VIBRANT ACADEMY (India) Private Limited

Regional Mathematical Olympiad-2014 Time : 3 Hours Instructions :

December 07, 2014

• Calculators (in any form) and protractors are not allowed. • Rulers and comes 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

∠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 Hence RD ⊥r BC

B

D

C

C

Q.2

Find all real numbers x and y such that x 2 + 2 y 2 +

Sol.

x 2 + 2y 2 +

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

2k (1155 − k ) λ =⇒ 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]

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

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