Solutions
-
171.
-
Let
be a positive integer. In a round-robin
match,
teams compete and each pair of teams plays exactly
one game. At the end of the match, the
th team has
wins and
losses. There are no ties. Prove that
Solution 1. Each game results in both a win and a loss,
so the total number of wins is equal to the total number of
losses. Thus
. For each team, the
total number of its wins and losses is equal to the number
of games it plays. Thus
for each i.
Accordingly,
from which the desired result follows.
Solution 2. Since
for each
and
(the number of games played), we find that
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172.
-
Let
,
,
,
.
,
be different
integers. Prove that
Solution 1. Since the sum of the differences is 0, an even
number, there must be an even number of odd differences, and
therefore an even number of odd squares. If the sum of the
squares is less than 18, then this sum must be one of the
numbers 6, 8, 10, 12, 14, 16. The only possibilities for
expressing any of these numbers as the sum of six nonzero squares is
Taking note that the sum of the differences is zero, the possible
sets of differences (up to order and sign) are
,
,
. Since the numbers are distinct,
the difference between the largest and smallest is
at least 5. This difference
must be the sum of differences between adjacent numbers; but checking
proves that in each case, an addition of adjacent differences must
be less than 5. Hence, it is not possible to achieve a sum of
squares less than 18. The sum 18 can be found with the set
.
Solution 2. [A. Critch] We prove a more general result.
Let
be a positive integer exceeding 1.
Let
be an
tple of distinct integers,
and suppose that the smallest of these is
. Define
,
and wolog suppose that
.
Suppose that, for
,
.
Let the largest integer be
; since the integers are distinct, we must have
and
Hence.
By the root-mean-square, arithmetic mean (RMS-AM)
inequality, we have that
so that
Thus,
Since the sum of the differences of consecutive
is zero, and
so even, the sum of the squares is even. Since
is even
and
, and since the sum exceeds
, we see that
How can this lower bound be achieved? Since it is equal to
, we can have
differences equal to
2 and 2 differences equal to 1. Thus, we can start by going up
the odd integers, and then come down via the even integers to 0.
In the case of
, this yields the
tple
.
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173.
-
Suppose that
and
are positive real numbers
for which
. Prove that
Determine when equality holds.
Remark. Before starting, we note that when
,
, then
. This is an immediate consequence
of the arithmetic-geometric means inequality.
Solution 1. By the root-mean-square, arithmetic mean inequality,
we have that
as desired.
Solution 2. By the RMS-AM inequality and the
harmonic-arithmetic means inequality, we have that
from which the result follows.
Solution 3.
Solution 4. [F. Feng] From the Cauchy-Schwarz and
arithmetic-geometric means inequalities, we find that
The desired result follows.
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174.
-
For which real value of
is the function
maximum? Determine its maximum value.
Solution 1. The function assumes negative values when
and
. Accordingly, we need only consider its
values on the interval
. Suppose, first, that
, in which case all factors of the function
are nonnegative. Then we can note, by the arithmetic-geometric
means inequality, that
with equality if and only if
. Thus, on the interval
,
the function takes its maximum value of 1 when
.
We adopt the same strategy to consider the situation when
. For convenience, let
, so that
the want to maximize
for
. In fact, we are going to maximize
where
,
and
are positive integers to be
chosen to make
(so that when we
calculate the airthmetic mean of the eight factors, the coefficient
of
will vanish), and
(so that we can actually find a case where equality will
occur). The latter conditions forces
or
. Plugging
into this yields that
Thus, let us take
.
Then, by the arithmetic-geometric means inequality, we obtain that
with equality if and only if
,
i.e.,
. Hence,
with equality if and only if
.
It remains to show whether the value of the function when
exceeds its value when
. Recall the Bernoulli inequality:
for
and positive integer
. This can be established by induction (do it!). Using this,
we find that
Thus, the function assumes its maximum value of
when
.
Solution 2. Let
.
Then
. We see that
if and only if
or
.
Hence
has relative maxima only when
and
. Checking these two candidates tells us that the
absolute maximum of
occurs when
.
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175.
-
is a triangle such that
. The point
is the midpoint of the arc with endpoints
and
of that arc of
the circumcircle of
that contains
. The foot of the
perpendicular from
to
is
. Prove that
.
Solution 1. Draw a line through
parallel to
that
intersects the circumcircle again at
. Let
be the foot
of the perpendicular from
to
. Then
is a rectangle.
Since arc
is equal to arc
, and since arc
is equal to
arc
(why?), arc
is equal to arc
. Therefore the
chords of these arcs are equal, so that
.
Hence
.
Solution 2. Note that the length of the shorter arc
is
less than the length of the shorter arc
. Locate a point
on the chord
so that
. Consider triangles
and
. We have that
,
and
. This is a case of SSA congruence, the
ambiguous case. Since angles
and
are both obtuse (why?),
they must be equal rather the supplementary, and the triangles
and
are congruent. (Congruence can also be established
using the Law of Sines.) In particular,
. Since
triangle
is isosceles,
is the midpoint of
, so that
.
Solution 3. Let
be a diameter of the circumcircle, so that
is the midpoint of one of the arcs
. Let
be that point on
the chord
for which
and let
be produced to meet
the circle again in
. Since
,
it follows that
.
Since
and
are both angle bisectors,
.
Because
is concyclic, triangles
and
are similar,
so that
. From the equality of angles
and
,
we deduce the equality of the arcs
and
, and so the
equality of the arcs
and
. Hence
.
Therefore
.
Solution 4. [R. Shapiro] Since
lies on the short arc
,
is obtuse. Hence the foot of the perpendicular
from
to
produced is outside of the circumcircle of
triangle
. In triangles
and
,
,
and
. Hence the triangles
and
are congruent, so that
and
.
Since the triangle
and
are right with a common
hypotenuse
and equal legs
and
, they are congruent
and
. Hence
, as desired.
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176.
-
Three noncollinear points
,
and
are given in
the plane. Construct the square such that one of its vertices is
the point
, and the two sides which do not contain this vertex are
on the lines through
and
respectively. [Note: In such
a problem, your solution should consist of a description of the
construction (with straightedge and compasses) and a proof in
correct logical order proceeding from what is given to what is
desired that the construction is valid. You should deal with
the feasibility of the construction.]
Solution 1. Construction.
Draw the circle with diameter
and centre
. This circle
must contain the point
, as
and
are to be perpendicular.
Let the right bisector of
meet the circle in
and
.
Join
and, if necessary, produce it to meet the circle at
.
Now draw the circle with diameter
and let it meet the right
bisector of
at
and
. Then
is the required
rectangle. There are two options, depending how we label the
right bisector
.
However, the construction does not work if
actually lies on the
circle with diameter
. In this case,
and
would coincide
and the situation degenerates. If
lies on the right bisector of
, then
can be the other point where the right bisector
intersects the circle, and
and
can be the other two vertices
of the square. If
is not on the right bisector, then there
is no square; all of the points
,
,
,
would have to
be on the circle, and
and
would have to subtend angles
of
at
, which is not possible.
Proof. If
and
are on the same side of
, then
, so that
makes an angle of
with
produced, and so
produced contains a side of the square. Similarly,
produced
contains a side of the square. If
and
are on opposite sides
of
, then
, and
still makes an
angle of
with
produced; the argument can be
completed as before.
Solution 2. Construction. Construct circle of diameter
. Draw
(with the segment
intersecting
the interior of the circle) and
. Construct the circle
.
Let this circle intersect the given circle at
. Then construct the
square with diagonal
. If
lies on the circle, then the candidates
for
are
and
. We cannot take
, as the situation
degererates; if we take
, then the angle
and segment
degenerate. We can complete the analysis as in the first
solution.
Proof. Since
is right isosceles,
. Hence the circle is the locus of points at which
subtends an angle equal to
or
.
Hence the lines
and
intersect at an angle of
.
Since
, the lines
and
also
intersect at
. It follows that the remaining points
on the square with diagonal
must lie on the lines
and
.
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177.
-
Let
,
,
,
be nonnegative
integers such that, whenever
,
,
, then
-
-
(a) Give an example of such a sequence which is not
an arithmetic progression.
-
-
(b) Prove that there exists a real number
such that
for
.
(a)
Solution. [R. Marinov] For positive integers
, let
when
and
when
, so
that the sequence is
.
Observe that
and
for positive integers
and
, whence we see that this
sequence satisfies the condition. The corresponding value of
is
.
Solution. [A. Critch] The assertion to be proved is that
all the semi-closed intervals
have a point in
common. Suppose, if possible, that this fails. Then there must
be a pair
of necessarily distinct integers for which
. This is equivalent to
. Suppose that the sum of these
two indices is as small as possible.
Suppose that
, so that
for some positive
.
Then
whence
and
. Thus
and
have the property of
and
and we get a contradiction
of the minimality condition.
Suppose that
, so that
for some positive
.
Then
so that
,
and
, once again contradicting the minimality
condition.