Difference between revisions of "1971 IMO Problems/Problem 4"

Revision as of 01:53, 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)

$\mathbf{Remark\ 1.}$ The problem is incorrect. More precisely, question (a) is correct, but both statements of question (b) are incorrect. The equality $\angle DAB + \angle BCD = \angle CDA + \angle ABC$ does not imply that there are any shortest polygonal paths, and when such shortest polygonal paths exist, their common length $\neq 2AC \sin(\alpha / 2)$ (where $\alpha = \angle BAC + \angle CAD + \angle DAB$ ).

We will give a counterexample to the problem, making it clear that (b) is wrong. Below, we will give a solution which proves statement (a), and which answers the following questions:

(b.1) Give conditions in terms of angles and sides of the tetrahedron when shortest polygonal paths exist, and show that in this case infinitely may such paths exist.

(b.2) Give a formula (in terms of angles and sides of the tetrahedron) for the length of the shortest polygonal path, when it exists.

$\mathbf{Remark\ 2.}$ The "solution" above is 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 is clearly incorrect, since it claims to prove something which is not true. But it is incomplete/incorrect even where it should make sense.

$\mathbf{Remark\ 3.}$ Below I will give the basic idea for solutions to this problem, and a counterexample to part (b) of the problem.

$\mathbf{Remark\ 4.}$ Then, I will give a correct solution, proving part (a) and the modified parts (b.1) and (b.2) of the problem. During the proof, I will take a break to highlight the likely error made by the author of the problem.

Basic idea of a solution, and counterexample

The basic idea in both the attempted solution above and in the solution below is to "fold out" the tetrahedron into a polygon in the plane. The path $XYZTX$ becomes a collection of connected segments with ends on two line segments.

Specifically, we rotate three faces around $BC$ so $\triangle BCD$ is in the same plane as $\triangle ABC$ . Then we rotate two faces around $CD$ so that $\triangle CDA$ is in the same plane with the previous triangles. Finally, we rotate $\triangle ABD$ around $AD$ , so that it is in the same plane with the other triangles. We denote $A', B', X'$ the new points corresponding to $A, B, X$ , to avoid confusion in the notation.

The polygonal path $XYZTX$ becomes $XYTZX'$ . It is clear that in order to minimize $XYZTX$ , we should make $XYTZX'$ be a segment o 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 mirror 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.]

@@ Line 62: / Line 62: @@
 below is to "fold out" the tetrahedron into a polygon in the plane.
 The path <math>XYZTX</math> becomes a collection of connected segments with
-one ends on two line segments.
+ends on two line segments.
 [[File:prob_1971_4_1.png|600px]]
+Specifically, we rotate three faces around <math>BC</math> so <math>\triangle BCD</math> is
+in the same plane as <math>\triangle ABC</math>.  Then we rotate two faces around
+<math>CD</math> so that <math>\triangle CDA</math> is in the same plane with the previous
+triangles.  Finally, we rotate <math>\triangle ABD</math> around <math>AD</math>, so that
+it is in the same plane with the other triangles.  We denote
+<math>A', B', X'</math> the new points corresponding to <math>A, B, X</math>, to avoid
+confusion in the notation.
+The polygonal path <math>XYZTX</math> becomes <math>XYTZX'</math>.  It is clear that in order
+to minimize <math>XYZTX</math>, we should make <math>XYTZX'</math> be a segment o a straight
+line.  Furthermore, to minimize the segment <math>XX'</math>, we want to choose
+<math>X \in AB</math> so that when we draw the line segment to its mirror image
+<math>X' \in A'B'</math>, the length of <math>XX'</math> is as short as possible.

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