Paper & Solutions - Career Point

RMO-2012 EXAMINATION
CAREER POINT
Regional Mathematical Olympiad-2012
Time : 3 Hours December 02, 2012
Instructions :
• Calculators (In any form) and protractors are not allowed.
• Rulers and compasses are allowed.
• Answer all the questions.
• All questions carry equal marks. Maximum marks : 102
• Answer to each question should start on a new page. Clearly indicate the question number.
1. Let ABCD be a unit square. Draw a quadrant of a circle with A as centre and B, D as end points of the arc.
Similarly, draw a quadrant of a circle with B as centre and A, C as end points of the arc. Inscribe a circle Γ
touching the arcs AC and BD both externally and also touching the side CD. Find the radius of the circle Γ .
⎛ 1 ⎞
Sol. Let centre of smaller circle is O ⎜ ,1− r⎟ with radius r
⎝ 2 ⎠
D(0,1)
⎛ 1 ⎞
M⎜
,1⎟
⎝ 2 ⎠
r
O(1/2, 1–r)
C(1,1)
⎛
⎜
1
,
⎝ 2
3 ⎞
⎟
2
⎠
1
A(0,0)
B(1,0)
Radius of quadrant of ABD is 1
∴ AO = 1 + r
AO 2 = (1 + r) 2
2
⎛ 1 ⎞
⎜ ⎟⎠ + (1 – r) 2 = (1 + r) 2
⎝ 2
1
r = 16
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RMO-2012 EXAMINATION
CAREER POINT
2. Let a, b, c be positive integers such that a divides b 5 , b divides c 5 and c divides a 5 . Prove that abc divides
(a + b + c) 31 .
Sol. (a + b + c) 31
⎡
5
b = k
⎥ ⎤
1a
⎢ 25 5
⎣b
= k1k3c⎦
⎡ c
⎢ 25
⎣c
5
= k ⎤
2b
5 ⎥
= k 2k1a
⎦
⎡
5
a = k ⎤
3c
⎢ 25 5 ⎥
⎣a
= k3k
2b
⎦
Case-I: abc will be common term when power is distributed among three terms
Case-II: when power is distributed between two terms
Let a b
Distribution of power 1 30 Then term will be
: : a . b 30 = a . b 5 . b 25
: : = a . b 5 (k 5 1 k 3 c)
6 25 → a 6 . b 25 = a . a 5 . b 25
= a . (k 3 c) (b 25 )
↓
soon
Same as for b & c
and c & a
Case-III: when power is distributed between one term
Let a 31 = a . a 5 . a 25 same as b 31 & c 31
= (a) (k 3 c) (k 5 3 k 2 b)
= (k 3 k 5 3 ) (a b c)
So (a + b + c) 31 is divisible by abc.
3. Let a and b be positive real numbers such that a + b = 1. Prove that a a b b + a b b a ≤ 1.
Sol.
By A.M. ≥ G.M.
a.a + b.b
a + b
=
2
a + b
a + b
2
≥
a
1
b a b
( a .b )
+
…..(i)
1
a
ab + ab b a
≥ ( a .b )
+ b
…..(ii)
a + b
add (i) & (ii)
(a + b) ≥ a a . b b + b a . a b
1 ≥ a a .b b + b a .a b
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RMO-2012 EXAMINATION
CAREER POINT
4. Let X = {1, 2, 3, …..., 10}. Find the number of pairs {A, B} such that A ⊆ X, B ⊆ X, A ≠ B and
Sol.
A ∩ B = {5, 7, 8}.
7 C 0 [ 7 C 1 + 7 C 2 + 7 C 3 + .......... + 7 C 7 ]
+ 7 C 1 [ 7 C 0 + ( 7 C 1 – 1) + 7 C 2 + 7 C 3 + .......... + 7 C 7 ]
+ 7 C 2 [ 7 C 0 + 7 C 1 + ( 7 C 2 – 2 C 2 ) + 7 C 3 + 7 C 4 + .......... + 7 C 7 ]
+ 7 C 3 [ 7 C 0 + 7 C 1 + 7 C 2 + ( 7 C 3 – 1) + 7 C 4 + .......... + 7 C 7 ]
+ 7 C 4 [ 7 C 0 + 7 C 1 + 7 C 2 + 7 C 3 + ( 7 C 4 – 1) + .......... + 7 C 7 ]
= 7 C 0 (2 7 – 1) + 7 C 1 (2 7 – 1) + 7 C 2 (2 7 – 1) +..........
= Σ 7 C r (2 7 – 1) = (2 7 – 1) 2 7 = 2 7 (2 7 – 1)
5. Let ABC be a triangle. Let D, E be a points on the segment BC such that BD = DE = EC. Let F be the
mid-point of AC. Let BF intersect AD in P and AE in Q respectively. Determine the ratio of the area of the
triangle APQ to that of the quadrilateral PDEQ.
Sol.
A(a)
r
P
Q
F
F r ≡
r r
(a + c)
2
∆APQ = 2
1 |PA × PQ| = 40
3
B(0,0)
P r ≡
r
⎛ c ⎞
D⎜
⎟
⎝ 3 ⎠
⎛
E⎜
⎝
r C(c
r )
2c ⎞
⎟
3 ⎠
a
r r + c
4
Q r ≡ 2 r
(a c
r +
5
)
r r
| a × c |
1 | a
r c r × |
∆ADE = |AD × AE| = 2 6
11
PQDE = 120
r r
| a × c |
So
∆APQ
9
=
PQDE
11
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RMO-2012 EXAMINATION
CAREER POINT
6. Find all positive integers n such that 3 2n + 3n 2 + 7 is a perfect square.
Sol. 3 2n + 3n 2 + 7
(a) If n is odd n = 2k + 1 n 2 is of form of 4λ + 1 Here λ ∈ I
3 4k + 2 + 3(4λ + 1) + 7
⇒ 9.81 k + 12λ + 10
(81) k ≡ 4λ + 1
9(81) k ≡ 4λ + 1
3 2n + 3n 2 + 7 ≡ 4λ + 3 and a perfect square can't be of form of 4λ + 3.
(b) If n is even n = 2k
9 2k + 12k 2 + 7
Now 9 2k < 9 2k + 12k 2 + 7 ≤ (9 k + 1) 2
where equality will hold only at k = 1
Rest it will be in between perfect square of 9 k and 9 k + 1
i.e. two consecutive numbers.
Hence n = 2 is only solution.
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