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RMO-2014 EXAMINATION
CAREER POINT
Regional Mathematical Olympiad-2014
Time : 3 Hours
Instructions :
December 07, 2014
• Calculators (in any form) and protractors are not allowed.
• Rulers and compasses are allowed.
• Answer all the questions.
• All question carry equal marks. Maximum marks : 102
• Answer to each questions should start on a new page. Clearly indicate the question number.
Q.1
Let ABC be an acute-angled triangle and suppose ∠ABC is the largest angle of the triangle. Let R be its
circumcentre. Suppose the circumcircle of triangle ARB cuts AC again in X. Prove that RX is perpendicular
to BC.
Sol.
Extend XR produced to cut BC at D .
A
Now consider various angles in the figure.
θ
Let ∠ABR = θ = ∠BAR
X
∠RAX = A – θ
R
θ
∠ARB = 180° – 2θ
∠AXR = 180° – θ
D
B
C
∠ARX = 180° – (∠RAX + ∠RXA)
= 180° – (A – θ + 180° – θ)
= 2θ – A
A
∠BRD = 180° – (∠ARB + ∠ARX)
= 180° – (180° – 2θ + 2θ – A)
X
180°–θ
=A
θ
∠CXR = 180° – (∠AXR) = 180° – (180° – θ) = θ
R
B
∠ACR = ∠CAR = A – θ
∠CRD = ∠CXR + ∠XCR
=θ+A–θ =A
R
Hence, in ΔBRC, (Note, ΔBRC is issosceles)
AA
∠BRD = ∠DRC = A
RD is angle bisector of ∠BRC
B
D
C
Hence RD ⊥r BC
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RMO-2014 EXAMINATION
Q.2
Find all real numbers x and y such that x 2 + 2 y 2 +
Sol.
x 2 + 2y 2 +
CAREER POINT
1
≤ x (2 y + 1) .
2
1
≤ 2xy + x
2
⇒ 2x2 + 4y2 – 4xy – 2x + 1 ≤ 0
⇒ x2 –4xy + 4y2 + x2 – 2x + 1 ≤ 0
⇒ (x – 2y)2 + (x – 1)2 ≤ 0
⇒ x – 2y = 0 and x – 1 = 0
⇒ x = 1, y =
1
2
Q.3
Prove that there does not exist any positive integer n < 2310 such that n(2310 – n) is a multiple of 2310.
Sol.
n < 2310
n(2310 – n) = 2310λ
'2310' has factors 2, 3, 5, 7, 11
all are prime numbers.
clearly, n ≠ odd
Now
if n = even = 2k
2k(2310 – 2k) = 2310 λ
4k(1155 – k) = 2310 λ
⇒ 2k (1155 – k) = 1155 λ
⇒ λ=
2k (1155 − k )
3 × 5 × 7 × 11
Since λ ∈ 'I'
∴ k must be multiple of '3' ⇒ k = 3μ1
& simililarly,
μ1 = 5μ2
μ2 = 7μ3
μ3 = 11μ4
∴ k = 3.5.7.11 μ4
in that case k > 1155
Which is not allowed.
[hence proved]
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RMO-2014 EXAMINATION
CAREER POINT
Q.4
Find all positive real numbers x, y, z, such that –
1
1
1
1
1
1
2x − 2 y + =
, 2 y − 2z + =
, 2z − 2 x + =
.
z 2014
x 2014
y 2014
Sol.
2x − 2 y +
1
1
=
z 2014
1
1
2 y − 2z + =
x 2014
1
1
2z − 2 x + =
y 2014
adding (1), (2) & (3)
1 1 1
3
+ + =
x y z 2014
...(1)
...(2)
...(3)
...(4)
Now,
z
2014
x
2xy − 2xz + 1 =
2014
y
2 yz − 2 xy + 1 =
2014
2xz − 2 yz + 1 =
...(5)
...(6)
...(7)
adding (5), (6) & (7)
x+y+z
=3
2014
x + y + z = 3 × 2014
...(8)
Now,
x+y+z
A.M. =
= 2014
3
3
H.M. =
= 2014
1 1 1
+ +
x y z
as A.M. = H.M. ⇒ x = y = z = 2014
Let ABC be a triangle. Let X be on the segment BC such that AB = AX. Let AX meet the circumcircle Γ of
triangle ABC again at D. Show that the circumcentre of ΔBDX lies on Γ .
Draw AP⊥BX
Sol.
A
AP ⊥r bisector of BX ⇒ BP = PX.
∠BAP = ∠PAD
α α
Γ
∴ BP= PD (equal chord subtend equal angle at same segment)
∴ In ΔBDX,
BP = XP = DP
X
B
C
∴ 'P' is circumcentre of ΔBDX which is lying on circumcircle of
ΔABC
D
P
Q.5
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RMO-2014 EXAMINATION
CAREER POINT
Q.6
For any natural number n, let S(n) denote the sum of the digits of n. Find the number of all 3-digit numbers n
such that S(S(n)) = 2.
Sol.
Let n = abc
where n = a×100 + b × 10 + c a three digit number
a + b + c = S(n); Here note that S(n) ≤ 27
Since S(S(n)) = 2,
It means sum of digits of S(n) is 2
Now, S(n) can be, S(n) = 2, 11, 20 only
Now
Case-1
a + b + c = 2; possible cases are {0, 1, 1} gives 2 number and {2,0, 0} gives 1 number (ex : 200)
Total number in case (1) are 3
Case-2
a + b + c = 11
If a, b, c are non zero
then
a + b + c = 11 {a ≥ 1, b ≥1, c ≥1}
⇒
a + b + c = 8 {give one to each}
⇒
number of solutions =
8+ 2
C 2 = 10 C 2 = 45
If b & c one can zero a + b + c = 11
ordered triplet of
{(a,b,c)} ≡ {(2,0,9); (3,0,9); (4,0,8); ....(9,0,2)}
Total 8 triplets.
Each triplet has 2 arrangements
∴ Total = 8 × 2 = 16 numbers
Total number in case (2) are 45 + 16 = 61
Case-3
a + b + c = 20
a + b + c = 20
2 9 9 ⎯→
3 9 8 ⎯→
4 9 7 ⎯→
5 9 6 ⎯→
4 8 8 ⎯→
5 7 8 ⎯→
6 6 8 ⎯→
7 6 7 ⎯→
(No. of ways)
3! / 2! = 3 ways
3! = 6 ways
6 ways
6 ways
3 ways
6 ways
3 ways
3 ways
Total in case (3) = (3 × 4 + 6 × 4) = 36 ways
Total numbers = 3 + 61 + 36 = 100
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-3040000
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