18 chuyên đề toán học theo xu hướng hội nhập quốc tế 2013

How to describe the two-variable function f x, y in the cases f x, y is even-odd in both variables x, y about the point p, q, i.e... The two-variable function f x, y in the cases f x, y

Mục lục

Lời nói đầu

Nguyen Van Mau

On the solutions of some classes of functional equations with transformed

Dam Van Nhi

Nguyen Dang Phat

Nguyễn Minh Tuấn

Nguyễn Bá Đang

Trần Xuân Đáng

Hoàng Minh Quân

Phương trình bậc bốn và hệ thức hình học trong tứ giác hai tâm 156 Nguyễn Thùy Trang

Nguyễn Đình Thức

Nguyễn Hoàng Cương

Nguyễn Thị Giang

Tính chia hết trong các bài toán về dãy số

Nguyễn Hữu Thiêm

ON THE SOLUTIONS OF SOME CLASSES OF FUNCTIONALEQUATIONS WITH TRANSFORMED ARGUMENT

NGUYEN VAN MAU

Abstract

We deal with some functional equations with transformed arguments

in real plane By an algbraic approach we solve some kinds of functionalequations with the reflection arguments

f (x, y)±f (2p−x, y)±f (x, 2q−y)+f (2p−x, 2q−y) = h(x, y), (x, y) ∈ Ω,

(0.1)where (p, q) is the symmetric center of the set Ω ⊂ R × R, h(x, y) isgiven

In applications, we formulate the necessary and sufficient conditionfor solvability of the following functional equations

f (xy)±f ((1−x)y±f (x(1−y))+f ((1−x)(1−y)) = h(xy), ∀x, y ∈ (0, 1).and

f (x + y) ± f (−x + y) ± f (x − y) + f (−x − y)) = h(x + y), ∀x, y ∈ (−1, 1)and describe the formulae of the general solution f (xy) and f (x + y), respectively.1

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Definition 1.1 Let be given a set Ω := P × Q ⊂ R × R and the point (p, q)

is the center (centeral point) of Ω Function f (x, y) defined in Ω is said to beeven-even (or even in both variables or skew symmetric) about the point (p, q)iff

f (2p − x, y) = f (x, y) and f (x, 2q − y) = f (x, y), ∀(x, y) ∈ Ω

Definition 1.2 Let be given a set Ω := P × Q ⊂ R × R and the point (p, q)

is the center (centeral point) of Ω Function f (x, y) defined in Ω is said to beeven-odd about the point (p, q) iff

f (2p − x, y) = f (x, y) and f (x, 2q − y) = −f (x, y), ∀(x, y) ∈ Ω

Remark 1 Similarly, we have the definitions of odd-even and odd-odd tions

fun-In a special case, we have

Definition 1.3 Function f (x, y) defined in R × R is said to be even-even iff

f (−x, y) = f (x, y) and f (x, −y) = f (x, y), ∀x, y ∈ R

The following natural questions arise:

∗

2000 Mathematics Subject Classification Primary 39B99, 39B62, 39B22, 39B32, 39B52.

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Problem 1.1 How to describe the two-variable function f (x, y) in the cases

f (x, y) is even-even in both variables (x, y) about the point (p, q), i.e

f (2p − x, y) = f (x, y) and f (x, 2q − y) = f (x, y), ∀(x, y) ∈ Ω (1.1)Solution Note that

f (2p − x, 2q − y) = f (x, 2q − y) = f (x, y), ∀x, y ∈ Ω,

so we can write

f (x, y) = 1

4[f (x, y) + f (x, 2q − y) + f (2p − x, y) + f (2p − x, 2q − y)] (1.2)Now we prove that the function f (x, y) is even in both variables (x, y) if andonly if there exists a function g(x, y) defined in R × R such that

f (x, y) = 1

4[g(x, y) + g(x, 2q − y) + g(2p − x, y) + g(2p − x, 2q − y)]. (1.3)Indeed, if f (x, y) is of the form (1.3) then it is easy to check the conditions(1.1) are saistified and if f (x, y) is even then it has the form (1.2) and thenthe form (1.3) with g = f

Corollary 1.1 The two-variable function f (x, y) is even in both variables(x, y), i.e

f (−x, y) = f (x, y) and f (x, −y) = f (x, y), ∀(x, y) ∈ R (1.4)iff it is of the form

f (x, y) = 1

4[g(x, y) + g(x, −y) + g(−x, y) + g(−x, −y)]. (1.5)where g(x, y) is an arbitrary function defined in R × R

Similarly, we have

Problem 1.2 How to describe the two-variable function f (x, y) in the cases

f (x, y) is even-odd in both variables (x, y) about the point (p, q), i.e

f (2p − x, y) = f (x, y) and f (x, 2q − y) = −f (x, y), ∀(x, y) ∈ Ω (1.6)Solution Note that

f (2p − x, 2q − y) = f (x, 2q − y) = −f (x, y), ∀x, y ∈ Ω,

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so we can write

f (x, y) = 1

4[f (x, y) + f (2p − x, y) − f (x, 2q − y) − f (2p − x, 2q − y)] (1.7)Now we prove that the function f (x, y) is even-odd in both variables (x, y) ifand only if there exists a function g(x, y) defined in R × R such that

f (x, y) = 1

4[g(x, y) + g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y)]. (1.8)Indeed, if f (x, y) is of the form (1.8) then it is easy to check the conditions(1.6) are saistified and if f (x, y) is even-odd then it has the form (1.7) andthen the form (1.8) with g = f

Similarly, we can formulate the following representations

Theorem 1.1 The two-variable function f (x, y) in the cases f (x, y) is even in both variables (x, y) about the point (p, q), i.e

even-f (2p − x, y) = even-f (x, y) and even-f (x, 2q − y) = even-f (x, y), ∀(x, y) ∈ Ω (1.9)

is of the form

f (x, y) = 1

4[g(x, y) + g(2p − x, y) + g(x, 2q − y) + g(2p − x, 2q − y)] (1.10)Theorem 1.2 The two-variable function f (x, y) in the cases f (x, y) is even-odd in both variables (x, y) about the point (p, q), i.e

f (2p − x, y) = f (x, y) and f (x, 2q − y) = −f (x, y), ∀(x, y) ∈ Ω (1.11)

is of the form

f (x, y) = 1

4[g(x, y) + g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y)] (1.12)Theorem 1.3 The two-variable function f (x, y) in the cases f (x, y) is odd-even in both variables (x, y) about the point (p, q), i.e

Theorem 1.4 The two-variable function f (x, y) in the cases f (x, y) is ood in both variables (x, y) about the point (p, q), i.e.

odd-f (2p − x, y) = −odd-f (x, y) and odd-f (x, 2q − y) = −odd-f (x, y), ∀(x, y) ∈ Ω (1.15)

is of the form

f (x, y) = 1

4[g(x, y) − g(2p − x, y) − g(x, 2q − y) + g(2p − x, 2q − y)] (1.16)Now we consider the special case of two-variable function f (x, y) defined inthe set Ω0 = (0, 1) × (0, 1) in the cases f (x, y) is even - even in both variables(x, y) about the point0,1

2

 So theorem 2.1 can be formulate in the followingform

Corollary 1.2 The two-variable function f (x, y) in the case when f (x, y) iseven-even in both variables (x, y) about the point 1

2,

12

, i.e

f (1 − x, y) = f (x, y) and f (x, 1 − y) = f (x, y), ∀x, y ∈ (0, 1) (1.17)

is of the form

f (x, y) = 1

4[g(x, y) + g(1 − x, y) + g(x, 1 − y) + g(1 − x, 1 − y)]. (1.18)Corollary 1.3 The two-variable function f (x, y) in the case when f (x, y) isodd-odd in both variables (x, y) about the point 1

2,

12

, i.e

f (1 − x, y) = −f (x, y) and f (x, 1 − y) = −f (x, y), ∀x, y ∈ (0, 1) (1.19)

is of the form

f (x, y) = 1

4[g(x, y) − g(1 − x, y) − g(x, 1 − y) + g(1 − x, 1 − y)]. (1.20)

fun-tions induced by involufun-tions

In this section we will solve the following functional equations

f (x, y) + f (2p − x, y) + f (x, 2q − y) + f (2p − x, 2q − y) = h(x, y), ∀(x, y) ∈ Ω

(2.1)

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f (x, y) − f (2p − x, y) − f (x, 2q − y) + f (2p − x, 2q − y) = h(x, y), ∀(x, y) ∈ Ω,

(2.2)where (p, q) is the symmetric center of the set Ω ⊂ R × R, h(x, y) is given.Denote by X the set of all functions defined on X and X = L0(X, X),where L0(X, X) denotes the linear space of all linear operators A : X → Xwith dom A = X It is easy to check that X is an algebra (linear ring) overfield R

Consider the following linear elements (operators) V and W in X as follows(V f )(x, y) = f (2p − x, y), (W f )(x, y) = f (x, 2q − y), f ∈ X (2.3)

It is easy to see that V and W are involution elements, i.e V2 = I and

W2 = I, where I is an identity element of X Moreover, they are commutative,i.e V W = W V

Now we rewrite (2.1) in the form

Kf := (I + V + W + V W )f = h (2.4)Lemma 2.1 Operator K defined by (2.4) is an algebraic element with chara-teristic polynomial

Proof Note that the following identities hold

V K = K, W K = K, V W K = K

So K2 = 4K and K is not a scalar operator, which toghether imply PK(t) =

t2− 4t, which was to be proved

Theorem 2.1 The general solution of the homogeneous equation Kf = 0 is

of the form

f (x, y) = 1

4[3g(x, y)−g(2p−x, y)−g(x, 2q−y)−g(2p−x, 2q−y)], g ∈ X (2.6)Proof By Lemma 2.1, from equality (K − K)f = 0, ∀f ∈ X, we find 14K2−4K



f = 0 ⇔ (K2 − 4K)f = 0 ⇔ K(K − 4I)f = 0 Hence (K − 4I)X ⊂ker K On the other hand, if ϕ ∈ ker K then Kϕ = 0 and K(K − 4I)ϕ =(K − 4I)Kϕ = 0 It follows ϕ ∈ =(4I − K), which was to be proved

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Theorem 2.2 The non-homogeneous equation (2.1) (Kf = h) is solvable ifand only if the following condition

If it is the case, then the general solution of (2.1) is of the form

f (x, y) = 3g(x, y) − g(2p − x, y) − g(x, 2q − y) − g(2p − x, 2q − y) (2.8)+1

4[3h(x, y) − h(2p − x, y) − h(x, 2q − y) − h(2p − x, 2q − y)], g ∈ X.Proof Suppose that the equation (2.1) is solvable and f0 is a solution Then

By Lemma 2.1 from equality Kf0 = h it follows K2f0 = Kh ⇔ 4Kf0 = Kh ⇔4h = Kh

Suppose that the condition (2.7) is satisfied Write the non-homogeneousequation (2.1) in the form Kf = 14Kh or in the equivalent form

Now we consider (2.2) Write it in the form

Lf := (I − V − W + V W )f = h (2.10)Lemma 2.2 Operator L defined by (2.4) is an algebraic element with chara-teristic polynomial

Proof The proof follows from the following identities

−V L = L, −W L = L, V W L = L

So L2 = 4L and L is not a scalar operator, which toghether imply PL(t) =

t2− 4t, which was to be proved

Theorem 2.3 The general solution of the homogeneous equation Kf = 0 is

of the form

f (x, y) = 1

4[3g(x, y) + g(2p − x, y) + g(x, 2q − y) − g(2p − x, 2q − y)], g ∈ X.

(2.12)9

Proof By the same method as for theorem2.1.

Theorem 2.4 The non-homogeneous equation (2.1) (Lf = h) is solvable ifand only if the following condition

If it is the case, then the general solution of (2.1) is of the form

f (x, y) = 3g(x, y) + g(2p − x, y) + g(x, 2q − y) − g(2p − x, 2q − y) (2.14)+1

4[3h(x, y) + h(2p − x, y) + h(x, 2q − y) − h(2p − x, 2q − y)], g ∈ X.Proof Suppose that the equation (2.2) is solvable and f0 is a solution Then

By Lemma 2.2 from equality Lf0 = h it follows L2f0 = Lh ⇔ 4Lf0 = Lh ⇔4h = Lh

Suppose that the condition (2.13) is satisfied Write the non-homogeneousequation (2.2) in the form Lf = 14Lh or in the equivalent form

Now we consider some special cases of equation when q = p = 1

2 In that case,the center point of Ω is 1

2,

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 and (2.1) is of the form

f (x, y)+f (1−x, y)+f (x, 1−y)+f (1−x, 1−y) = h(x, y), ∀x, y ∈ (0, 1) (3.1)and

f (x, y)−f (1−x, y)−f (x, 1−y)+f (1−x, 1−y) = h(x, y), ∀x, y ∈ (0, 1) (3.2)

In this case, the role of x and y in the left side of (3.1) are the same

Now return to the function f (t), we can formulate the following

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Theorem 3.1 The function f (t) satisfying the conditions

f ((1 − x)(1 − y)) = f (x(1 − y)) = f (xy), ∀x, y ∈ (0, 1)

So we can write f (t) in the form

f (xy) = 1

4[f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y))].Last equality gives us the proof of the theorem

Theorem 3.2 The functional equation

f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y))] = h(xy) (3.5)

is solvable if and only if there exists h(t) satisfying the following condition

h



t − 12



= h

12

, ∀t ∈

1

2, 1

.or

h(x) = h1

2

, ∀x ∈0,1

(3.7) and (3.8) together inply h(t) ≡ const

So, the necessary condition for solvability of (3.1) if

h(x, y) = h(y, x), ∀x, y ∈ (0, 1) (3.9)Now we formulate the similar results as in the previous section

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Theorem 3.3 The general solution of the homogeneous equation

f (x, y) + f (1 − x, y) + f (x, 1 − y) + f (1 − x, 1 − y) = 0, ∀x, y ∈ (0, 1) (3.10)

is of the form

f (x, y) = 1

4[3g(x, y) − g(1 − x, y) − g(x, 1 − y) − g(1 − x, 1 − y)], g ∈ X (3.11)Theorem 3.4 The non-homogeneous equation

f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y)) = h(xy), ∀x, y ∈ (0, 1)

Theorem 3.5 The function f (t) is a general solution of the homogeneousequation

f (xy) + f ((1 − x)y) + f (x(1 − y)) + f ((1 − x)(1 − y)) = 0, ∀x, y ∈ (0, 1) (3.14)

if and only if there exists a function g(t) such that

f (xy) = 1

4[3g(xy)−g((1−x)y)−g(x(1−y))−g((1−x)(1−y))], g ∈ X (3.15)Remark 2 The homogeneous equation (3.14) was posed by Sahoo and Sander

in 1990 (see [3]-[4]) The continuous solutions were found by Z Daroczy and

A Jarai in [5]

Now we deal with the equation induced by addition of arguments Weconsider the special cases of equation when q = p = 0 In that case, the centerpoint of Ω is (−1, 1) and (2.1) is of the form

f (x, y) + f (−x, y) + f (x, −y) + f (−x, −y) = h(x, y), ∀x, y ∈ (−1, 1) (3.16)Theorem 3.6 The function f (t) is a general solution of the homogeneousequation

f (x + y) + f (−x + y) + f (x − y) + f (−x − y)) = 0, ∀x, y ∈ (−1, 1) (3.17)

if and only if f (x) is an odd function in (−2, 2)

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Proof If f is odd in (−2, 2), then f (x − y) = −f (−x + y) and f (−x − y) =

−f (x + y) Hence,

f (x + y) + f (−x + y) + f (x − y) + f (−x − y)) = 0, ∀x, y ∈ (−1, 1).Conversely, suppose f is a solution of (3.17) Putting x = 0, y = 0 into (3.17)

we find f (0) = 0 Similarly, putting y = x, into (3.15) and ussing the equality

f (0) = 0, we find f (−2x) = −f (2x), i.e f is odd in (−2, 2)

Theorem 3.7 The non-homogeneous equation

f (x+y)+f (−x+y)+f (x−y)+f (−x−y)) = h(x+y), ∀x, y ∈ (−1, 1) (3.18)

is solvable iff h(t) ≡ const in (−2, 2) If it is the case, then the general solution

of (3.14) is of the form

f (t) = 1

where g is an arbitrary odd function in (−2, 2)

Proof Suppose equation (3.18) is solvable and f is its solution Putting y = xand y = −x into (3.18), we find h(2x) = h(0), x ∈ (−1, 1), i.e h(t) ≡ h(0) in(−2, 2)

If h(t) ≡ h(0) in (−2, 2) then we can reduce equation (3.18) to the equationϕ(x + y) + ϕ(−x + y) + ϕ(x − y) + ϕ(−x − y)) = 0, ∀x, y ∈ (−1, 1), (3.20)

where ϕ(t) = f (x)−1

4 Hence, the solution (3.19) follows from theorem 3.6.Now we return to the equation

f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y)) = 0 (3.21)

By the same way as previous equations, we have

Theorem 3.8 The function f (t) satisfying the conditions

Proof From (3.23) we find f (x(1 − y)) = −f (xy) and

f ((1 − x)(1 − y)) = −f (x(1 − y)) = f (xy), ∀x, y ∈ (0, 1)

So we can write f (t) in the form

f (xy) = 1

4[f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y))].Last equality gives us the proof of the theorem

Theorem 3.9 The functional equation

f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y))] = h(xy) (3.24)

is solvable if and only if there exists h(t) ≡ 0

Theorem 3.10 The non-homogeneous equation

f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y)) = h(xy), ∀x, y ∈ (0, 1)

f (xy) − f ((1 − x)y) − f (x(1 − y)) + f ((1 − x)(1 − y)) = 0, ∀x, y ∈ (0, 1) (3.27)

if and only if there exists a function g(t) such that

f (xy) = 1

4[3g(xy)+g((1−x)y)+g(x(1−y))−g((1−x)(1−y))], g ∈ X (3.28)Remark 3 The equation (3.27) was posed firstly by K Lajko in [8] for X = Rand then by Sahoo and Sander in 1990 (see [3]-[4]) The differentiable solutionswere found by C.J Eliezer in [6]

Theorem 3.12 The non-homogeneous equation

f (x+y)−f (−x+y)−f (x−y)+f (−x−y)) = h(x+y), ∀x, y ∈ (−1, 1) (3.29)

is solvable iff h(t) ≡ const in (−2, 2) If it is the case, then the general solution

of (3.27) is of the form

f (t) = c + 1

2(g(t) − g(−t)), t ∈ (−2, 2), (3.30)where g is an arbitrary function in (−2, 2), c = f (0)

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Proof Suppose equation (3.31) is solvable and f is its solution Putting y = xand y = −x into (3.31), we find h(2x) = −h(0), x ∈ (−1, 1), i.e h(t) ≡ −h(0)

in (−2, 2) Put x = 0 = y into (3.31) we find h(0) = 0

So (3.30) is of the form

f (x + y) − f (−x + y) − f (x − y) + f (−x − y)) = 0, ∀x, y ∈ (−1, 1) (3.31)and it has solution of the form (3.30)

Equa-[3] P.K Sahoo, T Riedel, Mean Value Theorems and Functional Equations,World Scientific, Singapore/New Jersey/London/HongKong, 1998.[4] B.R Ebanks, P.K Sahoo and W Sander, Determination of measurablesum from infomation measures satisfying (2, 2)−additivity of degree (α, β),Radovi Matematicki, vol.6, 77-96, 1990

[5] Z Daroczy and A Jarai, On the measurable solution of a functional tion of the information theory, Acta Math Acad Sci Hungaricae, vol.34,105-116, 1979

equa-[6] C.J Eliezer, A solution to f (1−x)(1−y))+f (xy) = f (x(1−y))+f (1−x)y,Aequationes Math., vol.46, 301, 1993

[7] Gy Maksa, Problem, Aequationes Math., vol.46, 301, 1993

[8] K Lajko, What is the general solution to the equation f (1 − x)(1 − y)) +

f (xy) = f (x(1 − y)) + f (1 − x)yAequationes Math., vol.10, 313, 1974.[9] D Przeworska - Rolewicz, Algebraic analysis, PWN - Polish ScientificPublishers and D Reided Publishing Company, Warszawa - Dordrecht,1988

[10] D Przeworska - Rolewicz and S Rolewicz, Equations in linear space,Monografie Matematyezne 47, PWN - Polish Scientific Publishers,Warszawa, 1968

[11] Ng V Mau, Boundary value problems and controllability of linear systemswith right invertible operators, Dissertationes Math., CCCXVI, Warszawa,1992

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INEQUALITY AND IDENTITY (M, N )

Dam Van Nhi Pedagogical University Ha Noi

136 Xuan Thuy Road, Hanoi, Vietnam.

Email: damvannhi@yahoo.com

2010 Mathematics subject classification: 26D05,26D15,51M16

Abstract In this paper we introduce the inequalty and identity (M, N ), of which Hayashi

is a spcial case: Inequality, Identity (M, N ) And after then we will present some esting applications

inter-Keywords Hayashi’s inequality, point, triangle, polygon

In Euclidian geometry, the Hayashi’s Inequality in R2 states: Suppose given a triangle4ABC of the lengths of sides a, b, c Then, with any point M, we have an inequality

aM B.M C + bM C.M A + cM A.M B > abc(see [3, pp 297, 311]) In this paper we propose to give a new inequality and it’sidentity, which is a generalization of the above inequality, and after then we want togive some interesting applications about triangle

2 Inequality and Identity (M, N )

Now we prove an inequality, of which Hayashi is a special case Use this result we cangive some new interesting inequalities We give the following:

Theorem 2.1 Let A1A2 An be a polygon, s be an integer, s < n, and arbitrarypoints N1, N2, , Ns, M in euclidean plane R2 we have the following inequality

i6=k

AkAi.M Ak, (M, N ).

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(i) If s = 0, we have Hayashi inequality.

(ii) If n = 3, s = 1, and A, B, C, N belong to the circle with the centere M we have theinequality aAN + bBN + cCN > 4SABC

Proof Suppose that Ak have affixe ak, M has affixe z and Nh affixe zh Using the

Lagrange interpolation formula, we have

Q

i=1

M Ai6

Q

i=1

M Ai6

n

P

k=1

1Q

i6=k

AkAi.M Ak.(ii) When n = 3, s = 1 and A, B, C, N belong to the circle with the centere M we havethe inequality abc

R 6 aAN + bBN + cCN or aAN + bBN + cCN > 4SABC .Remark 2.2 Denote N as the center of circumcircle From the iequality aAN +bBN +cCN > 4SABC by the inequality (M,N) (ii) we deduce R(a + b + c) > 2r(a + b + c) or

Proof (i) Applying the inequality (M,N) (ii) we obtain aOB.OC +bOC.OA+cOA.OB >abc or R2 > abc

abc3R, bGC.GA = R2

abc3RandcGA.GB = R2abc

3R therefore R1

abc3R + R2

abc3R + R3

abc3R > abc or R1+ R2+ R3 > 3R.(iii) Applying the inequality (M, N ) we have aIB.IC + bIC.IA + cIA.IB > abc SinceaIB.IC = 4raSIBC = 2rara = 4ra

ha

rabc4R =

ha

abc4R = R

0 1

y

hb

abc

R andcOA.OB = R03 z

Proposition 2.4 Suppose given a triangle ABC with the lengths of sides a, b, c tively and R is the radius of circumcircle of 4ABC Let’s I, Ja, Jb, Jc are the centres ofincircle and escribed circles of ∆ABC, respectively Then, with any point M, we have

√

c + a − b

√caM B +

√

a + b − c

√abM C .

CIabM C.

√

c + a − b

√caM B +

√

a + b − c

√abM C .

18

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(iii) Applying the inequality (M, N ) to n = 3, s = 1, we have the three inequalities

M Ja

M A.M B.M C 6 AJa

bcM A +

BJacaM B +

CJaabM C

M Ja

M A.M B.M C 6 AJb

bcM A +

BJbcaM B +

CJbabM C

CJa.CJcabM C

M Jb.M Jc

M A.M B.M C 6 AJb.AJc

bcM A +

BJb.BJccaM B +

CJb.CJcabM C

M Jc.M Ja

M A.M B.M C 6 AJc.AJa

bcM A +

BJc.BJacaM B +

CJc.CJaabM C .

On adding the three inequalities, we find the inequality M Ja.M Jb+ M Jb.M Jc+ M Jc.M Ja

b√4R2− b2

c√4R2− c2

CO.CHabM C .Thus, we obtain the inequality abcM O.M H

c√4R2− c2

19

Corollary 2.6 Suppose given the triangle ABC with the lengths of sides a, b, c,

respec-tively I, G, H the center of inscribed circle and the barycenter and the orthocenter of

∆ABC Then, with any point M, we always have the inequality

Proposition 2.7 Suppose the polygon A1A2 An is inscribed in the circle with the

center O and radius R Then, with any s < n points N1, , Ns in the plane A1A2 An,

we always have the inequality

Rn−1 follows from the inequality (M, N )

when M ≡ O With R = 1 we have

Example Giving the triangle ABC with the lengths of sides a, b, c and R is the dius of circumscribed circle; r1, r2, r3 are the radius of escribed circles Let’s da, db, dcthe distances from the center of circumscribed circle to the center of escribed circles.Then, with any point D belong to the circumscribed circle of ∆ABC we always have theinequality:

x√

b + c − a +

√ca

y√

c + a − b+

√ab

z√

a + b − c +

DJa.DJb.DJcxyz√

√

b + c − a +

√ca

√

c + a − b +

√ab

x√

b + c − a+

√ca

y√

c + a − b+

√ab

z√

a + b − c+

DJa.DJb.DJcxyz√

a + b + c.(ii) Since d2a = R2+ 2Rr1, d2b = R2+ 2Rr2, d2c = R2+ 2Rr3 therefore

s

(R + 2r1)(R + 2r2)(R + 2r3)

R3(a + b + c) 6

√bc

√

b + c − a+

√ca

√

c + a − b +

√ab

Without generality, we can assume that the radius R of the circle C equal to 1 Supposethat every point Ak has affixe ak = cos αk + i sin αk, and M has affixe z = cos u +

i sin u and every Nh has affixe zh = cos uh+ i sin uh Following the interpolation formula

21

We reduce all the factors 2i, e

t6=k

sinαk− αt

2e

2Q

t6=k

sinαk− αt2

sin(s + 1 − n)(αk− u)

2 = 0 With this result, we havejust built the identities under the form of trigonometry and geometry for the inequality(M, N ) as below:

Proposition 2.8 Assume that the polygon A1A2 An is inscribed in the circle withthe center O and radius R Taking s + 1 points N1, , Ns and M also belonging to thiscircle C Assuming that the coordinate Ak(cos αk; sin αk), k = 1, 2, , n; the coordinate

Nj(cos uj; sin uj), j = 1, 2, , s and the coordinate M (cos u; sin u) Then, we will havethese identities

t6=k

sinαk− αt

2when s = n − 1

t6=k

sinαk− αt2

DB Using the relationship

DA2.DB.DC.a + DC2.DA.DB.c = DB2.DC.DA.b

Hence DA2SDBC+ DC2SDAB = DB2SDCA [Feuerbach]

Proposition 2.10 Suppose the polygon A1A2 An is inscribed in the circle withthe radius R = 1 Taking s + 1 points N1, , Ns and M also belonging to this cir-cle C Assuming that the coordinate Ak(cos αk; sin αk), k = 1, 2, , n; the coordinate

Nj(cos uj; sin uj), j = 1, 2, , s and the coordinate M (cos u; sin u) Then, with the properchoices of + or − we will have the identities

Corollary 2.11 Asuming that the points A1, A2, , An, M in order belong to the circle

C with the center O We convent n + 1 := 1 Then, we have the identities

Proof These identities follow from the identity (M, N ) with s = 0

Corollary 2.12 Let the quadrilateral ABCD be inscribed in the circle C with the center

O Let a = BC, b = CA, c = AB Then, we have 2 identities:

(i) a cos(OD, OA)

(ii) a sin(OD, OA)

√

b + c − a

√ca

√

c + a − b

√ab

√

a + b − c

Open Problem 3.2 Giving the triangle ABC with the lengths of sides a, b, c and R

is the radius of circumscribed circle; r1, r2, r3 are the radius of escribed circles Let’s

da, db, dc the distances from the center of circumscribed circle to the center of escribedcircles Then, with any point D belong to the circumscribed circle of ∆ABC we alwayshave the inequality:

24

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b + c − a +

√ca

y√

c + a − b+

√ab

√

b + c − a +

√ca

√

c + a − b +

√ab

√

a + b − c.References

[1] T Andreescu and D Andrica, Proving some geometric inequalities by using complexnumbers, Eduatia Mathematica Vol 1 Nr 2 (2005),19-26

[2] T Hayashi, Two theorems on complex numbers, Thoku Math J., 4(1913/14), pp.68-70

[3] D.S Mitrinovic, J.E Pecaric and V Volenec, Recent Advances in Geometric equalities, Acad Publ., Dordrecht, Boston, London 1989

In-25

SOME NATIONAL OLYMPIAD PROBLEMS

Solution Let α = ∠CAB, β = ∠ABC, and γ = ∠BCA be the angles of triangle ABC Let the line through A and A 1 meet side BC at X Similarly, let the line through B and B 1 meet side CA at Y , and the line through C and C 1 meet side AB at Z (Fig 72) By the converse of Ceva’s Theorem, it suffices to prove that

BX XC

cotβ + 1 cotγ + 1.Similarly,

CY

Y A=

cotγ + 1 cotα + 1 and

AZ

ZB=

cotα + 1 cotβ + 1.Hence,

BX XC

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 A and 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◦.

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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 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 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

AG

BG=sin χ sin 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

sin β1sin β26 sin2β

2 and sin γ1sin γ26 sin2γ

γ 1 = γ 2 ; 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

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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 the intersection of F A and DE Find, with proof, the value of the sum.

DB

DD 0 + EC

EE 0 + F A

F F 0 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 ω 1 , ω 2 , ω 3 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

Problem 6 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.

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 the triangle 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− y 2 − 4y + 1 = 0, which can be factored as

2 + y − 1 (y + 1)(y 2 + 2y)=

= (y

2 + y − 1)y (y 2 + y)(y 2 + 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 A 1 lie on BC with OA 1 perpendicular to BC Define B 1 on CA and C 1 on AB similarly Prove that O is the circumcenter of ABC if and only if the perimeter

of A 1 B 1 C 1 is not less than any one of the perimeters of AB 1 C 1 , BC 1 A 1 , and CA 1 B 1

Solution If O is the circumcenter of 4ABC, then A 1 , B 1 and C 1 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 > P AB 1 C 1 , PBC1A1, PCA1B1 Let (Fig 76)

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Let A 2 be the point of intersection of the lines through B 1 and C 1 , which are parallel to AB and AC, respectively,

as shown in the figure above Assume that γ 1 > α and β 2 > α If one of these inequalities is strict, then A 1 is an interior point of 4B 1 C 1 A 2 Hence P A1B1C1 < P A2B1C1 = P AB1C1, which is a contradiction If γ 1 = α and β 2 = α, then A 1 = A 2 and therefore B 1 O⊥A 1 C 1 and C 1 O⊥A 1 B 1 Hence O is the orthocenter (intersection of the altitudes)

of 4A 1 B 1 C 1 , and thus OA 1 ⊥B 1 C 1 Hence B 1 C 1 //BC This implies that A 1 , B 1 and C 1 are the midpoints of

BC, CA and AB, respectively : i.e., triangles AB 1 C 1 ,

A 1 B 1 C 1 , A 1 B 1 C, and A 1 BC 1 are congruent Hence, O is the circumcenter of 4ABC Analogously, the same conclusion holds if α 1 ≥ β and γ 2 ≥ β, or β 1 ≥ γ and α 2 ≥ γ.

Suppose now that none of these cases are satisfied; i.e., it is not true that

γ1≥ α and β 2 ≥ α or

α1≥ β and γ2≥ β, or

β 1 ≥ γ and α 2 ≥ γ.

Suppose without loss of generality that γ 1 < α Then α 2 > γ,

since γ 1 + α 2 = π − β = α + γ Hence β 1 < γ, which implies that γ 2 > β Hence α 1 < β, implying that β 2 > α In conclusion,

γ 1 < α < β 2 , α 1 < β < γ 2 , and β 1 < γ < α 2 Since AC 1 OB 1 and A 1 CB 1 O are cyclic, we have ∠AOB 1 = γ 2 and

∠COB 1 = α 1 Hence, AO = OB1

cos γ 2

> OB1cos α 1

= CO In the same way, the inequalities γ 1 < β 2 and β 1 < α 2 imply that 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 either case, the assumption that BP P ” is nondegenerate leads BP = P P ” = P P0, thus to the conclusion that BP P0 is equilateral, 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 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

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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 Γ 1 and Γ 2 in the plane and ν is the set consisting of all circles

Λ touching Γ 1 and Γ 2 both internally or both externally and intersecting m ν 6= ∅ : Suppose Λ ∈ ν, touching Γ i

in A i and intersecting m in B i , i ∈ {1, 2}.

Prove, there exists a set W of four points in the plane, such that for every Λ ∈ ν every line A i B j , i, j ∈ {1, 2}

is incident with at least one point in W

Problem 10 Two circles Γ 1 and Γ 2 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 2 = KN.KM =

BK 2 , 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 Γ 1 and Γ 2 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

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 sin ∠BOD=

OD sin ∠OBD and

CD sin ∠COD=

OD sin ∠OCD.

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

BD

CD=

sin(180◦− 2B) sin(180 ◦ − 2C) =

sin 2B sin 2C.

Similarly,

CE

EA=

sin 2C sin 2A,

AF

F B=

sin 2A sin 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

D 0 Then ALA 0 L 0 is a rectangle symmetric with respect to the perpendicular bisector of BC (because A 0 L//BC) Consequently the points D and D 0 are situated symmetrically with respect to the midpoint of BC Hence BD = D 0 C and 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

A 1 A 2 A 3 A j + A 2 A 3 A 1 A j = A 1 A 3 A 2 A j

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 A 1 A 2 A n be cyclic Set

b 1 = −A 1 A 2 , b j = A 2 A j for j = 2, 3, , n; c j = A1Aj

A 1 A 2

for j = 1, 2, , n;

a definition suggested by Ptolemy’s theorem We now check that the numbers b j , c j 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− A 2 Ai.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 b j , c j give

A i A j = A 1 A j = b j 0−(−A 1 A 2 )c j = b j c 1 −b 1 c j Similarly, if i = 2 and 3 ≤ j ≤ n then A i A j = A 2 A j = b j 1−0.c j =

b j c 2 − b 2 c j The proof is complete.

Comment The numbers b j , c j are not uniquely determined Here is one more proof of the necessity Let A 1 , A 2 , , A n

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 A n and A 1 Let 2α j be the measure of the oriented angle U OA j (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 A j For any pair of indices i, j with i < j, we have AiAj = 2R sin1

2 ∠A i OAj, where ∠A i OAj is the respective central (non-oriented) angle Note that ∠A i OAj is equal to 2(α j − α i ), if 0 < 2(α j − α i ) ≤ 180◦, and to 360◦− 2(α j − α i ), if 180◦< 2(α j − α i ) < 360◦ In both cases,

A i A j = 2R sin(α j − α i ) = 2R sin α j cos α i − 2R cos α j sin α i

So we can define b j = u sin α j , c j = v cos α j 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

area ABT area 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 c 2 gives

cot∠ABQ = 2(a

2 + b 2 )

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

BT =

sin ∠AT B sin ∠BAT =

sin(α + β + ϕ) sin 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 ϕ sin ψ =

sin(β + ϕ) sin(β − ψ) This gives

sin βcotϕ + cos β = sin(β + ϕ)

sin 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 γ cos(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 a

maximum 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 Γ 1

and Γ 2 , with centres O 1 and O 2 , respectively Let W be the second point of intersection of the circumcircles of triangles ABZ and CDZ, named Γ 3 and Γ 4 , with centres O 3 and O 4 , respectively We are going to show that the point 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 angles DAX 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 Γ 3 and Γ 4 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⊥O 3 O4.

To this end, note that AZ⊥O1O3 Combined with XZ⊥O1O2, this implies that the angles O2O1O3 and AZX are 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 O 2 O 1 O 3 and O 1 O 2 O 3 are either equal or add up to 180◦ However, the latter case cannot occur because A and B are distinct points, and hence O 1 O 2 O 3 is a non-degenerate triangle Thus O 1 O 3 = O 2 O 3

Likewise, O 1 O 4 = O 2 O 4 These two equalities imply that O 1 O 2 ⊥O 3 O 4 As observed, this is enough to conclude that 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 W also belongs to the region AJ D In other words, the points Z and W lie on the same side of line AB, and also on the 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

∠AW B = ∠AZB = 360◦− (∠AZX + ∠BZX) = ∠ADX + ∠BCX = 2∠ADX.

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 , using inversion.

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 are similar and the orientations of their corresponding vertices (viewed as ordered triples of points) are opposite So they can be placed on the complex plane in such a way that b = λa, c = λd, where λ is a positive real number The point 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 of two 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 calculations are 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 − a d

 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)

ad − ad ,www.VNMATH.com

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 Without these 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 At twelve o’clock, when the church bells start chiming, each of them shoots at the one among the other nine gangsters who 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 we mark 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 P 1 , , P 6 , then one of the angles P i QP j would be not greater than 60◦, in contradiction

to P i P j being the longest side in the (non-isosceles) triangle P i QP j

Let AB be the shortest segment with endpoints A, B ∈ S, Clearly, A and B are both red We are going to show that there exists at least one more red point Assume the contrary, so that for each one of the remaining eight points, its closest neighbour is either A or B In view of the previous observation, A must be assigned to four points, 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 is greater than 60◦ Therefore each one of ∠M 1 AM 4 and ∠N 1 BN 4 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 the quadrilateral 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 (Fig 83a).

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 AN and 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 bigger circle 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 destroying the mentioned properties.

Problem 16 Let AH 1 , BH 1 , CH 3 be the altitudes of an acute-angled triangle ABC Its incircle touches the sides

BC, CA, AB at T 1 , T 2 , T 3 , respectively Consider the symmetric images of the lines H 1 H 2 , H 2 H 3 , H 1 H 3 with respect

to the lines T 1 T 2 , T 2 T 3 , T 1 T 3 Prove that these images form a triangle whose vertices lie on the incircle of the triangle ABC.

Solution Let M1, M2, M3 be the reflections of T1, T2, T3 across the bisectors of ∠A, ∠B, ∠C, respectively The points M1, M2, M3 obviously lie on the incirc1e of 4ABC We prove that they are the vertices of the triangle formed 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 H 2 with respect to T 2 T 3 lies on the line BI.

proof Let l⊥T 2 T 3 , H 2 ∈ l Denote by P and S the points of intersection of BI with l and BI with T 2 T 3 Note that S lies on both line segments T 2 T 3 and BP It is sufficient to prove that ∠P SH 2 = 2∠P ST 2

We have ∠P ST 2 = ∠BST 3 , and, by the external angle theorem, (in 4BST 3 )

∠BST 3 = ∠AT 3 S − ∠T 3 BS = (90◦− α) − β = γ.

Next, ∠BST 1 = ∠BST 3 = γ, by symmetry across BI Note that C and S are on the same side of IT 1 , since

∠BT 1 S = 90 ◦ + α > 90 ◦ Then, in view of the equalities ∠IST 1 = ∠ICT 1 (= γ), the quadrilateral SIT1C is cyclic,

so ∠ISC = ∠IT 1 C = 90 ◦ But then BCH2S is also a cyclic quadrilateral, because ∠BH 2 C = 90 ◦ It follows that

Suppose β 6= γ; let the line CB meet H 2 H 3 and T 2 T 3 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 ∠BDH 3 = 2|β − γ|, ∠BET 3 = |β − γ|, and so the line l 1 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 R and r are the circumradius and the inradius of 4ABC, respectively Consider the homothety h with centre J and coefficient −r/R By its definition, h takes the circumcircle of 4ABC to its incircle ω.

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