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.