Give me question paper for Regional Olympiad Mathematics examination in a PDF file format ?

Here I am giving you question paper for Regional Olympiad Mathematics examination......

1. Let ABC be an acute angled triangle. The circle _ with BC as diameter intersects AB and
AC again at P and Q, respectively. Determine \BAC given that the orthocenter of triangle
APQ lies on _.

Solution. Let K denote the orthocenter of triangle APQ. Since triangles ABC and AQP
are similar it follows that K lies in the interior of triangle APQ.
Note that \KPA = \KQA = 90__\A. Since BPKQ is a cyclic quadrilateral it follows that
\BQK = 180_ _\BPK = 90_ _\A, while on the other hand \BQK = \BQA_\KQA =
\A since BQ is perpendicular to AC. This shows that 90_ _ \A = \A, so \A = 45_.

2. Let f(x) = x3 + ax2 + bx + c and g(x) = x3 + bx2 + cx + a, where a; b; c are integers with
c 6= 0. Suppose that the following conditions hold:
(a) f(1) = 0;
(b) the roots of g(x) are squares of the roots of f(x).
Find the value of a2013 + b2013 + c2013.

Solution. Note that g(1) = f(1) = 0, so 1 is a root of both f(x) and g(x). Let p and q be the
other two roots of f(x), so p2 and q2 are the other two roots of g(x). We then get pq = _c and
p2q2 = _a, so a = _c2. Also, (_a)2 = (p+q+1)2 = p2+q2+1+2(pq+p+q) = _b+2b = b.
Therefore b = c4. Since f(1) = 0 we therefore get 1 + c _ c2 + c4 = 0. Factorising, we
get (c + 1)(c3 _ c2 + 1) = 0. Note that c3 _ c2 + 1 = 0 has no integer root and hence
c = _1; b = 1; a = _1. Therefore a2013 + b2013 + c2013 = _1.

3. Find all primes p and q such that p divides q2 _ 4 and q divides p2 _ 1.
Solution. Suppose that p _ q. Since q divides (p _ 1)(p + 1) and q > p _ 1 it follows that q
divides p + 1 and hence q = p + 1. Therefore p = 2 and q = 3.
On the other hand, if p > q then p divides (q _ 2)(q + 2) implies that p divides q + 2 or
q _ 2 = 0. This gives either p = q + 2 or q = 2. In the former case it follows that that q
divides (q+2)2_1, so q divides 3. This gives the solutions p > 2; q = 2 and (p; q) = (5; 3).

4. Find the number of 10-tuples (a1; a2; : : : ; a10) of integers such that ja1j _ 1 and
a2
1 + a2
2 + a2
3 + _ _ _ + a2
10 _ a1a2 _ a2a3 _ a3a4 _ _ _ _ _ a9a10 _ a10a1 = 2 :

Solution. Let a11 = a1. Multiplying the given equation by 2 we get
(a1 _ a2)2 + (a2 _ a3)2 + _ _ _ (a10 _ a1)2 = 4 :
Note that if ai _ ai+1 = _2 for some i = 1; : : : ; 10, then aj _ aj+1 = 0 for all j 6= i which
contradicts the equality P10
i=1(ai _ ai+1) = 0. Therefore ai _ ai+1 = 1 for exactly two values
of i in f1; 2; : : : ; 10g, ai _ ai+1 = _1 for two other values of i and ai _ ai+1 = 0 for all other
values of i. There are _10
2 __ _8
2_ = 45 _ 28 possible ways of choosing these values. Note
that a1 = _1; 0 or 1, so in total there are 3 _ 45 _ 28 possible integer solutions to the given
equation.
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Paper 1 Regional Mathematical Olympiad 2013 December 1, 2013

5. Let ABC be a triangle with \A = 90_ and AB = AC. Let D and E be points on the segment
BC such that BD : DE : EC = 3 : 5 : 4. Prove that \DAE = 45_.
Solution. Rotating the con_guraiton about A by 90_, the point B goes to the point C. Let
P denote the image of the point D under this rotation. Then CP = BD and \ACP =
\ABC = 45_, so ECP is a right-angled triangle with CE : CP = 4 : 3. Hence PE = ED.
It follows that ADEP is a kite with AP = AD and PE = ED. Therefore AE is the angular
bisector of \PAD. This implies that \DAE = \PAD=2 = 45_.

6. Suppose that m and n are integers such that both the quadratic equations x2 + mx _ n = 0
and x2 _ mx + n = 0 have integer roots. Prove that n is divisible by 6.