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Solutions.

290.
The School of Architecture in the Olymon University proposed two projects for the new Housing Campus of the University. In each project, the campus is designed to have several identical dormitory buildings, with the same number of one-bedroom apartments in each building. In the first project, there are 12096 apartments in total. There are eight more buildings in the second project than in the first, and each building has more apartments, which raises the total of apartments in the project to 23625. How many buildings does the second project require?
Solution. Let the number of buildings in the first project be n. Then there must be 12096/n apartments in each of them. The number of buildings in the second project is n +8 with 23625/(n+8) apartments in each of them. Since the number of apartments is an integer, n + 8, and so n, is odd. Furthermore, 12096 = 26 ·33 ·7 and 23625 = 33 ·53 ·7. Since n is an odd factor of 12096, n must take one of the values 1, 3, 7, 9, 21, 27, 63 or 189. Since n + 8 must be a factor of 23625, the only possible values for n are 1, 7 or 27. Taking into account that the number of apartments in each building of the second project is more than the number of apartments in each building of the first project, n must satisfy the inequality
 12096 n < 23625 n+8 ,
which is equivalent to n > 512/61. Thus, n ³ 9. Therefore, n = 27 and n + 8 = 35. The second project requires 35 buildings.

291.
The n-sided polygon A1, A2, ¼, An (n ³ 4) has the following property: The diagonals from each of its vertices divide the respective angle of the polygon into n-2 equal angles. Find all natural numbers n for which this implies that the polygon A1 A2 ¼An is regular.
Solution. Let the measures of the angles of the polygon at each vertex Ai be ai for 1 £ i £ n. When n = 4, the polygon need not be regular. Any nonsquare rhombus has the property.

Let n exceed 4. Consider triangle A1A2An. We have that

 a1 + a2 n-2 + an n-2 = 180° .
Since the sum of the exterior angles of an n-gon is a1 + a2 + ¼+ an = (n-2)180°, we find that
 (n-3)a1 = a3 + a4 + ¼+ an-1 .
Suppose, wolog, that a1 is a largest angle in the polygon. Then
 (n-3)a1 = a3 + a4 + ¼+ an-1 £ (n-3)a1
with equality if and only if a1 = a3 = ¼ = an-1. Suppose, if possible that one of a2 and an is strictly less than a1. We have an inequality for a4 analogous to the one given for a1, and find that
 (n-3)a4 = a1 + a2 + a6 + ¼+ an < (n-3)a1 = (n-3)a4 ,
a contradiction. Hence, all the angles ai must be equal and the polygon regular.

292.
1200 different points are randomly chosen on the circumference of a circle with centre O. Prove that it is possible to find two points on the circumference, M and N, so that:
· M and N are different from the chosen 1200 points;

· ÐMON = 30°;

· there are exactly 100 of the 1200 points inside the angle MON.

Solution. The existence of the points M and N will be evident when we prove that there is a central angle of 30° which contains exactly 100 of the given points. Construct six diameters of the circle for which none of their endpoints coincide with any of the given points and they divide its circumference into twelve equal arcs of 30°; such a construction is always possible. If one of the angles contains exactly 100 points, then we have accomplished our task. Assume none of the angles contains 100 points. Then some contain more and others, less. Wolog, we can choose adjacent angles for which the first contains d1 > 100 points and the second d2 < 100 points. Define a function d which represents the number of points inside a rotating angle with respect to its position, and imagine this rotating angle moves from the position of the first of the adjacent angles to the second. As the angle rotates, one of the following occurs: (i) a new point enters the rotating angle; (ii) a point leaves the rotating angle; (iii) no point leaves or enters; (iv) one point leaves while another enters. Thus, the value of d changes by 1 at a time from d1 to d2, as so at some point must take the value 100. The desired result follows.

293.
Two players, Amanda and Brenda, play the following game: Given a number n, Amanda writes n different natural numbers. Then, Brenda is allowed to erase several (including none, but not all) of them, and to write either + or - in front of each of the remaining numbers, making them positive or negative, respectively, Then they calculate their sum. Brenda wins the game is the sum is a multiple of 2004. Otherwise the winner is Amanda. Determine which one of them has a winning strategy, for the different choices of n. Indicate your reasoning and describe the strategy.
Solution. Amanda has a winning strategy, if n £ 10. She writes the numbers 1, 2, 22, 23, ¼, 2n-1. Recall that, for any natural number k, 2k > 1 + 2 + ¼+ 2k-1. Since 210 = 1024, it is clear that, regardless of Brenda's choice, the result sum but lies between -1023 and 1023, inclusive, and it is not 0, since its sign coincides with the sign of the largest participating number. Hence, the sum cannot be a multiple of 2004.

Let n ³ 11. Then it is Brenda that has a winning strategy. The set C of numbers chosen by Amanda has 2n - 1 > 2003 different non-empty subsets. By the Pigeonhole Principle, two of these sums must leave the same remainder upon division by 2004. Let A and B be two sets with the same remainder. Brenda assigns a positive sign to all numbers that lie in A but not in B; she assigns a negative sign to all numbers that lie in B but not in A; she erases all the remaining numbers of C. The sum of the numbers remaining is equal to the difference of the sum of the numbers in A and B, which is divisible by 2004. Thus, Brenda wins.

Therefore, Amanda has a winning strategy when n £ 10 and Brenda has a winning strategy when n ³ 11.

294.
The number N = 10101¼0101 is written using n+1 ones and n zeros. What is the least possible value of n for which the number N is a multiple of 9999?
Solution. Observe that N = 1 + 102 + ¼+ 102n, 9999 = 32 ·11 ·101 and 102 º 1 modulo 9 and modulo 11. Modulo 9 or modulo 11, N º n + 1, so that N is a multiple of 99 if and only if n º 98 (mod 99). N is a multiple of 101 if and only if n is odd. Hence the smallest value of n for which N is a multiple of 9999 is 197.

295.
In a triangle ABC, the angle bisectors AM and CK (with M and K on BC and AB respectively) intersect at the point O. It is known that
 |AO |¸|OM | = Ö6 + Ö3 + 1 2
and
 |CO |¸|OK | = Ö2 Ö3 - 1 .
Find the measures of the angles in triangle ABC.
Solution. Let AB = c, BC = a, AC = b, CM = x, AK = y, ÐABC = b, ÐACB = g and ÐCAB = a. In triangle AMC, CO is an angle bisector, whence
 AC : CM = AO : OM Û b x = Ö6 + Ö3 + 1 2 .
(1)
In triangle ABC, AM is an angle bisector, whence
 AB:AC = BM:CM Û c b = a - x x Û x a = b c + b .
(2)
Multiplying (1) and (2), we get
 b a = (Ö6 + Ö3 + 1)b 2(c+b) Ûa = 2(b+c) Ö6 + Ö3 + 1 .
(3)

Similarly, in triangle AKC, AO is an angle bisector. Hence

 CO:OK = AC:AK Û Ö3 - 1 Ö2 = y b .
(4)
In triangle ABC, CK is an angle bisector. Hence
 BC : AC = BK : AK Û a b = c - y y Û a+b b = c y .
(5)
Multiplying (4) and (5), we get
 c b = (a+b)(Ö3 - 1) b Ö2 Û c = (a + b)(Ö3 - 1) Ö2 .
(6)

Solve (3) and (6) to get b and c in terms of a. We find that

 (b, c) = æè æè Ö3 + 1 2 öø a, æè Ö6 2 öø a öø .
From the Law of Cosines for triangle ABC,
 cosg = a2 + b2 - c2 2ab = 1 2 Û g = 60° .
From the Law of Sines, we have that
 sina sing = a c Ûsina = 1 Ö2 Û a = 45° .
The remaining angle b = 75°.

296.
Solve the equation
 5 sinx + 5 2 sinx - 5 = 2 sin2 x + 1 2 sin2 x .
Solution 1. [G. Siu; T. Liu] Let u = sinx. For the equation to be meaningful, we require that u ¹ 0 (mod p). The equation is equivalent to
 0
 = 4u4 - 10u3 + 10u2 - 5u + 1 = (u - 1)(4u3 - 6u2 + 4u - 1)
 = (u - 1)[u4 - (1 - u)4] = (u - 1)[u2 - (u - 1)2][u2 + (u - 1)]2
 = u(u-1)(2u - 1)[u2 + (u-1)2] .
We must have that u = 1 or u = 1/2, whence x º p/6, p/2, 5p/6 (mod 2p).

Solution 2. We exclude x º 0 (mod p). Let y = sinx + (2 sinx)-1. The given equation is equivalent to

 0 = 2y2 - 5y + 3 = (2y - 3)(y - 1) .
Thus
 sinx + 1 2 sinx = 1
or
 sinx + 1 2 sinx = 3 2 .
The first equation leads to
 0 = sin2 x + (sinx - 1)2
with no real solutions, while the second leads to
 0 = 2 sin2 x - 3 sinx + 1 = (2sinx - 1)(sinx - 1) ,
whence it follows that x º p/6, p/2, 5p/6 (mod 2p).
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