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
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) = |
æ
è
|
|
æ
è
|
|
Ö3 + 1
2
|
|
ö
ø
|
a, |
æ
è
|
|
Ö6
2
|
|
ö
ø
|
a |
ö
ø
|
. |
|
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
|
|
|
= 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
or
The first equation leads to
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).
