MathDB

1

King Arthur fights the Dragon with 3 Heads and 3 Tails

King Arthur one day had to fight the Dragon with Three Heads and Three Tails. His task became easier when he managed to find a magic sword that could, with a single blow, do one (and only one) of the following things:

\bullet

cut off a head,

\bullet

cut off two heads,

\bullet

cut a tail,

\bullet

cut off two tails. Furthermore, Fairy Morgana revealed to him the dragon's secret:

\bullet

if a head is cut off, a new one grows,

\bullet

if two heads are cut off nothing happens,

\bullet

in place of a tail, two new tails are born,

\bullet

if two tails are cut off a new head grow,

\bullet

and the dragon dies if it loses its three heads and three tails. How many hits are needed to kill the dragon?

4

1

no participant solved all problems in final

To date, in each Mathematics Olympiad Final, no participant has been able to solve all the problems, but every problem has been solved by at least one participant. Prove that in each Final, there was a participant

A

who solved a problem

P_A

and another participant

B

who solved a problem

P_B

such that

A

did not solve

P_B

and

B

did not solve

P_A

.

3

1

11...11 - 2..2 is perfect square

Proce that number

\underbrace{11...11}_{2n \,\, digits}-\underbrace{22 ... 22}_{n \,\, digits}

is, for every natural

n

, a perfect square.

1

min natural with 1994 divisors.

Determine the smallest natural number that has exactly

1994

divisors.

5

1

max angle <OPQ, circle related 1994 Portugal p5

Consider a circle

C

of center

O

and its inner point

Q

, different from

O

. Where we must place a point

P

on the circle

C

so that the angle

\angle OPQ

is the largest possible?

2

1

EF = AE + FC wanted in square ABCD, <AED=<DEF, 1994 Portugal p2

Consider in a square

[ABCD]

a point

E

on the side

AB

, different from

A

and

B

. On the side

BC

consider the point

F

such that

\angle AED = \angle DEF

. Prove that

EF = AE + FC

.

1994 Portugal MO

Subcontests

King Arthur fights the Dragon with 3 Heads and 3 Tails

no participant solved all problems in final

11...11 - 2..2 is perfect square

min natural with 1994 divisors.

max angle <OPQ, circle related 1994 Portugal p5

EF = AE + FC wanted in square ABCD, <AED=<DEF, 1994 Portugal p2

1994 Portugal MO

Subcontests

King Arthur fights the Dragon with 3 Heads and 3 Tails

no participant solved all problems in final

11...11 - 2..2 is perfect square

min natural with 1994 divisors.

max angle &lt;OPQ, circle related 1994 Portugal p5

EF = AE + FC wanted in square ABCD, &lt;AED=&lt;DEF, 1994 Portugal p2

max angle <OPQ, circle related 1994 Portugal p5

EF = AE + FC wanted in square ABCD, <AED=<DEF, 1994 Portugal p2