Please answer the following questions of RMO:
Q1. Let ABC be an acute-angled triangle and suppose angle ABC is the largest angle of the triangle. let R be its circumcentre. Suppose the circumcircle of triangle ARB cuts BC again in X. Prove that RX is perpendicular to BC.
Q2. find all real numbers x and y such that
y( x^{2} + 2 y^{2} + \frac{1}{z} \leq x(2y + 1)
Q3. prove that there does not exist any positive integer n < 2310 such that n(2310-n) is a multiple of 2310.
Q4. find all positive real numbers x, y, z suh that
2x- 2y + \frac{1}{z} = \frac{1}{2014} \\ \\ 2y- 2z + \frac{1}{x} = \frac{1}{2014} \\ \\ 2z- 2x + \frac{1}{y} = \frac{1}{2014}[/tex]
Q5. Let ABC be a triangle. Let X be on the segment BC such AB=AX. Le AX meet the circumcircle T of triangle ABC again at D. Show that the circumcentre of triangle BDX lies on T.
Q6. 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

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Answers

2015-01-04T19:14:44+05:30
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.
Solution: Extend RX to meet BC in E. We show that ∠XEC = 90◦ . Join RA, RB and BX. Observe that ∠AXB = ∠ARB = 2∠C and ∠BXR = ∠BAR = 90◦ − ∠C. Hence ∠EXC = 180◦ − 2∠C − (90◦ − ∠C) = 90◦ − ∠C. This shows that ∠CEX = 90◦ . 

2. Find all real numbers x and y such that x 2 + 2y 2 + 1 2 ≤ x(2y + 1).
 Solution: We write the inequality in the form 2x 2 + 4y 2 + 1 − 4xy − 2x ≤ 0. Thus (x 2 − 4xy + 4y 2 ) + (x 2 − 2x + 1) ≤ 0. Hence (x − 2y) 2 + (x − 1)2 ≤ 0. Since x, y are real, we know that (x − 2y) 2 ≥ 0 and (x − 1)2 ≥ 0. Hence it follows that (x − 2y) 2 = 0 and (x − 1)2 = 0. Therefore x = 1 and y = 1/2. 

3. Prove that there does not exist any positive integer n < 2310 such that n(2310 − n) is a multiple of 2310.
 Solution: Suppose there exists n such that 0 < n < 2310 and n(2310−n) = 2310k. Then n 2 = 2310(n − k). But 2310 = 2 × 3 × 5 × 7 × 11, the product of primes. Hence n − k = 2310l 2 for some l. But n < 2310 and hence n − k < 2310. Hence l = 0. This forces n = k and hence n 2 = 2310(n − k) = 0. Thus n = 0 and we have a contradiction.

4. Find all positive real numbers x, y, z such that 2x − 2y + 1 z = 1 2014 , 2y − 2z + 1 x = 1 2014 , 2z − 2x + 1 y = 1 2014 .
Solution: Adding the three equations, we get 1 x + 1 y + 1 z = 3 2014 .We can also write the equations in the form 2zx − 2zy + 1 = z 2014 , 2xy − 2xz + 1 = x 2014 , 2yz − 2yx + 1 = y 2014 . Adding these, we also get 2014 × 3 = x + y + z. Therefore 1 x + 1 y + 1 z (x + y + z) = 3 2014 × (2014 × 3) = 9. Using AM-GM inequality, we therefore obtain 9 = 1 x + 1 y + 1 z (x + y + z) ≥ 9 × (xyz) 1/3 1 xyz1/3 = 9. Hence equality holds in AM-GM inequality and we conclude x = y = z. Thus 1 x = 1 2014 which gives x = 2014. We conclude x = 2014, y = 2014, z = 2014.

5. 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 4BDX lies on Γ.
 Solution: Draw perpendicular from A to BC and extend it to meet Γ in F. We show that F is the circumcentre of 4BDX. Since AB = AX, we observe that F lies on the perpendicular bisector of BX. Join CF and CD. We observe that ∠ABX = ∠CDX and ∠AXB = ∠CXD. Hence 4ABX is similar to 4CDX. In particular 4CDX is isosceles. Moreover, ∠BCF = ∠BAF and ∠DCF = ∠DAF. Since AF is the perpendicular bisector of BX, it also bisects ∠BAX. It follows that CF bisects ∠DCX and hence F lies on the perpendicular bisector of DX. Together F is the circumcentre of 4BXD. 

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.
Solution: Observe that S(S(n)) = 2 implies that S(n) = 2, 11 or 20. Hence we have to find the number of all all 3 digit numbers abc such that a + b + c = 2, 11 and 20. In fact we can enumerate all these: a + b + c = 2: abc = 101, 110, 200; a + b + c = 11; abc = 902, 920, 290, 209, 911, 191, 119, 803, 830, 308, 380, 812, 821, 182, 128, 218, 281, 731, 713, 317, 371, 137, 173, 722, 272, 227, 740, 704, 407, 470, 650, 605, 560, 506, 641, 614, 416, 461, 164, 146, 623, 632, 362, 326, 263, 236; a + b + c = 20; abc = 992, 929, 299, 983, 938, 398, 389, 839, 893, 974, 947, 794, 749, 479, 497, 965, 956, 659, 695, 596, 569, 884, 848, 488, 875, 875, 785, 758, 578, 587, 866, 686, 668, 776, 767, 677. There are totally 85 three digit numbers having second digital sum equal to 2.
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