Prize

class they are in, ... The Editor, Mathematics Medley, clo Department of Mathematics National University of Singapore ... Composite solution by Eric F...

Problem

J ~

Let [x] denote the largest integer which does not exceed

x. For example [3.21] define a"

=n +

= 3.

n,

For each positive integer

[19; 7]. Find the smallest number in

Prize

One \$ 700 book voucher

the sequence Ia,, a2, a1, .. .1.

CONTEST 1. Pri:es in the form of book Fouchers will be awarded to the first receiFed best solution( s) submitted by secondary school or junior college students in Singapore for each of these problems. 2. To qualify, secondary school or junior college students must include their full name, home address, telephone number, the name of their school and the class they are in, together with their solutions. 3.Solutions should be typed and sent to:

There are 2998 points inside a circle which has area 1 unit. Prove that it is possible to choose three points among them such that the triangle formed by using these three points as vertices has an area less than

Prize One \$ 700 book voucher

The Editor, Mathematics Medley, clo Department of Mathematics National University of Singapore Kent Ridge. Singapore 119260

and should arril'e before 31 January 1998. 4. The Editor's decision will be final and no correspondence will be entertained.

~98

1

.

Problem

4

Solutions to the problems in Volume 24 No. 1 March 1997

Problem 1 Let a and b be two non-zero integers such that Ia I .,:; 400 and I bl .,:; 399. Prove that la\1'3 + b..fBI > 1/1997.

Solution to Problem 1 by the Editor

SOlUTIONS

Since a and b are non-zero integers we have a\1'3 ± b..JB ~ 0. Therefore l(a\1'3 + b..fB)(a\1'3- b..fS)I is a positive integer and hence 13a2 Therefore la\1'3 + b..fBI

~

la\1'3

-

= l3a 2

-

8b 2 1

81J21 ~ 1.

~ b..JSI

~

> 21al

31bl > 800

~

1197

Editor's note: No solutions from students were received for this problem. Problem 2 In the figure ABCO is a square. Given that there is a point P on the same plane such that PA area of ABCD when

= 3,

PB

= 7 and

PO

= 5.

Find the

(i) P is inside ABCD; (ii) P is outside ABCO.

Solution to Problem 2 Composite solution by Eric Fang, RJ(, Class 2S01 C; Woo Sin Ai, Nanyang junior College, Class 2CT30 and Woo Liyi, Woodland Secondary School, Class 4C. The answers are (i) 58 and (ii) 16. Introduce co-ordinates so that 8 is at (0, 0), A is at (0, a), C is at (a, 0) and 0 is at (a, a) where

a is the length of the square.

Let (x, y) be the co-ordinates of P. Then we have the system of equations

yf = 25

2

(a - x) + (a 2

x + (a - y)

1

2

x +/

2

=9

= 49

and from the first two of these equations we find that x

= LJ..2.. and

2a Substituting these into the third equation we obtain the equation

a4

-

74a 2 + 928

Solving this equation we find that When a2 = 58 we have (x, y)

=

2

y

=a

+ 40 .

2a

= 0.

a = 58 or a2 = 16. 2

Wa ,:a )

which shows that P is inside the square and when a2 = 16 we have

(x, y) = (0, 7) which shows that P is outside the square.

Editor's note: The \$100 prize money was shared as follows: Eric Fang received \$50, Woo Liyi and Woo Sin Ai each received \$25 .

..w Mathematical

....,.

EDLEY

SEPTEMBER 1997

.. .... ....