Solutions
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73.
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Solve the equation:
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æ
ç
è
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Ö
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2 + Ö2
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ö
÷
ø
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x
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+ |
æ
ç
è
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Ö
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2 - Ö2
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ö
÷
ø
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x
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= 2x . |
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Solution 1. By inspection, we find that x = 2
satisfies the equation. We show that no other value of
x does so. Observe that Ö{2 - Ö2} < 1.
When x < 0, the second term of the left side exceeds 1
while the right side is less than 1, so the equation is
not satisfied. Henceforth, let x ³ 0 and let
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f(x) º 2x - |
æ
ç
è
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Ö
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2 + Ö2
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ö
÷
ø
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x
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and g(x) = (Ö{2 - Ö2})x.
Note that, if a > b > 1, then ax - bx = bx((a/b)x - 1) is an increasing function of x. Thus,
f(x) is increasing and g(x) is decreasing as x
increases. If 0 £ x < 2, then f(x) < f(2) = g(2) < g(x), while if x > 2, then f(x) > f(2) = g(2) > g(x). The desired result follows.
Solution 2. The equation can be rewritten in
the form
where 2y = x and a = 1/4(2 + Ö2).
Note that 0 < a < 1, so that each term is a
strictly decreasing function of y. Thus, f(y)
assumes each of its values at most once, and since
f(1) = 1, we find that x = 2 is the only solution.
Solution 3. Observe that
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1 + |
Ö2
2
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= 1 + cos |
p
4
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= 2cos2 |
p
8
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and
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1 - |
Ö2
2
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= 1 - sin |
p
4
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= 2sin2 |
p
8
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. |
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The equation becomes
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æ
ç
è
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cos |
p
8
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ö
÷
ø
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x
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+ |
æ
ç
è
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sin |
p
8
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ö
÷
ø
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x
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= 1 . |
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This holds for x = 2. If x > 2, then
x - 2 > 0 and so
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æ
ç
è
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cos |
p
8
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ö
÷
ø
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x
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= |
æ
ç
è
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cos |
p
8
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ö
÷
ø
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x-2
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æ
ç
è
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cos |
p
8
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ö
÷
ø
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2
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< |
æ
ç
è
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cos |
p
8
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ö
÷
ø
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2
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with a similar inequality for the sine function.
Thus, when x > 2, the left side is less than 1.
Similarly, it can be shown that when x < 2,
the left side exceeds 1. Hence the unique solution
is x = 2.
Comment. Generally, the solutions involved a function
similar to that used in Solution 2, and it was shown that
it was impossible for there to be more than one solution.
Some students came up with the use of trigonometry as in
Solution 3.
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74.
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Prove that among any group of
n + 2 natural numbers, there can be found two
numbers so that their sum or their difference is
divisible by 2n.
Solution 1. For 0 £ k £ n, let Sk
be the subset of numbers x among the n + 2 numbers
for which x differs from either k or 2n - k by a
multiple of 2n. Since there are n + 2 numbers and
only n + 1 subsets, the Pigeonhole Principle provides
that some subset must contain at least two numbers
u and v, say. Either u and v both leave the
same remainder upon division by 2n and so differ
by a multiple of 2n, or else one of them
differs from k by a multiple of 2n while the other
differs from 2n - k by a multiple of 2n. In the
latter case, u + v is a multiple of 2n.
Solution 2. [A. Fink] Consider an arbitrary set of
n + 2 natural numbers. If any two are congruent modulo
2n, then their difference is divisible by 2n and
the result follows. Suppose otherwise, that all numbers
have distinct residues modulo 2n. Apportion these
residues into the n+1 sets: { 0 },
{ 1, 2n-1 }, { 2, 2n - 2 }, ¼,
{ n - 1, n + 1 }, { n }. Since there are
n + 2 numbers, at least one of these sets must
contain two residues, and so the two numbers
involved must sum to a multiple of 2n.
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75.
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Three consecutive natural numbers, larger than
3, represent the lengths of the sides of a triangle. The
area of the triangle is also a natural number.
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(a) Prove that one of the altitudes ``cuts'' the triangle
into two triangles, whose side lengths are natural numbers.
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(b) The altitude identified in (a) divides the side
which is perpendicular to it into two segments. Find the
difference between the lengths of these segments.
Solution 1. Let the side lengths be x - 1, x,
x + 1. By Heron's formula, the area A of the triangle
is given by
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A2 = |
3
16
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x2(x+2)(x-2) = |
(3(x2 - 4))x2
16
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. |
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Since A is an integer, x must be even and 3(x2 - 4)
must be the square of a multiple of 2 ×3 = 6.
Hence for some integer y, we have that
3(x2 - 4) = (6y)2 = 36y2 or x2 - 12y2 = 4.
(Comment. This is a Pell's equation and it has
infinitely many solutions (x, y) = (xn, yn) given
by (xn, yn) = 2(7 + 2Ö[12])n and
(4 + Ö[12])(7 + 2Ö[12])n.
(a) The area A of the triangle is 1/4(6xy) = [3xy/2] where x is even and x2 - 12y2 = 4.
The altitude to the side of length x is 2A/x = 3y,
an integer. This is the desired altitude.
(b) The triangle is subdivided by the altitude in (a) into
two right triangles whose hypotenuses have lengths x-1
and x+1. Hence, the side of length x is split into
two parts of lengths
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| ___________
Ö(x-1)2 - (3y)2
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= |
| ______________
Öx2 - 2x + 1 - 9y2
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= |
| ____________
Ö(x2/4) - 2x + 4
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= |
1
2
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(x - 4) |
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and
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| ___________
Ö(x+1)2 - (3y)2
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= |
| ______________
Öx2 + 2x + 1 - 9y2
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= |
| ____________
Ö(x2/4) + 2x + 4
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= |
1
2
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(x + 4) . |
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The difference between the lengths of these segments is
4. (Note that the sum is x. as expected.)
(Exercise. Give some numerical examples.)
Solution 2. [L. Tchourakov] With the above notation,
we find that A = xÖ[(3(x2 - 4))]/4, so that the
length of the altitude to the side of length x is
1/2Ö[(3(x2 - 4))]. If x were odd, then
the numerator of the fraction for A would be odd and
A not an integer. Hence x is even, and so is the
altitude. Let u be one of the two parts of the side
cut off by the altitude. By the pythagorean theorem,
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u2 = (x + 1)2 - |
3(x2 - 4)
4
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= |
(x + 4)2
4
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, |
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so that u = 2 + x/2. Since x is even, u is an
integer. The altitude cuts the side into parts of
length u and x - u = u - 4, and so (b) follows.
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76.
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Solve the system of equations:
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logx + |
log(xy8)
log2 x + log2 y
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= 2 , |
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logy + |
log(x8/y)
log2 x + log2 y
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= 0 . |
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(The logarithms are taken to base 10.)
Solution 1. Let u = logx, v = logy and
w = u2 + v2. Note that w is nonzero.
The equations become
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u + (u + 8v)/w = 2 and v + (8u - v)/w = 0 . |
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Squaring and adding the equations u + 8v = (2 - u)w and
8u - v = -vw yields
65(u2 + v2) = (4 - 4w + u2 + v2)w2, or
65 = (4 - 4u + w)w. We can also write the system
as
which can be solved to yield
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u = |
2w(w-1)
w2 - 65
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v = |
16w
65 - w2
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Hence
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w = u2 + v2 = |
4w2(w-1)2 + 256w2
(65 - w2)
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Þ (65 - w2)2 = 4w(w - 1)2 + 256w |
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= w4 - 4w3 - 122w2 - 260w + 652 |
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The two relevant solutions are w = 13 and w = 5.
When w = 13, 17 - 4u = 5, which leads to (u, v) = (3, -2).
When w = 5, 9 - 4u = 13, so that (u, v) = (-1, 2).
The desired solutions are (x, y) = (103, 10-2),(10-1, 102).
Solution 2. With the same notation as (1), we can
write the given equations in terms of u and v.
Multiply the first equation by u and the second by v
and add them to obtain the equation 2uv + 8 = 2v,
whereupon u = 1 - (4/v). Eliminating u from the
second equation yields v4 - 16 = 0, whereupon
v = 2 or v = -2. The remaining part of the solution
is easily completed.
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77.
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n points are chosen from the circumference or
the interior of a regular hexagon with sides of
unit length, so that the distance between
any two of them is not less than Ö2. What is the largest
natural number n for which this is possible?
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78.
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A truck travelled from town A to town B
over several days. During the first day, it covered
1/n of the total distance, where n is a natural number.
During the second day, it travelled 1/m of the
remaining distance, where m is a natural number.
During the third day, it travelled 1/n of the distance
remaining after the second day, and during the fourth day,
1/m of the distance remaining after the third day.
Find the values of m and n if it is known that, by the
end of the fourth day, the truck had travelled 3/4
of the distance between A and B. (Without loss of
generality, assume that m < n.)
Solution. [R. Furmaniak, J. Rin]
Let d be the distance remaining at the
beginning of a two-day period. The distance remaining at the
end of the period is
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d - |
é
ê
ë
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d
n
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+ |
1
m
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æ
ç
è
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d - |
d
n
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ö
÷
ø
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ù
ú
û
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= d |
æ
ç
è
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1 - |
1
m
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ö
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ø
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æ
ç
è
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1 - |
1
n
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ö
÷
ø
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. |
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Thus, every two days the remaining distance is reduced by
a factor of r = (1 - (1/m))(1 - (1/n)). (Note that
this is symmetric in m and n.) After four days, the
distance remaining is reduced by a factor of r2; it
is given in the problem that this is 1/4. Hence r = 1/2.
Hence
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= |
æ
ç
è
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1 - |
1
m
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ö
÷
ø
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æ
ç
è
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1 - |
1
n
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ö
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ø
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Since m and n are positive integers with m < n,
we must have n - 2 = 2 and m - 2 = 1, i.e.,
(m, n) = (3, 4).
