolemy’s theorem, a sufficient condition for A1 A2 A3 Aj to be cyclic is A1 A2 A3 Aj + A2 A3 A1 Aj = A1 A3 A2 Aj In view of (*), this translates into (b2 c1 − b1 c2 )(bj c3 − b3 cj ) + (b3 c2 − b2 c3 )(bj c1 − b1 cj ) = (b3 c1 − b1 c3 )(bj c2 − b2 cj ) It is straightforward to verify that the last equality holds for each j = 4, , n Conversely, let A1 A2 An be cyclic Set b1 = −A1 A2 , bj = A2 Aj for j = 2, 3, , n; cj = A1 Aj for j = 1, 2, , n; A1 A2 a definition suggested by Ptolemy’s theorem We now check that the numbers bj , cj satisfy (*) If ≤ i < j ≤ n then A1 , A2 , Ai , Aj are the consecutive vertices of the cyclic quadrilateral A1 A2 Ai Aj Hence A1 A2 Ai Aj = A1 Ai A2 Aj − A2 Ai A1 Aj by Ptolemy’s theorem Dividing both sides by A1 A2 yields Ai Aj = A2 Aj A1 Ai A1 Aj − A2 Ai = bj ci − bi cj , A1 A2 A1 A2 34 as desired We are left with the cases i = and i = For i = and ≤ j ≤ n, the definitions of bj , cj give Ai Aj = A1 Aj = bj 0−(−A1 A2 )cj = bj c1 −b1 cj Similarly, if i = and ≤ j ≤ n then Ai Aj = A2 Aj = bj 1−0.cj = bj c2 − b2 cj The proof is complete Comment The numbers bj , cj are not uniquely determined Here is one more proof of the necessity Let A1 , A2 , , An be arranged counterclockwisely around a circle Γ with centre O and radius R Choose a system of rectangular coordinates with origin at O so that the positive x-axis intersects Γ at a point U between An and A1 Let 2αj be the measure of the oriented angle U OAj (j = 1, 2, , n), that is : the angle through which U has to be rotated in counterclockwise direction around O so that it coincides with Aj For any pair of indices i, j with i < j, we have Ai Aj = 2R sin ∠Ai OAj , where ∠Ai OAj is the respective central (non-oriented) angle Note that ∠Ai OAj is equal to 2(αj − αi ), if < 2(αj − αi ) ≤ 180◦ , and to 360◦ − 2(αj − αi ), if 180◦ < 2(αj − αi ) < 360◦ In both cases, Ai Aj = 2R sin(αj − αi ) = 2R sin αj cos αi − 2R cos αj sin αi So we can define bj = u sin αj , cj = v cos αj for j = 1, 2, , n, where u and v are arbitrary numbers such that uv = 2R Problem 13 The tangents at B and A to the circumcircle of an acute-angled triangle ABC meet the tangent at C at T and U respectively AT meets BC at P , and Q is the midpoint of AP ; BU meets CA at R, and S is the midpoint of BR Prove that ∠ABQ = ∠BAS Determine, in terms of ratios of side-lengths, the triangles for which this angle is a maximum Solution Let a = BC, b = CA, c = AB We have (Fig 81a) AB.BT sin(180◦ − C) BP perpendicular from B to AT area ABT = = = PC perpendicular from C to AT area ACT AC.CT sin(180◦ − B), c2 on using the "tangent-chord" angle theorem This ratio is equal to , by the sine rule and the fact that BT = CT b Hence BP = ac2 + c2 b2 Draw perpendiculars P M, QN from P, Q to AB Then cot∠ABQ = BN 1 BM + BA c + BP cos B , QN = P M = BP sin B, BN = = QN 2 2 35 For fixed A the minimum value of π π B = , C + A = and the minimum is 2 the left member is obtained when B − C has maximum value, that is 1 (sin A + + cos A) therefore it is greater than (1 + 1) = 2 Case ABC is obtuse-angled (Fig 85d)) Suppose for instance that m(B) > 90◦ Let P and Q be the intersections of the perpendicular bisectors of the sides [AB] and [BC] with AC For M ∈ [ADP ], min{M A, M B, M C} = M A and, as above, c a+b+c M A + M B ≤ P A + P B = 2.AP, M A + M C ≤ mc + < It is enough to prove that 4AP < a + b + c, 2 which is implied by 4AP < 2b < a + b + c A similar argument works in the case M ∈ [CEQ] For M ∈ [BEQP D], let BM ∩ AC = N Then min{M A, M B, M C} = M B and M B + M C ≤ N B + N C, M B + M A ≤ N B + N A, N B ≤ max{P B, QB}, therefore it is enough to prove that 2.P B < AB + BC Since P B < BD + DP < BD + P P , the needed relation is obvious Problem 18 A circle is called a separator for a set of five points in a plane if it passes through three of these points, it contains a fourth point inside and the fifth point is outside the circle Prove that every set of five points such that no three are collinear and no four are concyclic has exactly four separators Solution Let {A, B, C, D} be the inverses of four of the points of the set through an inversion having as pole the fifth point Notice that a separator which passes through the pole is transformed into a straight line which passes through two of the points A, B, C, D and separates the other two Notice also that a separator which does not pass through the pole is transformed into a circle which passes through three of the points A, B, C, D and contains the fourth point inside Considering K-the covex hull of the set {A, B, C, D}, two cases may appear Case K=quadrilateral (for instance ABCD) In this case we have as separators the two diagonals of the quadrilateral and one circle from each pair {(ABC), (ABD)}, {(CDA), (CDB)} (the one corresponding to the smaller angle from {ACB, ADB} and {CAD, CBD} respectively) Case K=triangle (for instance ABC) In this case we have as separator with obvious arguments, the straight lines DA, DB, DC and the circle (ABC) Another solution Consider coordinates and the points Ai (xi , yi ), i = 1, 2, 3, 4, Let dijkl be the value of the determinant |x2p + yp2 xp yp 1|p=i,j,k,l The circle (A1 A2 A3 ) is a separator iff d1234 and d1235 have different signs Let us look at the ten pairs (d1234 , d1235 ), (d1243 , d1245 ), etc corresponding to the ten circles which pass through three 42 of the five given points Denoting by an the number dijkl (i < j < k < l), ({i, j, k, l, n} = {1, 2, 3, 4, 5}), the ten pairs are (a5 , a4 ), (−a5 , a3 ), (−a1 , −a3 ), (a5 , a3 ), (a4 , −a2 ), (a3 , a2 ), (−a5 , a1 ), (−a4 , −a1 ), (−a3 , a1 ), (−a2 , −a1 ) We notice that the number of separators is equal to the number of pairs of terms with the same sign from the sequence S = (a1 , −a2 , a3 , −a4 , a5 ) We also remark (using the determinant |x2p + yp2 xp yp 11|p=1,2,3,4,5 ) that a1 − a2 + a3 − a4 + a5 = This shows that S cannot have all its terms of the same sign We claim also that S cannot have four terms of the same sign Indeed, it four terms have the same sign then all the six circles passing through one point of the given five are separators Taking this point as origin, the Ox axis through an other of the given points and passing to polar coordinates we would get, for instance, r1 r2 r3 r1 r3 r4 cos a cos b cos c r1 r2 r4 sin a sin b > 0, sin c cos a sin a cos c sin c > 0, and cos a cos b r2 r3 r4 sin a sin b < 0, cos b sin b cos c sin c < This would lead to r1 sin(c − b) + r2 sin(a − c) + r3 sin(b − a) > 0, −r1 sin b + r2 sin a + r4 sin(b − a) < 0, −r1 sin c + r3 sin a + r4 sin(c − a) > 0, −r2 sin c + r3 sin b + r4 sin(c − b) < Multiplying the last three ralation by r3 , −r2 and r1 respectively and adding the results we would get r4 [r1 sin(c − b) + r2 sin(a − c) + r3 sin(b − a)] < 0, impossible Hence three terms of S have the same sign and the other two have the other sign, therefore there are exactly four separators Problem 19 A set S of points from the space will be called completely symmetric if it has at least three elements and fulfils the condition : for every two distinct points A, B from S the perpendicular besector plane of the segment AB is a plane of simmetry for S Prove that if a completely symmetric set is finite then it cosists of the vertices of either of a regular polygon, or a regular tetrahedron or a regular octahedron Solution Denote by rP Q the reflection in the perpendicular bisector plane of a segment P Q and let G be the barycenter of S From rAB (S) = S we get rAB (G) = G for every A, B ∈ S, therefore all the points of S are at the same distance from G This shows that S is included in a sphere Σ Case S is included in a plane π In this case S is included in the circle Σ ∩ π and its points form a convex polygon A1 A2 An The reflection in the perpendicular bisector of A1 A3 transforms each half-plane bordered by A1 A3 into itshelf, therefore the point rA1 A3 (A2 ) can be only A2 Hence A1 A2 = A2 A3 In the same way A2 A3 = A3 A4 = · · · = An A1 Since S is included in a circle, this proves that A1 A2 An is a regular polygon Case the points of S are not coplanar In this case the points of S are the vertices of a convex polyhedron P (since they are on the sphere Σ) Each face A1 A2 Ak of P is invariant under every reflection rAi Aj , ≤ i < j ≤ k, therefore it is aregular polygon (from case 1) Notice also that every reflection rAB , A, B ∈ S transforms a face of P into a face of P Take now a vertex V of P and denote P ’s edges issuing from V by V V1 , V V2 , , V Vs , such that (V V1 , V V2 ), (V V2 , V V3 ), , (V Vs , V V1 ) are on the same face (Fig 86a)) Notice that the intersection of the falf-planes [V1 V3 , V2 ) and [V1 V3 , V ) with P are triangles V1 V2 V3 and V1 V3 V respectively The reflection rV1 V3 transforms each of these half-planes into itself, therefore it can transform V2 and V only into themselves This shows that rV1 V3 must transform the face containing (V V2 , V V3 ) into the face containing (V V2 , V V1 ) Hence these faces are congruent In the same way every two faces of P having a common side are congruent This shows that all P ’s faces are congruent, because every two faces can be "linked" by a chain of faces so that every two consecutive faces of the chain have a common edge It follows that P is a regular polyhedron (a similar argument shows that from each vertex emerges the same number of deges) It remains to rule out the cube, the regular dodecahedron and the regular icosahedron The cube (Fig 86b) is ruled out because of the reflection rAC” (the rectangle ACC A should be invariant, but it isn’t) The dodecahedron (Fig 86c) is excluded because of the reflection rA3 B1 (same argument for the rectangle A1 A3 B3 B1 ) Finally, the icosahedron (Fig 86d) is eliminated because of the reflection rAB (use rectangle AA1 BB1 ) AX = the point Y on the XB segment (CX) such that CY = 2Y X and, if possible, the point Z on the ray (CA such that CXZ = 180◦ − ABC We denote by Σ the set of all triangles T for which XY Z = 45◦ Problem 20 For a triangle T = ABC we take the point X on the side (AB) such that 43 Prove that all the triangle form Σ are similar and find the measure of their smallest angle Solution A convenient way to describe the position of the points using trigonometry is to employ the contangent We firstly prove the following Lemma (Fig 87a) In a triangle ABC, X ∈ (AB), XA : XB = m : n, CXB = α and ACX = β Then (m + n)cotα = ncotA − mcotB and mcotβ = (m + n)cotC + ncotA Proof of the lemma Let CF = h be the altitude from C Then, using oriented segments, AX = |AF + F X| = h.cotA − h.cotα and BX = |BF + F X| = h.cotB + h.cotα The first part of the conclusion follows now from n.XA = m.XB For the second part take XT //BC, T ∈ (AB) Then XT A = C and CT : T A = n : m The required result follows from the first part applied in AXC We have (Fig 87b) by the lemma and the hypothesis 4cotACX = 9cotC + 5cotA and cotACX − 2cotCXZ = 3cotXY Z = 9cotC + 5cotA + 2.cotB = 3, that is We also have CXZ = 180◦ − B, therefore 5cotA + 8cotB + 9cotC = 12 We will prove that this equation specifies the angles of the triangle ABC Denoting cotA = x, cotB = y, cotC = z we have 5x + 8y + 9z = 12 and the well-known relation xy + yz + zx = 44 Eliminating z we get (x + y)(12 − 5x − 8y) + 9xy = 9, that is 1 (4y + x − 3)2 + 9(x − 1)2 = This shows that x = 1, y = and z = , therefore all the triangles from Σ are similar and their smallest angle is A = 45◦ Problem 21 Let ABC be a triangle, Ω its incircle and Ωa , Ωb , Ωc three circles orthogonal toΩ passing through (B, C), (A, C) and (A, B) respectively The circles Ωa and Ωb meet again in C : in the same way we obtain the points B and A Prove that the circumradius of A B C is half the radius of Ω Solution Denote by I the incenter, by r the inradius, by D, E, F the contacts of the incircle with BC, CA, AB respectively and by P, Q, R the midpoints of the segments [EF ], [F D], [DE] (Fig 88) We will prove that Ωa is the circle (B, C, Q, R) We firstly notice that from the right-angled triangles IBD and IDQ we get IQ.IB = ID2 = r2 and in the same way IR.IC = r2 , therefore the points B, C, R, Q are on a circle Γa The points Q and R belong to the segments [IB] and [IC], so I is exterior to the circle (B, Q, R, C) and I’s power with respect to the circle (B, Q, R, C) is IB.IQ = r2 , which is the condition of Ω beeing orthogonal to the circle (B, Q, R, C) In the same way Ωb is the circle (C, R, P, A) and Ωc is the circle (A, P, Q, B) It follows that A , B , C coincide with P, Q, R and the required conclusion is now obvious 45 Another solution Let Oa be the center of Ωa and M be the midpoint of [BC] Denote, using oriented segments, −−→ −→ M Oa = x (the positive sence on the perpendicular bisector of [BC] beeing ID) a2 The radius of Ωa is x2 + and IOa2 = DM + (ID + M Oa )2 = a p−b − 2 +(r + x)2 The condition of Ω and Ωa beeing orthogonal is Oa I = Oa B + r2 , that is x2 + a2 b−c + r2 = 2 +(r + x)2 ⇔ x = (p − b)(p − c) A B C ⇔ x = 2R sin cos cos 2r 2 It follows that OOa = OM + M Oa = 2R sin A B C r cos cos = R + 2 2 r Therefore Oa , Ob , Oc are on the circle of center O and radius R + Let {N } = Oa Oc ∩ BC The angle Oa OOc has measure π − B (regardless of the position of B), so OOa Oc = M N = xtan B and B A B C p−c a p−c p−b = 2R sin sin cos = , BN = − = 2 2 2 2 BD and, because M N < BM, N ∈ (BM ) The same holds for the common point of AB and Oa Oc , therefore the reflection of B in Oa Oc is on DF This proves that B -the second common point of Ωa and Ωc - is the midpoint of [DF ] The conclusion follows now easily Hence, BN = Problem 22 Two circles Ω1 and Ω2 touch internally the circle Ω in M and N and the center of Ω2 is on Ω1 The common chord of the circle Ω1 and Ω2 intersects Ω in A and B M A and M B intersects Ω1 in C and D Prove that Ω2 is tangent to CD Solution Lemma The circle k1 touches internally circle k at A and touches one of k’s chords M N at B Let C be the mid-point of k’s arc M N which does not contain A Then the points A, B, C are collinear and CA.CB = CM Proof of the lemma (fig 89a) The homothety with center A which transforms k1 into k transforms M N into a tangent at k parallel to M N , i.e into the tangent at C to k, so A, B, C are collinear For the second part notice that N M C ≡ CAM , therefore ACM M CB, whence CA.CB = CM Solution of the problem Let O1 and O2 be the center of Ω1 and Ω2 respectively and t1 , t2 their common tangents (fig 89b)) Let α, β be the arcs cut from Ω by t1 and t2 , positioned like in the lemma Their midpoints have, according to the lemma, equal powers with respect to Ω1 and Ω2 , therefore they are on the radical axis of the two circles Thus A and B are the midpoints of α and β From the lemma we also conclude 46 that C and D are the points in which the tangents t1 and t2 touch Ω1 If H is the homothety with center M transforming Ω1 into Ω2 then H : CD → AB whence AB//CD Therefore CD⊥O1 O2 and O2 is midpoint of one of the arcs CD from Ω1 Let X be the point in which t1 touches Ω2 We get XCO2 ≡ CO1 O2 ≡ DCO2 , so O2 lies on the bisector of the angle XCD, therefore CD is tangent to the circle Ω2 Another solution (Fig 89c) An inversion of pole N transforms the figure as follows : Ω and Ω2 in parallel straight lines, Ω1 in a circle which passes through the reflection of N in Ω2 and is tangent to Ω, AB in the circle (congruent with Ω1 ) passing through N and through Ω1 ∩ Ω2 , AM and BM into the circle (congruent) passing through N, M and A, B respectively, therefore CD becomes circle (N CD) We have to prove that (N CD) is tangent to Ω2 Denote by d the distance between the centers U and V of the circles Ω1 and (N AB), by r their radii and by T the midpoint of the common chord Ω2 (Fig 89d) It follows that d2 d d d T U = , T S = r2 − , T M = r − , T N = r + 2 The circle (CDN ) has its center on M N and N D⊥BD, therefore it is enough to prove that B, D, T are collinear, that is the common point R of BT and Ω2 is the projection of N on BT This results from BR.2BT = BM = BN − N M = BN + BT − N T , the last equality being justified by M N = d2 , BT − N T = BU + U M − 2.U M.U T − N T d d d = r2 + −2 (d − r) − r + = −d2 2 Problem 23 The point M is inside the convex quadrilateral ABCD, such that M A = M C, AM B = M AD + M CD and CM D = M CB + M AB Prove that AB.CM = BC.M D and BM.AD = M A.CD (Fig 90a) Solution Construct the convex quadrilateral P QRS and the interior point T such that AM D and P T S CM D (Fig 90b) It follows that T R M D.M B MB M D.P T TS = = M D, = = MC TS M A.M D MC and ST R ≡ BM C, therefore RT S P T Q ≡ AM B, BM C The assumption on angles leads to QP S + RSP = QP T + T P S + T SP + T SR = P T S + T P S + T SP = 180◦ and RQP + SP Q = RQT + T QP + T P Q + T RS = QT P + T QP + T P Q = 180◦ , 47 QT R so P QRS is a parallelogram (Fig 90b) Hence P Q = RS and QR = P S, that is AB = BC.T S BC.M D AD.QT CD.T S = and = = CD MC MC AM MD The conclusion is now obvious Problem 24 Points A, B, C divide the circumcircle Ω of the triangle ABC into three arcs Let X be a variable point on the arc AB and O1 , O2 be the incenter of the triangles CAX and ABX Prove that the circumcircle of the triangle XO1 O2 intersects Ω in a fixed point Solution (Fig 91a) Let M be the midpoint of the arc BC It is a known result, (otherwise easy to prove) that triangle M O2 B is isosceles with M O2 = M B Therefore M O2 = M B = M C = M O, where O is the incenter of the triangle ABC In the same way, A, O1 , O, C are on a circle with center N , the midpoint of the arc AC Let P be on such that CP//M N and T be the second intersection of OP with Ω Then M O = M C = N P and M P = N C = N O, therefore M P N O is a parallelogram It follow that SM P T = SN P T , whence M P.M T = N P.N T Thus N C.M T = M C.N T , which shows that N T : N O1 = M T : M O2 But then N O1 T ∼ M O2 T (because O1 N T = XN T = XM T = O2 M T ), N T O1 = M T O2 , O1 XO2 = M T N = O1 T O2 therefore the quadrilateral XO1 O2 T is cyclic This proves that the circle (XO1 O2 ) passes through the fixed point T for every position of X Another solution (Fig 91b) It is easy to prove that if A, B are fixed points and C is a mobile point on one of the arcs AB of a fixed circle then the locus of the incenter of the triangle ABC is an arc with endpoints A, B Therefore O1 and O2 are on the circles with centers N and M which pass through A, C and B, C respectivlely This suggest a transformation of the figure using an inversion of pole C The circle (ABC) becomes the straight line AB, the straight line CO1 becomes the bisector of the angle ACX, the tangent at C to the the circle (ABC) becomes the parallel (p) from C to AB, the straight line CN becomes the bisector (q) of the angle (pCA), the circle (CAO1 ) becomes the perpendicular from A on (q), that is the external bisector of the angle BAC Hence O1 becomes an excenter for the triagle CAX In the same way O2 becomes an excenter for the triangle BCX It follows that O1 X⊥O2 X, therefore the circle (XO1 O2 ) becomes the circle of 48 diameter O1 O2 For X = A we have O1 = A, O2 = Ic (excenter for ABC), therefore the fixed point T must be the projection of Ic on AB This is implied by the following facts : i) O2 X passes through O1 = incenter of CAX; AT AO1 ii) AO1 T O1 O1 O2 ⇒ = ; O1 O2 O1 O1 CO1 O1 O2 = ; iii) CAIc CO1 O2 ⇒ AIc CA X X A C 4R sin sin 4R cos cos AO1 CO1 2 2 = AI sin A , iv) AT = AIc = AIc c C AC O1 O1 AC 4R cos where R and A, C, X are the usual notations meant in ACX Problem 25 A convex quadrilateral ABCD has perpendicular diagonals The perpendicular bisectors of AB and CD meet at a unique point P inside ABCD Prove that ABCD is cyclic if and only if triangles ABP and CDP have equal areas Solution Let AC and BD intersect at E (Fig 92a) Suppose by symmetry that P is in ABE Denote by M and N the respective feet of perpendiculars from P to AC and BD Without assuming that P A = P B and P C = P D, we express the areas [ABP ] and [CDP ] as follows1 : Throughout this booklet, [P ] stands for the area of the polygon P 49 2[ABP ] = 2[ABE] − 2[P AE] − 2[P BE] = (AM + P N )(BN + P M ) − (AM + P N )P M − (BN + P M )P N = AM.BN − P M.P N, 2[CDP ] = 2[CDE] + 2[P CE] + 2[P DE] = (CM − P N )(DN − P M ) + (CM − P N )P M + (DN − P M )P N = CM.DN − P M.P N Therefore 2([ABP ] − [CDP ]) = AM.BN − CM.DN (*) We now assume that P A = P B and P C = P D Suppose that ABCD is cyclic Then the uniqueness of P implies that it must be the circumcenter So M and N are the midpoints of AC and BD, respectively Hence AM = CM and BN = DN , so (*) implies [ABP ] = [CDP ] Conversely suppose that [ABP ] = [CDP ] Then, by (*), we have AM.BN = CM.DN If P A = P C, assume by symmetry that P A > P C Then AM > CM and also BN > DN , because P B > P D Thus AM.BN > CM.DN a contradiction Hence P A = P C which implies that P is equidistant from A, B, C and D We conclude then that ABCD is cyclic Alternative Solution Let AC and BD meet at E again Assume by symmetry that P lies in BEC and denote ∠ABE = ϕ and ∠ACD = ψ (Fig 92b) The triangles ABP and CDP are isosceles If M and N are the respective midpoints of their bases AB and CD, then P M ⊥AB and P N ⊥CD Note that M, N and P are not collinear due to the uniqueness of P Consider the median EM to the hypotenuse of the right triangle ABE We have ∠BEM = ϕ, ∠AM E = 2ϕ and ∠EM P = 90◦ − 2ϕ Likewise, ∠CEN = ψ, ∠DN E = 2ψ and ∠EN P = 90◦ − 2ψ Hence ∠M EN = 90◦ + ϕ + ψ, and a direct computation yields N P M = M EN = (90◦ + ϕ + ψ), ◦ ∠N P M = 360 − (∠EM P + ∠M EN + ∠EN P ) = 90◦ + ϕ + ψ = ∠M EN It turns out that, whenever AC⊥BD, the quadrilateral EM P N has a pair of equal opposite angles, the ones at E and P We now prove our claim Since AB = 2EM and CD = 2EN , we have [ABP ] = [CDP ] if and only if PN EM = On account of ∠M EN = ∠N P M , the latter is equivalent to EM N EM.P M = EN.P N , or EN PM P N M This holds if and only if ∠EM N = ∠P N M and ∠EN M = ∠P M N , and these in turn mean that EM P N is a parallelogram But the opposite angles of AM P N at E and P are always equal, as noted above So it is a parallelogram if and only if ∠EM P = ∠EN P ; that is, if 90◦ − 2ϕ = 90◦ − 2ψ We thus obtain a condition equivalent to ϕ = ψ or to ABCD being cyclic Another Solution Choose a coordinate system so that the axes lie on the perpendicular lines AC and BD, and so that the coordinates of A, B, C and D are (0, a), (b, 0), (0, c) and (d, 0), respectively By assumption, the perpendicular bisectors of AB and CD have a unique common point Hence the linear system formed by their equations 2bx − 2ay = b2 − a2 and 2dx − 2cy = −d2 − c2 has a unique solution, and it is given by xo = − c(b2 − a2 ) + a(d2 − c2 ) − d(b2 − a2 ) + b(d2 − c2 ) , yo = 2(ad − bc) 2(ad − bc) Naturally, (xo , yo ) are the coordinates of P Since it is interior to ABCD, orientation Then [ABP ] = [CDP ] if and only if b xo a yo 1 = d xo c yo ABP and CDP have the same 1 This is equivalent to axo + byo − ab = cxo + dyo − cd Inserting the expressions for xo and yo , after the inevitable algebra work we obtain the equivalent condition (ac − bd)[(a − c)2 + (b − d)2 ] = Now, the choice of the coordinate system implies that a and c have different signs, as well as b and d Hence the second factor is nonzero, so [ABP ] = [CDP ] if and only if ac = bd The latter is equivalent to AE.CE = BE.DE, where E is the intersection point of the diagonals However, it is a necessary and sufficient condition for ABCD to be cyclic Problem 26 Let ABCD be a cyclic quadrilateral Let E and F be variable points on the sides AB and CD, respectively, such that AE : EB = CF : F D Let P be the point on the segment EF such that P E : P F = AB : CD Prove that the ratio between the areas of triangles AP D and BP C does not depend on the choice of E and F 50 Solution We first assume that the lines AD and BC are not parallel and meet at S (Fig 93a) Since ABCD is cyclic, ASB and CSD are similar Then, since AE : EB = CF : F D, ASE and CSF are also similar, so that ∠DSE = ∠CSF Moreover, we have SA AB PE SE = = = , SF SC CD P F which implies that ∠ESP = ∠F SP Thus, ∠ASP = ∠BSP , and so P is equidistant from the lines AD and BC Therefore, [AP D] : [BP C] = AD : BC, which is a constant Next, assume that the lines AD and BC are parallel (fig 93b) Then ABCD is an isosceles trapezoid with AB = CD, and we have BE = DF Let M and N be the midpoints of AB and CD, respectively Then M E = N F and clearly, E and F are equidistant from the line M N Thus P , the midpoint of EF , lies on M N It follows that P is equidistant from AD and BC, and hence [AP D] : [BP C] = AD : BC Alternative Solution Let AE : EB = CF : F D = a : b, where a + b = Since P E : P F = AB : CD, we have CD AB d(E, AD) + d(F, AD), AB + CD AB + CD CD AB d(P, BC) = d(E, BC) + d(F, BC), AB + CD AB + CD d(P, AD) = where d(X, Y Z) stands for the distance from the point X to the line Y Z Thus, we obtain CD AB [AED] + [AF D] AB + CD AB + CD a.CD b.AB [ABD] + [ACD], = AB + CD AB + CD CD AB [BP C] = [BEC] + [BF C] AB + CD AB + CD b.CD a.AB = [BAC] + [BDC] AB + CD AB + CD [AP D] = Next, since A, B, C and D are concyclic, we have sin ∠BAD = sin ∠BCD and sin ∠ABC = sin ∠ADC Thus, [AP D] a.CD.[ABD] + b.AB.[ACD] = [BP C] b.CD.[BAC] + a.AB.[BDC] a.CD.AB.AD sin ∠BAD + b.AB.CD.AD sin ∠ADC = b.CD.AB.BC sin ∠ABC + a.AB.CD.BC sin ∠BCD AD a sin ∠BAD + b sin ∠ADC = BC b sin ∠ABC + a sin ∠BCD AD = BC 51 Problem 27 Let I be the incenter of triangle ABC Let K, L and M be the points of tangency of the incircle of ABC with AB, BC and CA respectively The line t passes through B and is parallel to KL The lines M K and M L intersect t respectively at the points R and S Prove that ∠RIS is acute Solution Since the lines KL and RS are parallel, we have In BKR (Fig 94a), A ∠BKR = 90◦ − ∠ , B ∠KBR = 90◦ − ∠ , C ∠BRK = 90◦ − ∠ , Hence, by the law of sine, A cos(∠ ) BK BR = C cos(∠ ) Similarly we have in (1) BLS, C ∠BLS = 90◦ − ∠ , A ∠BSL = 90◦ − ∠ , B ∠LBS = 90◦ − ∠ , so that C C cos(∠ ) cos(∠ ) BL = BK BS = A A cos(∠ ) cos(∠ ) 2 Notice now that BI⊥RS and IK⊥AB Then, on account of (1) and (2), we obtain (2) IR2 + IS − RS = (BI + BR2 ) + (BI + BS ) − (BR + BS)2 = 2(BI − BR.BS) = 2(BI − BK ) = 2IK > So, by the law of cosine, ∠RIS is acute Alternative Solution Let W be the midpoint of KL and Q the midpoint of KM Then Q ∈ AI, W ∈ BI, AI⊥KM and BI⊥KL We first prove that AW ⊥RI and CW ⊥SI (Fig 94b) 52 Since ∠RBI = ∠RQI = 90◦ , the points R, B, I, Q are concyclic, which implies ∠BRI = ∠BQI Also, in the right triangles AIK and BIK, we have IQ.IA = KI = IW IB, (= IK ), IQ IB so that = Hence AIW and BIQ are similar It follows that IW IA ∠BRI = ∠BQI = ∠AW I Since BR⊥IW , this implies that RI⊥AW By a similar argument, we can prove that SI⊥CW Thus, ∠RIS = 180◦ − ∠AW C It remains to prove that ∠AW C is obtuse −−→ −−→ −−→ −→ −−→ −→ −−→ Let T be the midpoint of AC Then 2W T = W C + W A = LC + KA Since LC and KA are not collinear; we have LC + KA CM + AM AC WT < = = 2 This implies that W is inside the circle with diameter AC, and so ∠AW C > 90◦ Therefore, ∠RIS is acute Comment Another proof for the fact AW ⊥RI is as follows Since ∠RBI = ∠RQI = 90◦ RBIQ is cyclic and its circumcircle is orthogonal to the diameter RI Consider the inversion J with respect to the incircle of ABC Since J takes B and Q into W and A respectively, it takes the circumcircle of RBIQ into the line AW Since J leaves the line RI invariant, we have AW ⊥RI Problem 28 Let M and N be points inside triangle ABC such that ∠M AB = ∠N AC and ∠M BA = ∠N BC Prove that AM.AN BM.BN CM.CN + + = AB.AC BA.BC CA.CB Solution Let K be the point on the ray BN such that ∠BCK = ∠BM A (Fig 95a) Note that K is outside ABC, because ∠BM A > ∠ACB Since ∠M BA = ∠CBK, we have ABM KBC; so AB BM AM = = BK BC CK (1) 12.3cm BM AB = , we see that Then, since ∠ABK = ∠M BC, and KB BC ABK 5.8cm M BC Hence AB BK AK = = (2) BM BC CM Now we have ∠CKN = ∠M AB = ∠N AC Consequently, the points A, N, C and K are concyclic By Ptolemy’s theorem, AC.N K = AN.CK + CN.AK, or AC(BK − BN ) = AN.CK + CN.AK 53 (3) From (1) and (2), we find CK = BK = AM.BC AB.CM , AK = and BM BM AB.BC Inserting these expressions in (3), we obtain BM AC AB.BC AN.AM.BC CN.AB.CM − BN = + , BM BM BM or AM.AN BM.BN CM.CN + + = AB.AC BA.BC CA.CB Alternative Solution Let the complex coordinates of A, B, C, M and N be a, b, c, m and n, respectively (Fig 95b) Since the lines AM, BM and CM are concurrent, as well as the lines AN, BN and CN , it follows from Ceva’s theorem that sin ∠BAM sin ∠M AC sin ∠BAN sin ∠N AC sin ∠CBM sin ∠M BA sin ∠CBN sin ∠N BA sin ∠ACM = 1, sin ∠M CB sin ∠ACN = sin ∠N CB (1) (2) By hypotheses, ∠BAM = ∠N AC and ∠M BA = ∠CBN Hence ∠BAN = ∠M AC and ∠N BA = ∠CBM Combined with (1) and (2), these equalities imply sin ∠ACM sin ∠ACN = sin ∠M CB sin ∠N CB Thus cos(∠N CM + 2∠ACM ) − cos ∠N CM = cos(∠N CM + 2∠N CB) − cos ∠N CM, and hence ∠ACM = ∠N CB Since ∠BAM = ∠N AC, ∠M BA = ∠CBN and ∠ACN = ∠M CB, the following complex ratios are all positive real numbers m−c a−c m−a c−a m−b c−b : , : and : b−a n−a a−b n−b b−c n−c Hence each of these equals its absolute value, and so AM.AN BM.BN CM.CN + + AB.AC BA.BC CA.CB (m − a)(n − a) (m − b)(n − b) (m − c)(n − c) + + = (b − a)(c − a) (a − b)(c − b) (b − c)(a − c) = Comment Contestants who are familiar with the notion of isogonal conjugate points may skip over the early part of the proof Problem 29 Let ABC be a triangle H its orthocenter, O its circumcenter, and R its circumradius Let D be the reflection of A across BC, E be that of B across CA, and F that of C across AB Prove that D, E and F are collinear if and only if OH = 2R Solution Let G be the centroid of ABC, and A , B and C be the midpoints of BC, CA and AB, respectively Let A”B”C” be the triangle for which A, B and C are the midpoints of B”C”, C”A” and A”B”, respectively Then G is the centroid and H the circumcenter of A”B”C” Let D , E and F denote the projections of O on the lines B”C”, C”A” and A”B”, respectively (Fig 96) −1 It maps A, B, C, A”, B” and C” into A , B , C , A, B and C, respectively Note that A D ⊥BC, which implies AD : A D = : = GA : GA and ∠DAG = ∠D A G We conclude that h(D) = D Similarly, h(E) = E and h(F ) = F Thus D, E and F are collinear if and only if D , E and F are collinear Now D , E and F are the projections of O on the sides B”C”, C”A” and A”B”, respectively By Simson’s theorem, they are collinear if and only if O lies on the circumcircle of A”B”C” Since the circumradius of A”B”C” is 2R, O lies on its circumcircle if and only if OH = 2R Alternative Solution Let the complex coordinates of A, B, C, H and O be a, b, c, h and o, respecrively Consequently aa = bb = cc = R2 and h = a + b + c Since D is symmetric to A with respect to line BC, the complex coordinates d and a satisfy Consider the homothety h with center G and ratio d−b a−b = , or (b − c)d − (b − c)a + (bc − bc) = c−b c−b 54 (1) Since R2 (b − c) R2 (b2 − c2 ) and bc − bc = , bc bc by inserting these expressions in (1), we obtain b−c=− − bc + ca + ab k − 2bc = , a a R2 (−a + b + c) R2 (h − 2a) d= = , bc bc d= where k = bc + ca + ab Similarly, we have e= k − 2ca R2 (h − 2b) k − 2ab R2 (h − 2c) , e= , f= and f = b ca c ab Since (b − a)(k − 2ab) e−d e−d ab = = (c − a)(k − 2ca) f −d f −d ca R2 (c − a)(a − b) −(ck − 2abc) (h − 2c) = × (bk − 2abc) −(h − 2b) a2 b2 c2 d = e f = d e f − R2 (b − c)(c − a)(a − b)(hk − 4abc) a2 b2 c2 and h = R2 k/abc, it follows that D, E and F are collinear ⇔ =0 ⇔ hk − 4abc = ⇔ hh = 4R2 ⇔ OH = 2R Problem 30 Let ABCDEF be a convex hexagon such that ∠B + ∠D + ∠F = 360◦ and AB CD EF · · = BC DE F A Prove that BC AE F D · · = CA EF DB 55 R2 (a − b)(h − 2c) abc R2 (a − c)(h − 2b) abc Solution Let P be the point such that ∠F EA = ∠DEP and ∠EF A = ∠EDP , where the angles are oriented Then F EA and DEP are similar (Fig 97) Hence FA = EF EF = ED DP , DE EA EP (1) (2) It follows from (1) and the given conditions that ∠ABC = ∠P DC and Therefore, ABC and DE.F A DP AB = = BC CD.EF CD P DC are similar, which gives ∠BCA = ∠DCP and CB CA = (3) CD CP Since ∠F ED = ∠AEP , (2) implies that F ED and AEP are similar Analogously, since ∠BCD = ∠ACP (3) yields the similarity of BCD and ACP Hence FD PA BC CA = and = EF AE DB PA Multiplying these together we have the desired result Alternative Solution Let the complex coordinates of A, B, C, D, E and F be a, b, c, d, e and f , respectively a−b c−d e−f Since ABCDEF is convex, ∠B, ∠D and ∠F are the arguments of the complex numbers , and , c−b e−d a−f ◦ respectively Then the condition ∠B + ∠D + ∠F = 360 implies that the product of these three complex numbers is AB CD EF AB CD EF a positive real number It is equal to the product of their absolute values , and Since = 1, BC DE FA BC DE F A we have a−b c−d e−f · · = c−b e−d a−f On the other hand, since (a − b)(c − d)(e − f ) − (c − b)(e − d)(a − f ) = = (b − c)(a − e)(f − d) − (a − c)(f − e)(b − d), we deduce immediately that b−c a−e f −d · · = a−c f −e b−d Taking absolute values on both sides gives BC AE F D · · = CA EF DB Comment Considering the arguments of the complex numbers on both sides of the equality b−c a−e f −d · · = 1, a−c f −e b−d it shows that ∠BDF = ∠AEF + ∠ACB 56 ... be cyclic is A1 A2 A3 Aj + A2 A3 A1 Aj = A1 A3 A2 Aj In view of (*), this translates into (b2 c1 − b1 c2 )(bj c3 − b3 cj ) + (b3 c2 − b2 c3 )(bj c1 − b1 cj ) = (b3 c1 − b1 c3 )(bj c2 − b2 cj )... ∠P SH2 = 2∠P ST2 We have ∠P ST2 = ∠BST3 , and, by the external angle theorem, (in BST3 ) ∠BST3 = ∠AT3 S − ∠T3 BS = (90◦ − α) − β = γ Next, ∠BST1 = ∠BST3 = γ, by symmetry across BI Note that... perpendicular bisector of A1 A3 transforms each half-plane bordered by A1 A3 into itshelf, therefore the point rA1 A3 (A2 ) can be only A2 Hence A1 A2 = A2 A3 In the same way A2 A3 = A3 A4 = · · · = An