Check - mate

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MATh.e .JEANS - “Bogdan Petriceicu Hasdeu” National College, Buzau, Romania Lycée François Arago, Perpignan, France

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Cet a ti le est dig pa des l es. Il peut o po te des ou lis et i pe fe tio s, auta t ue possi le sig al s pa os ele teu s da s les otes d' ditio .

Check - mate

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Students: MACOVEI Nicolae-Cristian, NICOLAE Mario-Alexandru, VOICILĂ Tristan-Ştefan (8^th Grade), MASSIF Damien, ANGLADE Cassandre, GOURBIERE Yohan (1^ère S), GIRARDIN Lila, PELTIER Maelys (1^èreL).

Schools: “Bogdan Petriceicu Hasdeu” National College, Buzau, Romania Lycée François Arago, Perpignan, France

Teachers: Melania NICOLAE, Marie DIUMENGE, Christelle QUERU

Researchers: Robert BROUZET (Université de Perpignan), Bogdan ENESCU

The subject:

When you play the chess, a king can attack any piece placed in an adjacent square, either horizontally, vertically, or in diagonal. Find the number of ways in which non-attacking kings can be placed on a chess board.

The solutions found:

We have to prove that we can place ^� + , for ≥ non-attacking kings on a chess board, were „non-attacking kings” means that the kings are not attacking each other.

I. First method(1).

We divide the chess board in n squares of the type 2x2.

First, we place the kings on the first column for each square. For every square we have two possibilities to put the kings in this way. The number of the cases is · ∙ ∙ … .∙ = ^�

• • ……… • •

• • ………. • •

Then, we move the king which is placed on the last square on the second column.

We have · ∙ ∙ … .∙ = ^� possibilities, too.

• • ……… • •

• • ………. • •

After we have done that we also move the king from the (n-1)^th square on the second column.

Again, we will have · ∙ ∙ … .∙ = ^� possibilities

• • ……… • •

• • ………. • •

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We will repeat this method until every king from each square will be placed on the second column. We will also have ^�solutions.

• • ……… • •

• • ………. • •

So, for every case we have ^�possibilities. There are n+1 cases, so we have ^�∙ � + ways to place n non-attacking kings on a chess board.

II. Equation of induction(2)

We name _�the number of possibilities for placing kings on a board.

We want to find an equation between _�, _�+ _�+ .

The total number of cases for + kings is _�+ ∙ , including the cases in which the kings attack each other, because for each way we place the first + kings we can place the + ^th in 4 different ways:

We have to consider the cases in which the (n+1)^th king and (n+2)^th king attack each other.

For each way we place the first n kings in _�cases, we have 4 ways to place the last two ones so that they attack each other, because the (2n+2)^thking will attack the (2n+3)^thking(3).

1 2 ….. 2n+2 2n+3 2n+4

1 … •

2 …

1 2 ….. 2n+2 2n+3 2n+4

1 … •

2 …

1 2 ….. 2n+2 2n+3 2n+4

1 …

2 … •

1 2 ….. 2n+2 2n+3 2n+4

1 …

2 … •

1 2…..2n+1 2n+2 2n+3 2n+4

1 … • •

2 …

1 2…..2n+1 2n+2 2n+3 2n+4

1 … •

2 … •

1 2…..2n+1 2n+2 2n+3 2n+4

1 … •

2 … •

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So, we have to substract ∙ _� (the number of cases in which the kings attack each other).

Therefore, the number of ways we can place n+2 kings is

�+ = ∙ �+ − ∙ � ⇒ ��+ = � ∙ ��+ − ��

III. Second method(4)

We can place a king in any row: (1,1), (1,2), (2,1), (2,2) ⇒ � � � �

We can place two kings in 12 ways, and no attack: (1,1) and (1,3), (1,1) and (1,4), (1,1) and (2,3), (1,1) and (2,4), (1,2) and (1,4), (1,2) and (2,4), (2,1) and (1,3), (2,1) and (1,4), (2,1) and (2,3), (2,1) and (2,4), (2,2) and (1,4),(2,2) and (2,4).(5)

We want to prove that : � = ^�∙ + is true, where _� is the number of possibilities which n kings are not attacking each other, can be placed on a chess board ∙ :

: = ∙ + = possibilities : = ∙ + = possibilities

and are true We suppose that and + are true, ∈ ℕ\{ }

: _� = ^�∙ + possibilities (1)

+ : _�+ = ^�+ ∙ + possibilities (2) We want to demonstrate that + is true, ∈ ℕ\{ }

+ ∶ _�+ = ^�+ ∙ +

We know that _�+ = _�+ − _� ∙ (**) We are replacing (1) and (2) in (**) ⇒

�+ = [ ^�+ ∙ + − ^�∙ + ] ∙ ⇒

�+ = ^�∙ + − − ∙ ⇒

�+ = ^�∙ + ∙ ⇒

�+ = ^�+ ∙ + , ∈ ℕ\{ } ⇒

⇒ + � ∈ ℕ\{ } ⇒ � ∈ ℕ\{ }

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IV. Third method(11)

We know that: =4 and =12 and _�+ = ∙ _�+ − ∙ _�, ≥

⇔ _�+ − ∙ _�+ = ∙ _�+ − ∙ _�

� ∶ ��+ − ∙ �� = ��, ≥ ⇒ �+ − ∙ �+ = �+

⇒ ��+ = ∙ ��, ≥

If = ⇒ = − ∙ = − = ⇒ =

_� = ∙ _�− ⇒ _� = ∙ _�−

_�− = ∙ _�− ⎸ ∙ ⇒ ∙ _�− = ∙ _�−

.

1 2 1 2

1 2 3 4 1

2

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page = ∙ ⎸ ∙ . ^�− ⇒ ^�− ∙ = ^�− ∙

Summing these equations, we will have:

� = ^�− ∙ , where = ⇒

� = ^�− ∙ ⇒ �� = ^�+ , ≥

So , �+ − ∙ � = ^�+ ⇒ �+ = ^�+ + ∙ �

⇒ _� = ∙ _�− + ^� ⇒ _� = ∙ _�− + ^�

�− = ∙ �− + ^�− ⎸ ∙ 2¹ ⇒ ∙ �− = ∙ �− + ^�

�− = ∙ �− + ^�− ⎸ ∙ 2² ⇒ ∙ �− = ∙ �− + ^� ..

= ∙ + ⎸ ∙ ^�− ⇒ ^�− ∙ = ^�− ∙ + ^� Summing these equations, we will have:

� = ^�− ∙ + ^�∙ − ⇒ � = ^�− ∙ + − ⇒

� = ^�− ∙ + ⇒ �_� = ^�∙ � + , � ≥

Notes d’édition

‘athe : Fi st o putatio ‘athe : I du tio elatio ship ‘athe : + a d +

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O e ould oti e that the o putatio ade i pa t I do ot sho that all the ases a e des i ed this a , so e a ot e su e that the pa t I o pute all ases.

The pa t II ill e useful to get a o plete a s e of this p o le .

As a o lusio , o e e pe t that the alue of a is the sa e as the u e hi h is o puted i pa t I. This is the goal of pa t III.

This ase is ot e useful, a d does ot e plai h all the ases a e des i ed. The i e e pla atio i pa t ould e used i stead.

So e eda tio al e a k : e ha e to e a little it o e a eful : fi st hoose k i N*, the suppose pk a d p k+ .

This is useless

These p e eedi g sig s ⇒ a e useless

We should ite: the p k+ is t ue. The efo e, e ha e sho that p is t ue fo all i N*

A o lusio ould ha e ee i e. The p o le i t odu ed at the egi i g. It should e i te esti g to thi k a out o e ted uestio s.

Fo e a ple, hat happe s if e ha ge the fo of the hess oa d, a d take a s ua e? What happe s ith k ights i stead of ki gs? …

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