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Difference between revisions of "1971 IMO Problems/Problem 4"
Line 72: Line 72:
  
 
The idea of the solution to the problem is now easy to explain.
 
The idea of the solution to the problem is now easy to explain.
−
First of all, <math>ZYXTZ'</math> needs to be a segment on a straight line.
+
First of all, <math>ZYXTZ'</math> needs to be a segment on a straight line,
−
Clearly <math>ZYXTZ'</math> has a lower bound (after all, it is <math>> 0</math>), so we
+
in which case its length is the length of the segment <math>ZZ'</math>..
−
can think about its lower limit.  If there is a position of <math>Z</math> in
+
Clearly <math>ZYXTZ' = ZZ'</math> has a lower bound (after all, it is <math>> 0</math>),
−
which this lower limit is achieved, then this lower limit is a
+
so we can think about its lower limit.  If there is a position of
−
minimum.  Otherwise, there is no minimum value for <math>ZYXTZ'</math>.
+
<math>Z</math> in which this lower limit is achieved, then this lower limit
 +
is a minimum.  Otherwise, there is no minimum value for <math>ZYXTZ'</math>.
 +
(Pay attention to the subtle distinction between lower bound,
 +
lower limit, and minimum.)
  
 
In the picture above, <math>CD</math> and <math>C'D'</math> are not parallel.  The segment
 
In the picture above, <math>CD</math> and <math>C'D'</math> are not parallel.  The segment
Line 88: Line 91:
 
Indeed, in this case <math>CC' \parallel ZZ' \parallel DD'</math>, so the only
 
Indeed, in this case <math>CC' \parallel ZZ' \parallel DD'</math>, so the only
 
requirement is for all the points <math>X, Y, Z, T</math> to be inside the
 
requirement is for all the points <math>X, Y, Z, T</math> to be inside the
−
respective segments.
+
respective segments. (<math>ZZ' = DD'</math> is a lower limit, but not a
 +
minimum.)
  
 
We will prove that <math>\angle DAB + \angle BCD = \angle CDA + \angle ABC</math>
 
We will prove that <math>\angle DAB + \angle BCD = \angle CDA + \angle ABC</math>
Line 107: Line 111:
 
<math>CD \parallel C'D'.</math>
 
<math>CD \parallel C'D'.</math>
  
 +
Now it is obvious that when
 +
<math>\angle DAB + \angle BCD \neq \angle CDA + \angle ABC</math> a lower limit
 +
exists (as expected), but it is not a minimum, and a minimum does not
 +
exist.
  
−
 
+
On the other hand, when <math>\angle DAB + \angle BCD = \angle CDA + \angle ABC</math>,
−
 
+
in the typical, general case, <math>ZZ' = DD'</math> is the minimum, and there are
−
 
+
infinitely many segments like this (see Figure 2 below).
  
 
[[File:prob_1971_4_3.png|800px]]
 
[[File:prob_1971_4_3.png|800px]]
  
 +
There is one delicate point we need to worry about: we have to be sure
 +
that <math>ZZ'</math> intersects <math>CB, BA, AD'</math>.  See Figure 3 for an example when
 +
this does not happen; note that in this example we have some obtuse
 +
angles.  Formally, <math>A</math> has to be on the same side of <math>DD'</math> as <math>C, C'</math>
 +
and <math>B</math> has to be on the same side of <math>CC'</math> as <math>D, D'</math>.
  
 +
Let us concentrate on the position of <math>A</math> vs. <math>DD'</math> (the statement
 +
about <math>B</math> is similar).  We want <math>\angle DAB + \angle BAD' < \pi</math>.  We
 +
have <math>\angle DAB < \angle CAD</math> and
 +
<math>\angle BAD' = \text{the original } \angle BAD \text{ from the tetrahedron}</math>.
 +
Since both <math>\angle CAD, \angle BAD</math> are acute, it follows that
 +
<math>\angle DAB + \angle BAD' < \pi</math>.
  
 +
The last thing to do is to compute <math>ZZ'</math>.  We have <math>ZZ' = CC'</math>, and
 +
from the isosceles triangle <math>\triangle CAC'</math> we have
 +
<math>CM = \sin \angle CAM</math>, where <math>M</math> is the midpoint of <math>CC'</math>.  This gives
 +
<math>ZZ' = CC' = 2 AC \sin \frac{\alpha}{2}</math>, the formula we wanted.
  
−
 
+
[Solution by pf02, December 2024]
−
 
 
−
 
 
−
[TO BE CONTINUED.  SAVING MID WAY SO I DON"T LOSE WORK DONE SO FAR.]
 
  
  

Revision as of 01:20, 30 December 2024

Problem

All the faces of tetrahedron $ABCD$ are acute-angled triangles. We consider all closed polygonal paths of the form $XYZTX$ defined as follows: $X$ is a point on edge $AB$ distinct from $A$ and $B$; similarly, $Y, Z, T$ are interior points of edges $BC, CD, DA$, respectively. Prove:

(a) If $\angle DAB + \angle BCD \neq \angle CDA + \angle ABC$, then among the polygonal paths, there is none of minimal length.

(b) If $\angle DAB + \angle BCD = \angle CDA + \angle ABC$, then there are infinitely many shortest polygonal paths, their common length being $2AC \sin(\alpha / 2)$, where $\alpha = \angle BAC + \angle CAD + \angle DAB$.


Solution

Rotate the triangle $BCD$ around the edge $BC$ until $ABCD$ are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting $X$ and $Z$. Therefore, $XYB=ZYC$. Summing the four equations like this, we get exactly $\angle ABC+\angle ADC=\angle BCD+\angle BAD$.

Now, draw all four faces in the plane, so that $BCD$ is constructed on the exterior of the edge $BC$ of $ABC$ and so on with edges $CD$ and $AD$.

The final new edge $AB$ (or rather $A'B'$) is parallel to the original one (because of the angle equation). Call the direction on $AB$ towards $B$ "right" and towards $A$ "left". If we choose a vertex $X$ on $AB$ and connect it to the corresponding vertex $X'$ on A'B'. This works for a whole interval of vertices $X$ if $C$ lies to the left of $B$ and $D$ and $D$ lies to the right of $A$. 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 $ACA'$ which gives the result with the angles around $C$ instead of $A$, but the role of the vertices is symmetric.


Remarks (added by pf02, December 2024)

The solution above is incomplete and/or 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 $AB \parallel A'B'$) 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 $XYZTX$ becomes a collection of connected segments $XY, YZ, ZT, TX'$ with the unconnected ends $X, X'$ on the two line segments $AB, A'B'$.

Prob 1971 4 1.png

Specifically, we rotate the solid consisting of three faces around $BC$ so $\triangle BCD$ is in the same plane as $\triangle ABC$. Then we rotate the solid consisting of two faces around $CD$ so that $\triangle CDA'$ is in the same plane with the previous triangles. (We denoted $A'$ the new "copy" of $A$, since the original $A$ is being used in the picture.) Finally, we rotate $\triangle A'BD$ around $A'D$, so that it is in the same plane with the other triangles. (We denote $B'$ the new "copy" of $B$, and $X'$ the new copy of $X$.)

The polygonal path $XYZTX$ becomes $XYZTX'$. It is clear that in order to minimize $XYZTX$, we should make $XYZTX'$ be a segment on a straight line. Furthermore, to minimize the segment $XX'$, we want to choose $X \in AB$ so that when we draw the line segment to its corresponding image $X' \in A'B'$, the length of $XX'$ 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 $\triangle BCD$ around $BC$, then we rotate $\triangle ABD$ around $AB$ to obtain $\triangle ABD'$, and finally, we rotate $\triangle AD'C$ around $AD'$ to obtain $\triangle AD'C'$. Now $Z \in CD$ becomes $Z' \in C'D'$. The polygonal path becomes $ZYXTZ'$ going from $CD$ to $C'D'$. We need to make this as small as possible, to find its minimum, if it exists.

Prob 1971 4 2.png

The idea of the solution to the problem is now easy to explain. First of all, $ZYXTZ'$ needs to be a segment on a straight line, in which case its length is the length of the segment $ZZ'$.. Clearly $ZYXTZ' = ZZ'$ has a lower bound (after all, it is $> 0$), so we can think about its lower limit. If there is a position of $Z$ in which this lower limit is achieved, then this lower limit is a minimum. Otherwise, there is no minimum value for $ZYXTZ'$. (Pay attention to the subtle distinction between lower bound, lower limit, and minimum.)

In the picture above, $CD$ and $C'D'$ are not parallel. The segment $ZZ'$ would be shortest when $Z = D$ (and $Z' = D'$). But this is not an acceptable position for $ZYXTZ'$, because the problem stated that $Z$ is between $C$ and $D$, not equal to any of them. So in this picture, there is no minimum for the polygonal path.

On the other hand, if $CD \parallel C'D'$ (see pictures below) then there are lots of points $Z \in CD$ yielding a minimum value for $ZYXTZ'$. Indeed, in this case $CC' \parallel ZZ' \parallel DD'$, so the only requirement is for all the points $X, Y, Z, T$ to be inside the respective segments. ($ZZ' = DD'$ is a lower limit, but not a minimum.)

We will prove that $\angle DAB + \angle BCD = \angle CDA + \angle ABC$ if and only if $CD \parallel C'D'$. After this it will be easy to deduce all the statements of the problem.

Let $M, N, P$ be on $CA, BD', AC'$ be such that $CD \parallel MB \parallel AN \parallel PD'$. The equality $\angle DAB + \angle BCD = \angle CDA + \angle ABC$ becomes $\angle D'AB + \angle BCD = \angle C'D'A + \angle ABC$, then $\angle D'AN + \angle NAB + \angle BCD = \angle C'D'P + \angle PD'A + \angle ABM + \angle MBC$. Because of the parallelism of $CD, MB, AN, PD'$ we have several equal angles on the two sides of this equality. This equality becomes $\angle C'D'P = 0$.

So, the original equality $\angle DAB + \angle BCD = \angle CDA + \angle ABC$ is true if and only if $\angle C'D'P = 0$, which is true if and only if $CD \parallel C'D'.$

Now it is obvious that when $\angle DAB + \angle BCD \neq \angle CDA + \angle ABC$ a lower limit exists (as expected), but it is not a minimum, and a minimum does not exist.

On the other hand, when $\angle DAB + \angle BCD = \angle CDA + \angle ABC$, in the typical, general case, $ZZ' = DD'$ is the minimum, and there are infinitely many segments like this (see Figure 2 below).

Prob 1971 4 3.png

There is one delicate point we need to worry about: we have to be sure that $ZZ'$ intersects $CB, BA, AD'$. See Figure 3 for an example when this does not happen; note that in this example we have some obtuse angles. Formally, $A$ has to be on the same side of $DD'$ as $C, C'$ and $B$ has to be on the same side of $CC'$ as $D, D'$.

Let us concentrate on the position of $A$ vs. $DD'$ (the statement about $B$ is similar). We want $\angle DAB + \angle BAD' < \pi$. We have $\angle DAB < \angle CAD$ and $\angle BAD' = \text{the original } \angle BAD \text{ from the tetrahedron}$. Since both $\angle CAD, \angle BAD$ are acute, it follows that $\angle DAB + \angle BAD' < \pi$.

The last thing to do is to compute $ZZ'$. We have $ZZ' = CC'$, and from the isosceles triangle $\triangle CAC'$ we have $CM = \sin \angle CAM$, where $M$ is the midpoint of $CC'$. This gives $ZZ' = CC' = 2 AC \sin \frac{\alpha}{2}$, the formula we wanted.

[Solution by pf02, December 2024]


See Also

1971 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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