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Difference between revisions of "1971 IMO Problems/Problem 4"
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The solution above is incomplete and/or incorrect.  The first part
 
The solution above is incomplete and/or incorrect.  The first part
 
(which claims to prove part (a) of the problem) is incomplete at best,
 
(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
+
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
+
incorrect.) The second part, which claims to prove part (b) of the
−
problem skips too many steps.  Some of the arguments in the proof
+
problem, skips too many steps.  Some of the arguments in the proof
−
(e.g. the fact that <math>AB \parallel A'B'</math> are true but need proof.
+
(e.g. the fact that <math>AB \parallel A'B'</math>) are true but need proof.
−
I could make no sense of other arguments in the proof, so I can
+
I could not make any sense of other arguments in the proof, so I
−
not judge whether they are correct or not (I suspect not).
+
can not judge whether they are correct or not (I suspect not).
  
−
I will give a solution along the same basic idea.
+
I will give a robust solution below.  It goes along the same basic
 +
idea.
  
  
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[[File:prob_1971_4_1.png|600px]]
 
[[File:prob_1971_4_1.png|600px]]
  
−
Specifically, we rotate three faces around <math>BC</math> so <math>\triangle BCD</math> is
+
Specifically, we rotate the solid consisting of three faces around
−
in the same plane as <math>\triangle ABC</math>. Then we rotate two faces around  
+
<math>BC</math> so <math>\triangle BCD</math> is in the same plane as <math>\triangle ABC</math>.
−
<math>CD</math> so that <math>\triangle CDA</math> is in the same plane with the previous
+
Then we rotate the solid consisting of two faces around <math>CD</math> so
−
triangles.  Finally, we rotate <math>\triangle ABD</math> around <math>AD</math>, so that
+
that <math>\triangle CDA'</math> is in the same plane with the previous
−
it is in the same plane with the other triangles.  We denote
+
triangles. (We denoted <math>A'</math> the new "copy" of <math>A</math>, since the
−
<math>A', B', X'</math> the new points corresponding to <math>A, B, X</math>, to avoid
+
original <math>A</math> is being used in the picture.) Finally, we rotate
−
confusion in the notation.
+
<math>\triangle A'BD</math> around <math>A'D</math>, so that it is in the same plane
 +
with the other triangles.  (We denote <math>B'</math> the new "copy" of <math>B</math>,
 +
and <math>X'</math> the new copy of <math>X</math>.)
  
 
The polygonal path <math>XYZTX</math> becomes <math>XYZTX'</math>.  It is clear that in order
 
The polygonal path <math>XYZTX</math> becomes <math>XYZTX'</math>.  It is clear that in order

Revision as of 03:04, 28 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.




[TO BE CONTINUED. SAVING MID WAY SO I DON"T LOSE WORK DONE SO FAR.]


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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