MathDB

1

\sqrt{2 - x}+\sqrt{2 + 2x} =\sqrt{(x^4 + 1)/(x^2 + 1)}+ (x + 3)/(x + 1)

x

such that

-1 < x \le 2

and

\sqrt{2 - x}+\sqrt{2 + 2x} =\sqrt{\frac{x^4 + 1}{x^2 + 1}}+ \frac{x + 3}{x + 1}.

.

7

1

perpendicular bisector of BC bisects segment AD

Let

M

be the midpoint of side

BC

of acute triangle

ABC

. The circle centered at

M

passing through

A

intersects the lines

AB

and

AC

again at

P

and

Q

, respectively. The tangents to this circle at

P

and

Q

meet at

D

. Prove that the perpendicular bisector of

BC

bisects segment

AD

.

6

1

among any k positive integers, one may select even number with sum s, n|s

Let a positive integer

n

be given. Determine, in terms of

n

, the least positive integer

k

such that among any

k

positive integers, it is always possible to select a positive even number of them having sum divisible by

n

.

3

2

x^2 + y^2 = 2, \frac{x^2}{2 - y}+\frac{y^2}{2 - x}= 2

Find all real numbers

x

and

y

such that

x^2 + y^2 = 2

\frac{x^2}{2 - y}+\frac{y^2}{2 - x}= 2.

resiprocal sums differ by less than 0.01

Let

S

be a set of

10

positive integers. Prove that one can find two disjoint subsets

A =\{a_1, ..., a_k\}

and

B = \{b_1, ... , b_k\}

of

S

with

|A| = |B|

such that the sums

x =\frac{1}{a_1}+ ... +\frac{1}{a_k}

and

y =\frac{1}{b_1}+ ... +\frac{1}{b_k}

differ by less than

0.01

, i.e.,

|x - y| < 1/100

.

5

2

5 times in round-robin tournament

A round-robin tournament is one where each team plays every other team exactly once. Five teams take part in such a tournament getting:

3

points for a win,

1

point for a draw and

0

points for a loss. At the end of the tournament the teams are ranked from first to last according to the number of points. (a) Is it possible that at the end of the tournament, each team has a different number of points, and each team except for the team ranked last has exactly two more points than the next-ranked team? (b) Is this possible if there are six teams in the tournament instead?

max m, x_n(x_n - 1)(x_n - 2) . . . (x_n - m + 1)/{m!} never multiple of 7

The sequence

x_1, x_2, x_3, . . .

is defined by

x_1 = 2022

and

x_{n+1}= 7x_n + 5

for all positive integers

n

. Determine the maximum positive integer

m

such that

\frac{x_n(x_n - 1)(x_n - 2) . . . (x_n - m + 1)}{m!}

is never a multiple of

7

for any positive integer

n

.

4

2

an empty bag and a chessboard and unlimited supply of tokens

On a table, there is an empty bag and a chessboard containing exactly one token on each square. Next to the table is a large pile that contains an unlimited supply of tokens. Using only the following types of moves what is the maximum possible number of tokens that can be in the bag?

\bullet

Type 1: Choose a non-empty square on the chessboard that is not in the rightmost column. Take a token from this square and place it, along with one token from the pile, on the square immediately to its right.

\bullet

Type 2: Choose a non-empty square on the chessboard that is not in the bottommost row. Take a token from this square and place it, along with one token from the pile, on the square immediately below it.

\bullet

Type 3: Choose two adjacent non-empty squares. Remove a token from each and put them both into the bag.

AQ = BF wanted, touchpoints of incircle of right triangle ABC

Triangle

ABC

is right-angled at

B

and has incentre

I

. Points

D

,

E

and

F

are the points where the incircle of the triangle touches the sides

BC

,

AC

and AB respectively. Lines

CI

and

EF

intersect at point

P

. Lines

DP

and

AB

intersect at point

Q

. Prove that

AQ = BF

.

2

product of the 31 numbers is 5th power, from pairs of 0-61

Is it possible to pair up the numbers

0, 1, 2, 3,... , 61

in such a way that when we sum each pair, the product of the

31

numbers we get is a perfect f ifth power?

a^2 + b^2 + c^2 = 1 , a(2b - 2a - c) >= 1/2

Find all triples

(a, b, c)

of real numbers such that

a^2 + b^2 + c^2 = 1

and

a(2b - 2a - c) \ge \frac12

.

1

2

<PCB =? ABCD rectangle 1x2, MBP equilateral. AM=//MD

ABCD

is a rectangle with side lengths

AB = CD = 1

and

BC = DA = 2

. Let

M

be the midpoint of

AD

. Point

P

lies on the opposite side of line

MB

to

A

, such that triangle

MBP

is equilateral. Find the value of

\angle PCB

.

a^2 + b = b^{2022} diophantine

Find all integers

a, b

such that

a^2 + b = b^{2022}.

2022 New Zealand MO

Subcontests

\sqrt{2 - x}+\sqrt{2 + 2x} =\sqrt{(x^4 + 1)/(x^2 + 1)}+ (x + 3)/(x + 1)

perpendicular bisector of BC bisects segment AD

among any k positive integers, one may select even number with sum s, n|s

x^2 + y^2 = 2, \frac{x^2}{2 - y}+\frac{y^2}{2 - x}= 2

resiprocal sums differ by less than 0.01

5 times in round-robin tournament

max m, x_n(x_n - 1)(x_n - 2) . . . (x_n - m + 1)/{m!} never multiple of 7

an empty bag and a chessboard and unlimited supply of tokens

AQ = BF wanted, touchpoints of incircle of right triangle ABC

product of the 31 numbers is 5th power, from pairs of 0-61

a^2 + b^2 + c^2 = 1 , a(2b - 2a - c) >= 1/2

<PCB =? ABCD rectangle 1x2, MBP equilateral. AM=//MD

a^2 + b = b^{2022} diophantine

2022 New Zealand MO

Subcontests

\sqrt{2 - x}+\sqrt{2 + 2x} =\sqrt{(x^4 + 1)/(x^2 + 1)}+ (x + 3)/(x + 1)

perpendicular bisector of BC bisects segment AD

among any k positive integers, one may select even number with sum s, n|s

x^2 + y^2 = 2, \frac{x^2}{2 - y}+\frac{y^2}{2 - x}= 2

resiprocal sums differ by less than 0.01

5 times in round-robin tournament

max m, x_n(x_n - 1)(x_n - 2) . . . (x_n - m + 1)/{m!} never multiple of 7

an empty bag and a chessboard and unlimited supply of tokens

AQ = BF wanted, touchpoints of incircle of right triangle ABC

product of the 31 numbers is 5th power, from pairs of 0-61

a^2 + b^2 + c^2 = 1 , a(2b - 2a - c) &gt;= 1/2

&lt;PCB =? ABCD rectangle 1x2, MBP equilateral. AM=//MD

a^2 + b = b^{2022} diophantine

a^2 + b^2 + c^2 = 1 , a(2b - 2a - c) >= 1/2

<PCB =? ABCD rectangle 1x2, MBP equilateral. AM=//MD