Solutions

269.

Prove that the number
N = 2 ×4 ×6 ×¼×2000 ×2002+ 1 ×3 ×5 ×¼×1999 ×2001 

is divisible by 2003.
Solution 1. We will start with more general observations.
Let k be a natural number, A = 2 ×4 ×6 ×
¼×(2k), B = 1 ×3 ×5 ×
¼×(2k
1), C = 2k + 1 and M = A + B. Since 1 = C
 2k,
3 = C
 (2k
 2) and so on,
B = (C
 2k)(C
 (2k
 2))
¼(C
 2). Upon expansion, we find
that the only term in the right side that does not contain C is
(
1)
^{k}×2 ×4 ×
¼×(2k). Thus
M = C × natural number + (1 + (1)^{k})×A , 

so that, when k is odd (for example, when k = 1001), M
is divisible by C. The result follows.
Solution 2. [T. Yue] Modulo 2003,


 
 
º (2001) ×(1999) ×(1997) ×¼×3 ×1 
 
 
= (2001 ×1999 ×1997 ×¼×3 ×1) . 

Therefore, N
º 0 (mod 2003),
i.e., N is divisible
by 2003.

270.

A straight line cuts an acute triangle into two
parts (not necessarily triangles). In the same way, two other lines
cut each of these two parts into two parts. These steps repeat
until all the parts are triangles. Is it possible for all the
resulting triangle to be obtuse? (Provide reasoning to support
your answer.)
Solution 1. It is clear that if in the final step there are
k cuts, made as required, they form k + 1 triangles. Assume,
if possible, that all of these triangles be obtuse.
Note the total number of acute or right angles in the configuration
after each cut. When the cutting line intersects an existing
side of a triangle, it forms two new angles with a sum of 180
^{°},
so that at least one of them is acute or right. When the cutting line
passes through a vertex of a triangle, it forms two new angles, dividing
the existing angle (smaller than 180
^{°}) into smaller angles,
so that there is one more acute or right angle than
before. Hence at each step,
the total number of acute and right angles in the configuration
increases at least by 2. Starting from a configuration with three
such angles, after k steps, we get at least 2k + 3 acute
or right angles. On the other hand, in k + 1 obtuse triangles,
there must be exactly 2(k + 1) nonobtuse angles. This contradicts
our assumption, so that the answer to the question of the problem is
"no".
Solution 2. Suppose that there were a way to cut the given
triangle into t obtuse triangles. According to the required
procedure of cutting, if two triangles with a common vertex appear
after one cut, then they will lie on the same side of the plane
with respect to another line segment (say, a side of the triangle
or a previous cut). Denote by n the number of points that are
vertices of the obtuse triangles but not vertices of the given
triangle. On the one hand, the sum of the interior angles in
all the triangles is 180t^{°}. On the other hand, for each
of the n points, the sum of all triangular angles at a vertex
there is
180^{°}. So the sum of all the interior angles of the
triangles will be (180n + 180)^{°} (we must add the sum of
the angles of the original triangle). Hence t = n + 1. However,
only the n interior vertices can be vertices of an obtuse angle,
and each of them can be the vertex of at most one obtuse angle.
Hence t £ n, yielding a contradiction. Thus, it is impossible
to cut the original triangle into obtuse triangles only.

271.

Let x, y, z be natural numbers, such that the
number
is rational.
Prove that


(a) xz = y^{2};


(b) when y ¹ 1, the numbers x^{2} + y^{2} + z^{2}
and x^{2} + 4z^{2} are composite.
Solution. (a) Since the given number is rational, it can be
represented as a reduced fraction p/q, where p and q
¹ 0
are two coprime integers. This yields
xq  yp = (yq  zp)  Ö

2003

. 

Since the left side is rational, the right must be as well.
Since
Ö{2003} is irrational, both sides must vanish.
Thus xq
 yp = yq
 zp = 0, whence x/y = y/z = p/q, so that
xz = y
^{2}.
(b) Let M = x^{2} + y^{2} + z^{2} and N = x^{2} + 4z^{2}. We will prove
that M and N are both composite, provided that y ¹ 1.
Since xz  y^{2},
M = x^{2} + y^{2} + z^{2} = x^{2} + 2xz + z^{2}  y^{2} = (x + z)^{2}  y^{2} = (x + z  y)(x + z + y) . 

For M to be composite, the smaller factor, x + z
 y
must differ from 1. (It cannot equal
1. Why?) Since
y is a natural number distinct from 1, y > 1. As xz = y
^{2},
at least one of x and z is not less than y. Say that x
³ y.
If x = y, then z = y and x + z
 y = y > 1; if x > y,
then x + z
 y
³ z > 1. Thus in all possible cases,
x + y
 z > 1 and M is the product of two natural numbers
exceeding 1.
Similarly,
N = x^{2} + 4z^{2} = x^{2} + 4xz + 4z^{2}  4y^{2} = (x + 2z)^{2}  (2y)^{2} = (x + 2z  2y)(x + 2z + 2y) . 

To prove that N is composite, it suffices to show that the
smaller factor x + 2z
 2y exceeds 1. (Why cannot
this factor equal
1?) We prove this by contradiction.
Suppose, if possible, that x + 2z
 2y = 1. Then
x + 2z = 2y + 1, whence
x^{2} + 4xz + 4z^{2} = 4y^{2} + 4y + 1Û x^{2} + 4z^{2} = 4y + 1 . 

However, it is clear that x
^{2} + 4z
^{2} ³ 4xz = 4y
^{2}, from
which it follows that 4y + 1
³ 4y
^{2}. But this inequality is
impossible when y > 1. Thus, we conclude that x + 2z
 2y
¹ 1
and so N is composite.

272.

Let ABCD be a parallelogram whose area is 2003 sq.
cm. Several points are chosen on the sides of the parallelogram.


(a) If there are 1000 points in addition to A, B, C, D,
prove that there always exist three points among these 1004
points that are vertices of a triangle whose area is less that
2 sq. cm.


(b) If there are 2000 points in addition to A, B, C, D, is
it true that there always exist three points among these 2004 points
that are vertices of a triangle whose area is less than 1 sq. cm?
Solution. (a) Since there are 1000 points on the sides of
a parallelogram, there must be at least 500 points on one
pair of adjacent sides, regardless of the choice of points. Wolog,
let these points be on the sides AB and BC of the parallelogram.
and let m of the points P
_{1}, P
_{2},
¼, P
_{m} be on AB
and k of the points Q
_{1}, Q
_{2},
¼, Q
_{k} be on BC.
Let P
_{1} and Q
_{1} be the points closest to B. Connect the
vertex C to P
_{1}, P
_{2},
¼, P
_{m} and the point P
_{1}
to Q
_{1}, Q
_{2},
¼, Q
_{k} to get m + k + 1 triangles the sum
of whose areas equals the area of ABC. Thus [ABC] =
^{1}/
_{2}[ABCD] = 1001.5 sq cm. Let us assume that each of these m + k + 1
triangles has an area that exceeds 2 sq cm. Then [ABC]
³ 501 ×2 = 1002 > 1001.5, a contradiction. Therefore, at least one of these
triangles must have an area of less than 2 sq cm.
(b) No, this is not always true. We will construct a counterexample
to justify this answer. Let us choose 2000 points on the sides of
ABCD so that 1000 of them are on AB and 1000 of them are on
CD. We will consider the first set of 1000 points, and then do
symmetrical constructions and considerations for the second set.
Using the notation from (a), let m = 1000, k = 0 and select the
points so that BP_{1} = P_{1}P_{2} = P_{2}P_{3} = ¼ = P_{1000}A.
Then the triangle CBP_{1}, CP_{1}P_{2}, ¼, CP_{1000}A have
the same area, say s sq cm. However, s = [ABC]/1001 = (1001.5)/(1000) > 1; thus, this choice of the first 1000 points
allows a construction of triangles such that the area of each of
them exceeds 1 sq cm. Similarly, all triangles formed symmetrically
with vertices among the other set of 1000 points have an area which
exceeds 1 sq cm. So it is not true that there always exists three
points among the chosen 2000 points and the points A, B, C, D that
are vertices of a triangle whose area is less than 1 sq cm.
Comments. (1) It was not specified in the text of the question
that the three points chosen to be the vertices of a triangle
have to be noncollinear. Otherwise, we get the trivial case of
a "triangle" with an area of 0, which is not interesting, because
0 < 1, 0 < 2 and such a triangle will be an example of
existence in both cases. However, it is expected that candidates
will make a reasonable interpretation of the problem that renders
it nontrivial.
(2) Looking into possible interpretations of this problem, Michael
Lipnowski came up with a different, but very similar, and interesting
problem. Let ABCD be a parallelogram whose area is 2003
sq cm. Several points are chosen inside the parallelogram.
(a) If there are 1000 points in addition to A, B, C, D, prove that
there are always three points among these 1004 points that are vertices
of a triangle whose area is less than 2 sq cm. (b) If there are
2000 points in addition to A, B, C, D, is it true that there are
always three points among the 2004 points that are vertices of a
triangle whose area is less than 1 sq cm. We provide a solution
to this problem. Please note that the answer to (b) differs from
the answer of the corresponding part of the original question.
(a) Let AB  = x, AD  = y. Let P and
Q lie on AB and CD respectively, so that PQ  AD and
AP  = DQ  = (4/2003)x. This way, we have
a parallelogram "cut" from ABCD. Construct analogous parallelograms
with respect to the sides AB, BC and CD, drawing lines
parallel to these sides, so that each of them has a width
of (4x)/2003 or (4y)/2003 respectively. (1) If at least one of
the points lies within the parallelograms "cut", say, R, is within
APQD, then [ARD] < (1/2)(4/2003)[ABCD] = (1/2)(4/2003)(2003) = 2,
so this proves what is required. (2) Let us assume that all 1000 points
(without the vertices of course) lie within the interior parallelogram
KLMN whose vertices are the intersection points of the four lines
drawn before. Clearly, it is similar to ABCD, and the coefficient
of proportionality is 1995/2003, so its area is
(1995/2003)^{2} ·(2003) = (1995^{2})/(2003). Divide KLMN into 499
congruent parallelograms (for example, by drawing 498 equally spaced
lines parallel to KL). Then, since 1000 = 2 ×499 + 2 points
lie inside KLMN, at least one of the 499 parallelograms contains
at least three of them, according to the extended pigeonhole principle.
Consider the triangle formed by them. Since each of these
parallelograms has an area equal to (1/499)[KLMN] = (1/499)(1995^{2}/2003) < (1995 ·1996)/(499 ·2003) = 4 ·(1995/2003) < 4, then the area of the triangle will not
exceed half of 4, namely 2. So there must be at least one triangle
inside ABCD of area less than 2.
(b) Yes, it is always true that there exists three among the 2004
points that are vertices of a triangle with area less than 1. Proceed
as in (a) except for the following differences: (1) Construct the
parallel lines so that the width of the "cut" parallelograms is
(2x)/2003 or (2y)/2003, respectively. Now, the parallelogram
KLMN is similar to ABCD, with a coefficient of proportionality
1999/2003 and an area of (1999^{2})/2003. (2) Divide KLMN into
999 congruent parallelograms. Since 2000 = 2 ×999 + 2 points
lie within 999 regions, at least one region contains at least three
of the points. Similar calculations show that in this case, the area
of the triangle formed by these three points has area less than 1.
The result holds.

273.

Solve the logarithmic inequality
log_{4} (9^{x}  3^{x}  1) ³ log_{2} Ö5 . 

Solution. Let 3
^{x} = y. Then y > 0 and the given
inequality is equivalent to log
_{4} (y
^{2}  y
 1)
³ log
_{2} Ö5. Since the logarithmic function is defined
only for positive numbers, we must have y
^{2}  y
 1 > 0.
In this domain, the inequality is equivalent to
y
^{2}  y
 1
³ 5 or (y + 2)(y
 3)
³ 0. The solution of
the last inequality consists of all numbers not less than 3 (since
y > 0). Hence 3
^{x} ³ 3 or x
³ 1. Thus, the inequality
is satisfied if and only if x
³ 1.
Comment. It is very important before starting to solve the
inequality to determine the domains so that all functions are
welldefined. It is mandatory to take these restrictions into
consideration for the final answer as well as along the way in
making transformations.

274.

The inscribed circle of an isosceles triangle
ABC is tangent to the side AB at the point T and
bisects the segment CT. If CT = 6Ö2, find the
sides of the triangle.
Solution. Denote the midpoint of CT by K, and the tangent
point of the inscribed circle and BC by L. Then, from the given
information,
We will use the standard notation a, b, c for the lengths
of BC, CA and AB, respectively. It is not specified which
two sides of the isosceles triangle are equal, so there are
two possible cases.
Case 1. AC = BC or a = b. Then T is also the midpoint of
AB. By the tangentsecant theorem, CL^{2} = CK ·CT, which
together with (1) implies that
(a  (c/2))^{2} = CL^{2} = (1/2)CT^{2} = 36. Hence a = 6 + (c/2)
(2).
On the other hand, from the Pythagorean theorem applied to triangle
BCT, we get that a^{2} = (c^{2}/4) + 72. Using (2), we obtain that

æ è

6 + 
c
2


ö ø

2

= 
c^{2}
4

+ 72Û 36 + 6c = 72 Û c = 6 , 

whence a = b = 9. So the lengths of the triangle are
(a, b, c) = (9, 9, 6).
Case 2. AB = AC or c = b. Now L is the midpoint of the
side BC so that
CL^{2} = CK ·CT = (1/2)CT^{2}Û (a^{2}/4) = (1/2)(6 Ö2)^{2} = 36Û a = 12 . 
 (3) 
Next we have to calculate the lengths of AB and AC. From the cosine
law, applied to triangle BCT with
b =
ÐABC,


= BT^{2} + BC^{2}  2BT ·BC cosb 
 
 
Û (6 Ö2)^{2} = (a^{2}/4) + a^{2}  a^{2} cosb 
 
 
Û 72 = 36 + 144  144 cosbÛcosb = 3/4 . 

On the other hand, the cosine law for triangle ABC leads to


= c^{2} + a^{2}  2ca cosb = b^{2} + a^{2}  2ba cosb 
 
 

This, with (3), implies that c = b = 8. Therefore,
(a, b, c) = (8, 8, 12).

275.

Find all solutions of the trigonometric equation
sinx  sin3x + sin5x = cosx  cos3x + cos5x . 

Solution 1. [M. Lipnowski] Note that, if x =
q
satisfies the equation, then so does x =
q+
p. Thus,
it suffices to consider 0
£ x
£ p. A simple computation
shows that x =
p/2 is not a solution, so that we may assume
that cosx
¹ 0. Multiplying both sides of the equation
by 2 cosx
¹ 0 yields that


 sin3x + sin5x = cosx  cos3x+ cos5x 
 
 
Û2 sinx cosx  2 sin3x cosx + 2 sin5x cosx = 2 cos^{2} x  2 cos3x cosx + 2 cos5x cosx 
 
 
Û sin2x  (sin4x + sin2x) +sin6x + sin4x) = 1 + cos2x  (cos4x + cos2x)+ (cos6x + cos4x) 
 
 

Squaring both sides of the last equation, we get
sin^{2} 6x  2 sin6x cos6x + cos^{2} 6x = 1Û sin12x = 0 . 

This equation has as a solution x = k
p/12 for k an integer.
Checking each of these for validity, we find that the solutions are
x =
p/12, 2
p/12, 5
p/12, 9
p/12, 10
p/12,
and the general solution is obtained by adding a multiple of
p to
each of these.
Solution 2. The given equation is equivalent to
2 sin3x cos2x  sin3x = 2 cos3x cos2x  cos3x 

Û (2 cos2x  1)(sin3x  cos3x) = 0 . 

Thus, either cos2x =
^{1}/
_{2} in which case x =
±(
p/6) + k
p for some integer k, or cos2x
¹ ^{1}/
_{2}.
In the latter case, we must have cos3x
¹ 0 (why?), so that
tan3x = 1 and x = (
p/12) + (k
p/3). Thus, all solutions
of the equation are x = (
p/12) + k
p, (
p/6) + k
p,
(5
p/12) + k
p, (3
p/4) + k
p and (5
p/6) + k
p
where k is an integer.