1971 IMO Problems/Problem 4
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[hide]Problem
All the faces of tetrahedron are acute-angled triangles. We consider all closed polygonal paths of the form
defined as follows:
is a point on edge
distinct from
and
; similarly,
are interior points of edges
, respectively. Prove:
(a) If , then among the polygonal paths, there is none of minimal length.
(b) If , then there are infinitely many shortest polygonal paths, their common length being
, where
.
Solution
Rotate the triangle around the edge
until
are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting
and
. Therefore,
.
Summing the four equations like this, we get exactly
.
Now, draw all four faces in the plane, so that is constructed on the exterior of the edge
of
and so on with edges
and
.
The final new edge (or rather
) is parallel to the original one (because of the angle equation). Call the direction on
towards
"right" and towards
"left". If we choose a vertex
on
and connect it to the corresponding vertex
on A'B'. This works for a whole interval of vertices
if
lies to the left of
and
and
lies to the right of
. It is not hard to see that these conditions correspond to the fact that various angles are acute by assumption.
Finally, regard the sine in half the isosceles triangle which gives the result with the angles around
instead of
, but the role of the vertices is symmetric.
Remarks (added by pf02, December 2024)
The solution above is incomplete, and very likely, incorrect. The first
part (which claims to prove part (a) of the problem) is incomplete at
best, since it is not clear how it leads to the result. (Very likely
it is incorrect.) The second part, which claims to prove part (b) of
the problem, skips too many steps. Some of the arguments in the proof
(e.g. the fact that ) are true but need proof.
I could not make any sense of other arguments in the proof, so I
can not judge whether they are correct or not (I suspect not).
I will give a robust solution below. It goes along the same basic idea.
Solution 2
The basic idea is to "fold out" the tetrahedron into a polygon in
the plane. The path becomes a collection of connected
segments
with the unconnected ends
on
the two line segments
.
Specifically, we rotate the solid consisting of three faces around
so
is in the same plane as
.
Then we rotate the solid consisting of two faces around
so
that
is in the same plane with the previous
triangles. (We denoted
the new "copy" of
, since the
original
is being used in the picture.) Finally, we rotate
around
, so that it is in the same plane
with the other triangles. (We denote
the new "copy" of
,
and
the new copy of
.)
The polygonal path becomes
. It is clear that in order
to minimize
, we should make
be a segment on a straight
line. Furthermore, to minimize the segment
, we want to choose
so that when we draw the line segment to its corresponding
image
, the length of
is as short as possible.
This was the "folding out" done in the solution above. To continue,
we will do a different "folding out", which will serve better for
solving the problem. Specifically, we rotate around
, then we rotate
around
to obtain
, and finally, we rotate
around
to obtain
. Now
becomes
. The polygonal path becomes
going from
to
. We need to make this as small as possible, to
find its minimum, if it exists.
The idea of the solution to the problem is now easy to explain.
First of all, needs to be a segment on a straight line,
in which case its length is the length of the segment
.
Clearly
has a lower bound (after all, it is
),
so we can think about its lower limit. If there is a position of
in which this lower limit is achieved, then this lower limit
is a minimum. Otherwise, there is no minimum value for
.
(Pay attention to the subtle distinction between lower bound,
lower limit, and minimum.)
In the picture above, and
are not parallel. The segment
would be shortest when
(and
). But this is
not an acceptable position for
, because the problem stated
that
is between
and
, not equal to any of them. So in
this picture, there is no minimum for the polygonal path. (In this
case there is a lower limit for
, namely
, which is not a
minimum.)
On the other hand, if (see pictures below) then
generally there are lots of points
yielding a minimum
value for
. Indeed, in this case
, so the only requirement is for
all the points
to be inside the respective segments.
(In this case,
is a lower limit, and it is a
minimum.)
We will prove that
if and only if
. After this it will be easy to
deduce all the statements of the problem.
Let be on
be such that
. The equality
becomes
, and then it
becomes
. Because of the
parallelism of
we have several equal angles on the
two sides of this equality. This equality becomes
.
So, the original equality
is true if and only if
, which is true if and only if
Now it is obvious that when
a lower limit
exists (as expected), but it is not a minimum, and a minimum does not
exist.
On the other hand, when ,
in the typical, general case,
is the minimum, and there are
infinitely many segments
of the same size (see Figure 2 below).
There is one delicate point we need to worry about: we have to be sure
that intersects
. See Figure 3 for an example when
this does not happen; note that in this example we have some obtuse
angles. Formally,
has to be on the same side of
as
and
has to be on the same side of
as
.
Let us concentrate on the position of vs.
(the statement
about
is similar). We want
. We
have
and
.
Since both
are acute, it follows that
.
The last thing to do is to compute . We have
, and
from the isosceles triangle
we have
, where
is the midpoint of
. This gives
, which is the formula we wanted.
[Solution by pf02, December 2024]
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1971 IMO (Problems) • Resources | ||
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1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
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