1971 IMO Problems/Problem 4

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 $AB, A'B'$ .

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.

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. Clearly $ZYXTZ'$ 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 $ZYXTZ'$ .

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.

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.

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

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

Page

Toolbox

Search

1971 IMO Problems/Problem 4

Contents

Problem

Solution

Remarks (added by pf02, December 2024)

Solution 2

See Also