Bài 3 (tr 26 56) nguyễn đăng phất

Let A2be the point of intersection of the lines through B1and C1, which are parallel to AB and AC, respectively,as shown in the figure above.. Both these image lines are parallel to CR;h

SOME NATIONAL OLYMPIAD PROBLEMS

Solution Let α =∠CAB, β = ∠ABC, and γ = ∠BCA be the angles of triangle ABC Let the line through Aand A1 meet side BC at X Similarly, let the line through B and B1 meet side CA at Y , and the line through Cand C1meet side AB at Z (Fig 72) By the converse of Ceva’s Theorem, it suffices to prove that

BXXC

cotβ + 1cotγ + 1.Similarly,

CY

Y A=

cotγ + 1cotα + 1 and

AZ

cotα + 1cotβ + 1.Hence,

BXXC

CY

Y A

AZ

ZB= 1completing the proof

Problem 2 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 Aand P , respectively, across the perpendicular bisector of BC Let R denote the circumradius of 4ABC, 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◦

It follows from this and from OA = OK = R that KA ≥ R and QP ≥ R Therefore, using the TriangleInequality, we have

OP + R = OQ + OC > QC = QP + P C ≥ R + P C

It follows that OP > P C, and hence in 4COP, 4P CO > δ Now since

α = 1

2∠BOC = 12(180◦− 2∠P CO) = 90◦− ∠P CO,

it indeed follows that α + β < 90◦

Problem 3 Let ABC be a triangle with centroid G Determine, with proof, the position of the point P in theplane of ABC such that AP.AG + BP.BG + CP.CG is a minimum, and express this minimum value in terms ofthe 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 showthat 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

AG

sin χsin ϕ.

Also BK = 2R sin θ, CK = 2R sin ϕ, BC = 2R sin χ where R is the radius of S Hence

P K.AG 6 BP.BG + CG.CP

Addition of AP.AG to both sides gives

(AP + P K).AG 6 AP.AG + BP.BG + CP.CG

Since AK 6 AP + AK (by the Triangle Inequa1ity, we have

Solution Let α, β, γ denote the angles A, B, C respectively Also let

sin β1sin β26 sin2β

2 and sin γ1sin γ26 sin2γ

γ1= γ2; in other words, p(M ) achieves its maximum value when M is the center of the inscribed circle of triangleABC and this maximum value is

Problem 5 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 D0 be the intersection of lines DB and EF , let E0 be the intersection of EC and DF , and let F0 be theintersection of F A and DE Find, with proof, the value of the sum

DB

DD0+ EC

EE0+ F A

F F0.Solution Note that∠ADC, ∠BEA, ∠CF B < π since ABC is an acute triangle

Thus the polygon DAEBF C is convex and

∠ADC + ∠BEA + ∠CF B = 2(∠BAC + ∠ABC + ∠ACB) = 2π

Let ω1, ω2, ω3be 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 Awith respect to DE and the reflection of B with respect to EF Let [XY Z] denote the area of triangle XY Z Wehave

Problem 6 Let ABC be a triangle and P an exterior point in the plane of the triangle Suppose AP, BP, CP meetthe 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

Solution 1 Let D, E, and F divide the sides BC, CA, and AB in the signed ratios z

us assume that [ABC] = 1, where [U V W ] denotes the signed area of 4U V W Note that for P to lie outside thetriangle at least one of x, y, z must be positive and at least one must be negative Also,

(z + x)(x + y + z) and [P AF ] =

yz(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 = 1.This yields

y(y + 1) = 1 + x = x(x + y)

So x = y2+ y − 1 from the first equation, and hence we have

(y2+ y − 1)2+ (y2+ y − 1)y = y2+ y Simplification gives y4+ 3y3− y2− 4y + 1 = 0, which can be factored as

2+ y − 1(y + 1)(y2+ 2y)=

2+ y − 1)y(y2+ y)(y2+ 2y)=

P BD, P CE, and P AF are oriented opposite to 4ABC

Problem 7 Let O be an interior point of acute triangle ABC Let A1lie 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 A1B1C1 is not less than any one of the perimeters of AB1C1, BC1A1, and CA1B1

Solution If O is the circumcenter of 4ABC, then A1, B1 and C1 are the midpoints of BC, CA, and AB, tively, and hence

respec-PA1B1C1 = PAB1C1 = PBC1A1 = PCA1B1, where PXY Z denotes the perimeter of 4XY Z

Conversely, suppose that PA1B1C1 > PAB 1 C 1, PBC1A1, PCA1B1 Let (Fig 76)

Let A2be the point of intersection of the lines through B1and 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 aninterior point of 4B1C1A2 Hence PA1B1C1 < PA2B1C1 = PAB1C1, which is a contradiction If γ1= α and β2= α,then A1= A2and therefore B1O⊥A1C1and C1O⊥A1B1 Hence O is the orthocenter (intersection of the altitudes)

of 4A1B1C1, and thus OA1⊥B1C1 Hence B1C1//BC This implies that A1, B1 and C1 are the midpoints of

BC, CA and AB, respectively : i.e., triangles AB1C1,

A1B1C1, A1B1C, and A1BC1 are congruent Hence, O is the circumcenter of 4ABC Analogously, the sameconclusion 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≥ αor

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 Inconclusion,

γ1< α < β2, α1< β < γ2, and β1< γ < α2.Since AC1OB1 and A1CB1O are cyclic, we have ∠AOB1= γ2and

∠COB1= α1 Hence, AO = OB1

cos γ2

> OB1cos α1

= CO In the same way, the inequalities γ1< β2 and β1 < α2 implythat CO > BO and BO > AO, a contradiction

Problem 8 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 P0 so that BP0 = BP , and construct

P ” on AQ so that AP ” = AP0 Then BP0P is an isosceles triangle with base angle β

2 Since

AQ + QP ” = AB + BP0 = AB + BP = AQ + QB,

it follows that QP ” = QB Since AP0P ” is equilateral and AP bisects the angle at A, we have P P0 = 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 =

∠P P0B = ∠P P ”Q = β2 Thus the diagram appears as below, or else with P is on the other side of BP ” In eithercase, the assumption that BP P ” is nondegenerate leads BP = P P ” = P P0, thus to the conclusion that BP P0 isequilateral, and finally to the absurdity β

2= 60

◦ so

α + β = 60◦+ 120◦= 180◦

Thus points B, P, P ” are collinear, and P ” = C as claimed.

Since triangle BCQ is isosceles, we have 120◦− β = γ = β

P and let CB meet the circle (BXY ) again at Q Since C lies on XY , we have by the power of a point theoremCA.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 twogiven circles One of these homotheties transforms line CR to the line tangent at P to one of these circles, and theother 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 requiredproperty

Comment In each one of the two remaining cases (out of the three mentioned in the beginning) one can also use

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 thetwo 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 Γ1and Γ2in 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 6= ∅ : 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 AiBj, 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 CDmeet 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, AK2= KN.KM =

BK2, 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 Γ1and Γ2 Thus the trianglesACM 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

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

BDsin ∠BOD=

ODsin ∠OBDand

CDsin ∠COD=

ODsin ∠OCD.

The right-hand sides of these two equations are equal Division yields

BD

sin(180◦− 2B)sin(180◦− 2C) =

sin 2Bsin 2C.Similarly,

CE

sin 2Csin 2A,

AF

F B=

sin 2Asin 2B.Thus

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 A0 and L0 and let the circumdiameter AA0 cut BC at

D0 Then ALA0L0 is a rectangle symmetric with respect to the perpendicular bisector of BC (because A0L//BC).Consequently the points D and D0are situated symmetrically with respect to the midpoint of BC Hence BD = D0Cand DC = BD0

Defining analogously points E0 and F0 on sides CA and AB we get analogous equalities Thus the product in(*) is equal to the product D

A1A2.A3Aj+ A2A3.A1Aj= A1A3.A2Aj

In view of (*), this translates into

(b2c1− b1c2)(bjc3− b3cj) + (b3c2− b2c3)(bjc1− b1cj) = (b3c1− b1c3)(bjc2− b2cj)

It is straightforward to verify that the last equality holds for each j = 4, , n

Conversely, let A1A2 An be cyclic Set

b1= −A1A2, bj = A2Aj for j = 2, 3, , n; cj = A1Aj

A1A2

for j = 1, 2, , n;

a definition suggested by Ptolemy’s theorem We now check that the numbers bj, cj satisfy (*)

If 3 ≤ i < j ≤ n then A1, A2, Ai, Aj are the consecutive vertices of the cyclic quadrilateral A1A2AiAj Hence

A1A2.AiAj = A1Ai.A2Aj− A2Ai.A1Aj by Ptolemy’s theorem Dividing both sides by A1A2 yields

as desired We are left with the cases i = 1 and i = 2 For i = 1 and 2 ≤ j ≤ n, the definitions of bj, cj give

AiAj = A1Aj= bj.0−(−A1A2)cj = bjc1−b1cj Similarly, if i = 2 and 3 ≤ j ≤ n then AiAj= A2Aj = bj.1−0.cj =

bjc2− b2cj The proof is complete

Comment The numbers bj, cjare 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 ordinates with origin at O so that the positive x-axis intersects Γ at a point U between An and A1 Let 2αj bethe measure of the oriented angle U OAj (j = 1, 2, , n), that is : the angle through which U has to be rotated

co-in counterclockwise direction around O so that it coco-incides with Aj For any pair of indices i, j with i < j, wehave AiAj = 2R sin1

2∠AiOAj, where ∠AiOAj is the respective central (non-oriented) angle Note that ∠AiOAj isequal to 2(αj− αi), if 0 < 2(αj− αi) ≤ 180◦, and to 360◦− 2(αj− αi), if 180◦< 2(αj− αi) < 360◦ In both cases,

AiAj= 2R sin(αj− αi) = 2R sin αjcos αi− 2R cos αjsin 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 themidpoint of BR

Prove that ∠ABQ = ∠BAS Determine, in terms of ratios of side-lengths, the triangles for which this angle is

area ABTarea ACT =

1

2.AB.BT sin(180

◦− C)1

on using the sine and cosine rules Similarly (note the symmetry in a and b), cot∠BAS has the same value, so

∠ABQ = ∠BAS Using the cosine rule again for c2 gives

cot∠ABQ = 2(a

2+ b2)

ab sin C − 3cotC ≥ 4

sin C− 3cotC = y,with equality when a = b (note that sin C > 0) We have

7 This shows that cot∠ABQ ≥√7, so ∠ABQ ≤ tan−1(√1

7), with equality if C = θ = cos

−1(3

4).

Therefore the maximum angle is tan−1(√1

7), occurring for isosceles triangles with a = b, c

4 , cos C =

3

4.Another Solution Assume the standard angle notation for 4ABC and denote by R its circumradius, Let

∠BAT = ϕ Since ∠T BC = ∠BAC = α by the tangent-chord angle theorem, the isosceles 4BCT gives

2 cos α= Rtanα (Fig 81b) In 4ABT , we have ∠ABT = α + β,

∠BAT = ϕ, so by the law of sines,

AB

sin ∠AT Bsin ∠BAT =

sin(α + β + ϕ)

In view of the equalities AB = 2R sin γ, BT = Rtanα and α + β + γ = 180◦, this yields

cotϕ = cotγ + 2cotα

Next, we set ∠ABQ = ψ and apply the law of sines to 4BAQ and 4BP Q This gives BQAQ = sin ϕ

sin ψ,

BQ

P Q =sin ∠AP B

sin(β − ψ) Since BQ is a median, the ratios in the left-hand sides are equal, so

sin ϕsin ψ =

sin(β + ϕ)sin(β − ψ) This gives

sin βcotϕ + cos β = sin(β + ϕ)

sin(β − ψ)sin ψ = sin βcotψ − cos β,leading to cotψ = 2cotβ + cotϕ Therefore

By symmetry, cot∠BAS has the same value The cotangent function is strictly decreasing in the open interval(0◦, 180◦) and so ∠ABQ = ∠BAS

Maximizing ψ is equivalent to minimizing cotψ and this can be done in many ways We can use formula (2).Since

2(cotα + cotβ) = 2 sin γ

sin α sin β =

4 sin γcos(α − β) + cos γ≥ 4 sin γ

Now, applying the AM-GM inequality gives

7

For equality to occur, it is necessary that α = β = 90◦−γ

2 and7

4 , sin α = sin β =

√14

4 So, by the law of sines, the angle ψ is amaximum when the ratios of side lengths of 4ABC are a : b : c =√

Solution Let Z be the second point of intersection of the circumcircles of triangles ADX and BCX, named Γ1

and Γ2, with centres O1 and O2, respectively Let W be the second point of intersection of the circumcircles oftriangles ABZ and CDZ, named Γ3and Γ4, with centres O3 and O4, respectively We are going to show that thepoint W coincides with Y The lines AB and CD extended meet at a point J so that the quadrilateral ABCD lies

in the convex angular region AJ D We claim that the point Z also lies in that region Assume that X lies closer

to line AB than Z does (as in the diagram) (Fig 82a) Then Z lies on the same side of line AB as the segment

CD Let XC0and XD0 be diameters of Γ2and Γ1, respectively The points C0, Z, D0 are collinear Since the anglesDAX and CBX are acute, the points C and D lie on the opposite side of line C0D0than X Thus Z, being a point

of the segment C0D0, lies in the region AJ D Now, depending on which one of the points X, Z lies closer to line

AB, we have one of two alternatives

or

In each case it follows that the line ZX bisects the angle AZB By symmetry, it also bisects the angle CZD.Hence it meets the circles Γ3 and Γ4 at the midpoints of their arcs AB and CD situated outside the region

AJ D Denote these midpoints by P and Q, respectively To prove that W coincides with Y , we have to show that

W A = W B and W C = W D This is equivalent to showing that W P and W Q are diameters of circles Γ3 and Γ4;i.e., to proving that W Z⊥XZ As XZ⊥O1O2 and W Z⊥O3O4, it will be enough to prove that O1O2⊥O3O4

To this end, note that AZ⊥O1O3 Combined with XZ⊥O1O2, this implies that the angles O2O1O3 and AZXare either equal or add up to 180◦ By symmetry, the same conclusion holds for the angles O1O2O3 and BZX.And since

∠AZX = ∠BZX, we get that the angles O2O1O3and O1O2O3 are either equal or add up to 180 However, thelatter case cannot occur because A and B are distinct points, and hence O1O2O3 is a non-degenerate triangle.Thus O1O3= O2O3.

Likewise, O1O4= O2O4 These two equalities imply that O1O2⊥O3O4 As observed, this is enough to concludethat W = Y

Consider the triangle P QW Its vertices P and Q lie outside the region AJ D, whereas Z, the foot of its altitude

W Z, lies within that region Its sides P W and QW are respectively perpendicular to lines J A and J D Thus Walso belongs to the region AJ D In other words, the points Z and W lie on the same side of line AB, and also onthe same side of line CD

We again refer to the alternatives (1), (2) By the properties of angles inscribed in the circles Γ3, Γ1and Γ2,

∠AW B = ∠AZB = ∠AZX + ∠BZX = ∠ADX + ∠BCX = 2∠ADX,or

And since W = Y , we get the result

Comment The proposer has also supplied one more proof of the fact that the point W coincides with Y , usinginversion

Another Solution

Toss the picture on the complex plane Let the points X, A, B,

C, D and Y be represented by the complex numbers 0, a, b, c, d and y (Fig 82b) The triangles XAD and XBC aresimilar and the orientations of their corresponding vertices (viewed as ordered triples of points) are opposite Sothey can be placed on the complex plane in such a way that b = λa, c = λd, where λ is a positive real number Thepoint Y is characterised by the conditions |y − a| = |y − λa| and |y − d| = |y − λd The first condition rewrites as(y − a)(y − a) = (y − λa)(y − λa); i.e

(λa − a)y + (λa − a)y = (λ2− 1)|a|2.The second condition leads to the analogous equation, with a and a replaced by d and d So we get a system oftwo linear equations with unknowns y and y, which is easily solved

y = λad(a − d) + ad(d − a)

(The condition that the lines AB and CD are not parallel is equivalent to λ26= 1 and ad 6= ad, so the calculationsare legitimate.)

The claim is that the angular argument of the ratio a − y

b − y equals twice the argument of the ratio

a − d

0 − d This isequivalent to showing that

a − y

b − y =

d − ad

2

.µ, where µ is a positive real number

Now, having an expression for y we simply calculate

a − y = d(a − d)(a − λa)

ad − ad , b − y = λa − y =

d(a − d)(a − λa)

and hence

a − y

b − y =

d(a − d)d(a − d =

the result follows

Comment In its original formulation, the problem statement allowed X to be an arbitrary point of the plane,not necessarily inside ABCD, and did not require that the angles ADX, BCX, DAX, CBX be acute Withoutthese assumptions the conclusion of the problem requires considering oriented angles, possibly exceeding 360◦.Problem 15 Ten gangsters are standing on a flat surface, and the distances between them are all distinct Attwelve o’clock, when the church bells start chiming, each of them shoots at the one among the other nine gangsterswho is the nearest, At least, how many gangsters will be killed?

Solution The problem can be restated using mathematical terminology as follows:

A set S or ten points in the plane is given, with all the mutual distances distinct For each point P ∈ S wemark red the point Q ∈ S (Q 6= P ) nearest to P Find the least possible number of red points

Note that every red point can be assigned (as the closest neighbour) to at most five points from S Otherwise, if

a point Q were assigned to P1, , P6, then one of the angles PiQPjwould be not greater than 60◦, in contradiction

to PiPj being the longest side in the (non-isosceles) triangle PiQPj

Let AB be the shortest segment with endpoints A, B ∈ S, Clearly, A and B are both red We are going toshow that there exists at least one more red point Assume the contrary, so that for each one of the remainingeight points, its closest neighbour is either A or B In view of the previous observation, A must be assigned to fourpoints, M1, M2, M3, M4, and B must be assigned to the remaining four points, N1, N2, N3, N4, Choose labelling

so that the angles MiAMi+1 (i = 1, 2, 3) are successively adjacent, angles NiBNi+1 are so too, the points M1, N1lie on one side of line AB, and M4, N4 lie on the opposite side, As before, each angle MiAMi+1 and NiBNi+1 isgreater than 60◦ Therefore each one of ∠M1AM4 and ∠N1BN4 is less than 180◦, and hence

of the quadrilateral ABN M is less than 180◦ Similarly, its internal angle N M A is less than 180◦ Thus ABN M is

a convex quadrilateral Choose points U, V, X, Y arbitrarily on the rays M A, N B, AM, BN produced beyond thequadrilateral The previous condition ∠M AB + ∠N BA < 180◦ implies the inequalities ∠U AB + ∠ABV > 180◦and ∠XM N + ∠M N Y < 180◦ Denote the angles : α = ∠N AB, β = ∠ABM, γ = ∠BM N, δ = ∠M N A In

triangle N AB we have AB < N B, so that ∠AN B < ∠N AB = α, and thus ∠ABV = ∠N AB + ∠AN B < 2α (Fig83a)

In triangle BM N we have M N > BN , so that ∠M BN > ∠BM N = γ and consequently ∠M N Y = ∠BM N +

∠M BN > 2γ Analogously, ∠U AB < 2β and ∠XM N > 2δ Hence

2α + 2β > ∠ABV + ∠U AB > 180◦> ∠M N Y + ∠XM N > 2γ + 2δ

This yields the desired contradiction because α + β = γ + δ (= ∠AZM , where Z is the point of intersection of ANand BM )

Thus, indeed, there exists a third red point The following example shows that a fourth red point need not exist,

so that three is the minimum sought

Example The two tangent circles in the figure differ slightly in size The acute central angles are greater than 60◦.Six points are just a bit outside the circles The length of the vertical segment is equal to the radius of the biggercircle Each point has a unique (hence well-defined) closest neighbour, which has to be marked red (Fig 83b).The only three points which will be marked red are the two centres and the point of tangency If some of the(irrelevant) distances happen to be equal, one can slightly perturb the positions of any points without destroyingthe mentioned properties

Problem 16 Let AH1, BH1, CH3 be the altitudes of an acute-angled triangle ABC Its incircle touches the sides

BC, CA, AB at T1, T2, T3, respectively Consider the symmetric images of the lines H1H2, H2H3, H1H3with respect

to the lines T1T2, T2T3, T1T3 Prove that these images form a triangle whose vertices lie on the incircle of the triangleABC

Solution Let M1, M2, M3 be the reflections of T1, T2, T3 across the bisectors of ∠A, ∠B, ∠C, respectively Thepoints M1, M2, M3 obviously lie on the incirc1e of 4ABC We prove that they are the vertices of the triangleformed by the images in question, which settles the claim (Fig 84) By symmetry, it suffices to show that the

reflection l1 of H2H3in T2T3passes through M2 Let I be the incentre of 4ABC Note that T2and H2are always

on the same side of BI, with T2 closer to BI than H2 We consider only the case when C is on this same side of

BI, as in the figure (minor modifications are needed if C is on the other side)

Let ∠A = 2α, ∠B = 2β, ∠C = 2γ

Lemma 1 The mirror image of H2 with respect to T2T3 lies on the line BI

proof Let l⊥T2T3, H2∈ l Denote by P and S the points of intersection of BI with l and BI with T2T3 Notethat S lies on both line segments T2T3 and BP It is sufficient to prove that ∠P SH2= 2∠P ST2

We have ∠P ST2= ∠BST3, and, by the external angle theorem, (in 4BST3)

∠BST3= ∠AT3S − ∠T3BS = (90◦− α) − β = γ

Next, ∠BST1 = ∠BST3 = γ, by symmetry across BI Note that C and S are on the same side of IT1, since

∠BT1S = 90◦+ α > 90◦ Then, in view of the equalities ∠IST1= ∠ICT1(= γ), the quadrilateral SIT1C is cyclic,

so ∠ISC = ∠IT1C = 90◦ But then BCH2S is also a cyclic quadrilateral, because ∠BH2C = 90◦ It follows that

Suppose β 6= γ; let the line CB meet H2H3and T2T3 at D and E, respectively (Note that D and E lie on line

BC on the same side of the segment BC.) An easy angle computation gives ∠BDH3= 2|β − γ|, ∠BET3= |β − γ|,and so the line l1 is indeed parallel to BC The proof is complete

Another Solution Let J be the point dividing the line segment OI internally in ratio OJ : J I = R : r, where Rand r are the circumradius and the inradius of 4ABC, respectively Consider the homothety h with centre J andcoefficient −r/R By its definition, h takes the circumcircle of 4ABC to its incircle ω