1973 IMO Problems/Problem 1
Contents
[hide]Problem
Point lies on line
are unit vectors such that points
all lie in a plane containing
and on one side of
Prove that if
is odd,
Here
denotes the length of vector
Solution
We prove it by induction on the number of vectors. The base step (when we have one vector) is clear, and for the induction step we use the hypothesis for the
vectors obtained by disregarding the outermost two vectors. We thus get a vector with norm
betwen two with norm
. The sum of the two vectors of norm
makes an angle of
with the vector of norm
, so their sum has norm
, and we're done.
The above solution was posted and copyrighted by grobber. The original thread for this problem can be found here: [1]
Remarks (added by pf02, May 2025)
The "solution" given above is somewhat incomplete. It simply
shoves the difficulty of the problem into the phrase "... so their
sum has norm , and we're done." Indeed, while true, this
is not obvious, and it needs a proof.
Below, I will fill in the gaps in the first "solution". Then, I will give two more solutions.
Solution with gaps filled in
Let . We give a proof by induction on
.
For , we have one vector
whose
norm (i.e. length) is
, so we are done.
Assume the statement is true for and prove
that it is true for
. For ease of visualization, assume the
plane to have
axes,
to be the line
, and
. Let
. Assume the vectors are determined
by
, such that
.
Let and
be determined by
(like in the solution above). Since
is the
sum of
unit vectors on one side of
, its norm
is
(because of the induction hypothesis). Let us say that
. We want to show that the norm
.
Since are unit vectors
whose angles with the positive direction of the
-axis are
the direction of
is
.
Since
, the angle between
and
is
.
Now use that the norm of is
, and the
general fact that if the angle between two vectors
is
, then
.
It follows that
.
To finish the proof, let us show this general fact about
forming an angle
.
Let
and
.
Let be the lengths of the segments
. Consider
the triangle
. The cosine formula in
yields
. Since
is
obtuse, we have
, so
.
This implies
.
Solution 2
This solution starts exactly like the previous one, till the line
. We want to show that the norm
.
Then, we continue as follows:
The end point of
,
and the square of the norm of this vector is
.
After some computations, we have
.
Since is a decreasing function from
to
on the interval
, it follows that
.
Thus .
It follows that the norm
.
[Solution by pf02, May 2025]
Solution 3
For ease of visualization, assume the plane to have axes,
to be the line
, and
. Assume the
vectors are determined by
,
such that
.
We are asked to show that
. Before
embarking on the computations which prove this inequality, let
us give the idea of the proof. Let
. We will prove
formally that by "moving"
to their limiting positions
on
(i.e. replacing
by
and
by
)
the norm of the sum of vectors will be decreased. Repeat this
argument for
, etc. (for a total of
times). We
are left with one vector (the middle vector), which has norm
.
Now let us turn this idea into a formal proof.
The point determining the sum of vectors is
.
The square of the norm of the sum of vectors is
.
After some computations, we get
.
Denote
.
When
, we define
. (This makes sense both formally, and
as an accommodation to
.)
We have .
Since is a decreasing function from
to
on the interval
, it follows that
.
But
In short, we showed that
.
Now, if , and we apply this argument
times, we get
.
Thus, .
[Solution by pf02, May 2025]
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1973 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |