Download Regional Mathematical Olympiad 1995 Sol

Q. 1 RMO 1995
BC. KL = BK. CL

BC CL

BK KL
since AL bisects
CL AC

KL AK

BC AC

BK AK
This equation holds if f AB bisects the exterior an angle of KAC There
fore
BAK + KAL
so
=
BAL
=
1
(1800)
2
=
900
AL ┴ AB.
Q. 2 1995
Case I
n = 4k + 1
n = 4k + 1 = (4k+1) x (1)2k x (-1)2k
= (4k + 1) +
1  1  ..  1  
2k times
+
[( 1)  (1)  ..(1)]
2ktimes
case II =
4 l here there are two cases
(a) l is even with l > 2
and
(b) l is add with l > 3
sub case (a)
n = 4l, l is even
consider integers w and v such that,
n = 4l = 2l x 2 x (1)w x (-1)v
= 2l + 2 +
(1  1  ...  1) [( 1)  (1)  (1)]
+
vtimes
wtimes
Now by the definition of good integer we have
2 + w + v = 4l
(there are 2+w+v factors)

w + v = 4l – 2
………(i)
and again since 4l = 2l + 2 + w – v
we get w – v = 2l – 2
………(ii)
w = 3l – 2
form (i) & (2) we get
v=l
case (b)
l is odd with l > 3
choose w and v such that
n = 4l = (2l) x (-2) x (1)w x (-1)v
= 2l + (-2) +
(1  1  1  .  1) [( 1)  (1)....( 1)]
+
4times
vtimes
Again since There are w+u+2 factors
We have
w + v + 2 = 4l
w + v = 4l – 2
and
4l = 2l – 2 two – v
……….(iii)
[by definition of good integers]
 w – v = 2l + 2
………..(iv)
since l > 3 and l in add
l > 2 >1
Now n = 4l = 2l x (-2) x (1)3 x (-1)1-2
= 2l + (-2) +
(1  1  ....  1) [( 1)  (1)  ....(1)]
+
3ltimes
l  2times
= 2l – 2 + 3l
= 4l
Q. 3 RMO 1995
Among 18 consecutive integers there are two numbers which in
divisible by I.
The sum of the digits of these two numbers must be 9, 18 & 27.
If the sum of the digit in 9, then the number is divisible by the sum
of the digits, so there is nothing to prove.
If, the sum of the digits is 27, then the three digit number should be
999 = 9 x 11 = 9 x 3 x 37 and hence the result.
Let both the numbers have 18 the some of digit let those numbers be
a and b. If a is odd and sum of its digit is 18,
However, the other number b is also divisible by 9 and b should be
a + 9, so a is odd and hence, b is even and sum of its digits is, and
hence, b is an even number divisible by 9 and b is divisible by 18.
For example, 6 + 7 is divisible by 9 and 6 + 5 + 9 = 684 is divisible
by is.
Q. 4 Assume its contrary n be an integer root
x2 + 7x – 14 (a2 + 1) = 0
of
then n2 + 7n – 14 (a2 + 1) = 0

n2 = -7 (n + 2a2 + 2)

7
7
where 7 in prime no.
2 and
n
n
Let n = 7n1
Then equation (i) can be written as
49 n12 + 49 n1 = 14 (a2 + 1)

so
7 n12 + 7 n1 = 2 (a2 + 1)
7
(a2+1)  a2 + 1 = 0 (mod 7)
2
or a2 = 6 (mod 7)
then a = 0, 12, 3, 4, 5, 6 (mod 7)
a2 = 0, 1, 4, 2, 2, 4, 1 (mod 2)
respectively
Hence
a2 ≠ 6 (mod 7) for any integer
Therefore, these exists no interval
So of for the gives
Quadratic equation.
Q. 5 RMO 1995
Let a>, b>, c so that
c2 – a2 = a2 – c2 is the maximum of
a2 – b2, b2 – c2 and c2 – a2
It is enough to prove that
a2 + b2 + c2 – 3 (a2 – c2) > 0
Now
a2 + b2 + c2 – 3 (a2 – c2)
> a2 + (a - c) 2 + c2 – 3 (a2 – c2)
(as b> a-c by triangle inoculating)
> 2a2 + 2c2 – 2ac – 3 a2 + 3 c2
> (2 - 3 ) a2 + (2 + 3 ) c2 – 2ac
But
( 3 - 1)2 = 2 (2 - 3 )
and
( 3 + 1)2 = 2 (2 - 3 )
so
a2 + b2 + c2 - 3 = (a2 – c2)
>
[( 3  1)a]  4ac  [( 3  1)c]2
2
=
1
2
>
0

hence the result
 
3 1 a 

3 1 c
2
Q. 6 Let us pick a vertex and draw a line passing through 0 and the
vertex. Label that vertex as 1. Label the verity in clockwise as 2, 3
…11. For anticlockwise direction, label it as -2, -3, … -11.
Join 1 and 11 and let it be one of the side of the triangle. Now we
join it with any other vertices from 2 to 10, we will have triangle
without centre in them. So there are a total of I ways. Now we have
join 1 with 10 and similarly there are 8 ways to form a triangle
without centre in them. So we proceed and get the number of triangle
to be
1 + 2 + 3 + … + 9 = 45
Symmetrically we can use vertex 2, 3… 11, -2, -3… -11 also hence
it is 45 x 21 = 945 triangle without the centre.
Hence those with centre is
=
21c3 – 945
=
385.
Let x, y, z E N be the ‘are count side lengths of the triangle so for
example it triangle is A3 A9 A15, the ‘are count’ side triangle are 6, 6,
9 because it takes 6 arcs from A3 to A9, to A15 and it takes 9 arcs
from A15 to A3. Notice that x + y + z = 2.
From that 0 is inside triangle iff it is acute, therefore x, y, z, < 11
amume
0 < x < y < z < 11
we have few configuration
7, 7, 7,
equilateral triangle has symmetric c sides so we have 21/3 = 7
triangles here containg 0.
6, 7, 8
5, 6, 10
5, 7, 9
4, 7, 10
4, 8, 9
3, 8, 10
2, 9, 10
These are scalene triangles and each configuration has 2 different
rotational combinations i.e. 6, 78 and 687, as in the preview case we
can rotate through her Ai and we have in total 21 x 2 x 7 = 29
So total number of triangle containing 0 is 294 + 84 + 7 = 385
Q. 7 Let f (x)
= x2 sinx + x cos x + x2 + ½
2

1 cos x  1 

  0
= (1 + sinx)   x 
2
1

sin
x

 4

This holds because 1 + sinx > 0
for any real x ≠ -

+2n
2
otherwise it in zero
and
2
1 cos x  1 1

x 
  
2 1  sin x  4 4

If
 
 1
 2n    0
 2
 2
x = 
Then f > 0 for all reals x.
Method II
t = tan
x
on substitute of
2
(1 + sinx) x2 + xcos +
=
1
2
1 t2  1
2t  2



x

x
1 

2
2 
 1 t 
1 t  2
=
1  t 2 x2 
=
1  t 2  x 
1 t
2



1  t 2 


2
2

 1t 2  1 1
1 2
1 t2

x




2
2
2
2
(1  t )
2 4 (1  t )(1  t ) 
 2(1  t ) 

2
1 t2 
1
 (1  t ) 2  0
2
2(1  t ) 
4