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

Latest revision as of 00:25, 31 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 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 $AB \parallel A'B'$ ) 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, 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. (In this case there is a lower limit for $ZZ'$ , namely $DD'$ , which is not a minimum.)

On the other hand, if $CD \parallel C'D'$ (see pictures below) then generally 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. (In this case, $ZZ' = CC' = DD'$ is a lower limit, and it is 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 $\angle D'AB + \angle BCD = \angle C'D'A + \angle ABC$ , and then it becomes $\angle D'AN + \angle NAB + \angle BCD = \angle C'D'P + \angle PD'A + \angle ABM + \angle MBC$ . 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 $ZZ'$ of the same size (see Figure 2 below).

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 $ZZ' = CC' = 2 AC \sin \frac{\alpha}{2}$ , which is the formula we wanted.

[Solution by pf02, December 2024]

@@ Line 1: / Line 1: @@
 ==Problem==
 All the faces of tetrahedron <math>ABCD</math> are acute-angled triangles. We consider all closed polygonal paths of the form <math>XYZTX</math> defined as follows: <math>X</math> is a point on edge <math>AB</math> distinct from <math>A</math> and <math>B</math>; similarly, <math>Y, Z, T</math> are interior points of edges <math>BC, CD, DA</math>, respectively. Prove:
@@ Line 20: / Line 21: @@
 ==Remarks (added by pf02, December 2024)==
-<math>\mathbf{Remark 1.}</math> The problem is incorrect.  More precisely,
+The solution above is incomplete, and very likely, incorrect.  The first
-question (a) is correct, but both statements of question (b) are
+part (which claims to prove part (a) of the problem) is incomplete at
-incorrect.  The equality
+best, since it is not clear how it leads to the result.  (Very likely
-<math>\angle DAB + \angle BCD = \angle CDA + \angle ABC</math> does not imply
+it is incorrect.)  The second part, which claims to prove part (b) of
-that there are any shortest polygonal paths, and when such shortest
+the problem, skips too many steps.  Some of the arguments in the proof
-polygonal paths exist, their common length
+(e.g. the fact that <math>AB \parallel A'B'</math>) are true but need proof.
-<math>\neq 2AC \sin(\alpha / 2)</math> (where
+I could not make any sense of other arguments in the proof, so I
-<math>\alpha = \angle BAC + \angle CAD + \angle DAB</math>).
+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 <math>XYZTX</math> becomes a collection of connected
+segments <math>XY, YZ, ZT, TX'</math> with the unconnected ends <math>X, X'</math> on
+the two line segments <math>AB, A'B'</math>.
-We will give a counterexample to the problem, making it clear that
+[[File:prob_1971_4_1.png|600px]]
-it 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
+Specifically, we rotate the solid consisting of three faces around
-when exactly one shortest polygonal path exists.
+<math>BC</math> so <math>\triangle BCD</math> is in the same plane as <math>\triangle ABC</math>.
+Then we rotate the solid consisting of two faces around  <math>CD</math> so
+that <math>\triangle CDA'</math> is in the same plane with the previous
+triangles.  (We denoted <math>A'</math> the new "copy" of <math>A</math>, since the
+original <math>A</math> is being used in the picture.)  Finally, we rotate
+<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>.)
-(b.2) Give conditions in terms of angles and sides of the tetrahedron
+The polygonal path <math>XYZTX</math> becomes <math>XYZTX'</math>.  It is clear that in order
-when infinitely many shortest polygonal paths exist.
+to minimize <math>XYZTX</math>, we should make <math>XYZTX'</math> be a segment on 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 corresponding
+image <math>X' \in A'B'</math>, the length of <math>XX'</math> is as short as possible.
-(b.3) Show that (b.1) and (b.2) cover all possible cases.
+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 <math>\triangle BCD</math> around
+<math>BC</math>, then we rotate <math>\triangle ABD</math> around <math>AB</math> to obtain
+<math>\triangle ABD'</math>, and finally, we rotate <math>\triangle AD'C</math> around
+<math>AD'</math> to obtain <math>\triangle AD'C'</math>.  Now <math>Z \in CD</math> becomes
+<math>Z' \in C'D'</math>.  The polygonal path becomes <math>ZYXTZ'</math> going from
+<math>CD</math> to <math>C'D'</math>.  We need to make this as small as possible, to
+find its minimum, if it exists.
-(b.4) Give a formula (in terms of angles and sides of the tetrahedron)
+[[File:prob_1971_4_2.png|300px]]
-for the length of the shortest polygonal path, when it exists.
-<math>\mathbf{Remark 2.}</math> The "solution" above is incorrect.  The first part
+The idea of the solution to the problem is now easy to explain.
-(which claims to prove part (a) of the problem) is incomplete at best,
+First of all, <math>ZYXTZ'</math> needs to be a segment on a straight line,
-since it is not clear how it leads to the result.  Very likely it is
+in which case its length is the length of the segment <math>ZZ'</math>.
-incorrect.  The second part, which claims to prove part (b) of the
+Clearly <math>ZYXTZ' = ZZ'</math> has a lower bound (after all, it is <math>> 0</math>),
-problem is clearly incorrect, since it claims to prove something which
+so we can think about its lower limit.  If there is a position of
-is not true.  But it is incomplete/incorrect even where it should make
+<math>Z</math> in which this lower limit is achieved, then this lower limit
-sense.
+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.)
-<math>\mathbf{Remark 3.}</math> Below I will give the basic idea of the solution
+In the picture above, <math>CD</math> and <math>C'D'</math> are not parallel.  The segment
-from above, and a counterexample to part (b) of the problem.
+<math>ZZ'</math> would be shortest when <math>Z = D</math> (and <math>Z' = D'</math>).  But this is
+not an acceptable position for <math>ZYXTZ'</math>, because the problem stated
+that <math>Z</math> is between <math>C</math> and <math>D</math>, 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 <math>ZZ'</math>, namely <math>DD'</math>, which is not a
+minimum.)
-<math>\mathbf{Remark 4.}</math> Then, I will give a correct solution, proving part
+On the other hand, if <math>CD \parallel C'D'</math> (see pictures below) then
-(a) and the modified parts (b.1) - (b.4) of the problem.  During the
+generally there are lots of points <math>Z \in CD</math> yielding a minimum
-proof, I will take a break to highlight the likely error made by the
+value for <math>ZYXTZ'</math>.  Indeed, in this case
-author of the problem.
+<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 respective segments.
+(In this case, <math>ZZ' = CC' = DD'</math> is a lower limit, and it is a
+minimum.)
+We will prove that <math>\angle DAB + \angle BCD = \angle CDA + \angle ABC</math>
+if and only if <math>CD \parallel C'D'</math>.  After this it will be easy to
+deduce all the statements of the problem.
-==Basic idea of a solution, and counterexample==
+Let <math>M, N, P</math> be on <math>CA, BD', AC'</math> be such that
+<math>CD \parallel MB \parallel AN \parallel PD'</math>.  The equality
+<math>\angle DAB + \angle BCD = \angle CDA + \angle ABC</math> becomes
+<math>\angle D'AB + \angle BCD = \angle C'D'A + \angle ABC</math>, and then it
+becomes <math>\angle D'AN + \angle NAB + \angle BCD =
+\angle C'D'P + \angle PD'A + \angle ABM + \angle MBC</math>.  Because of the
+parallelism of <math>CD, MB, AN, PD'</math> we have several equal angles on the
+two sides of this equality.  This equality becomes <math>\angle C'D'P = 0</math>.
+So, the original equality <math>\angle DAB + \angle BCD = \angle CDA + \angle ABC</math>
+is true if and only if <math>\angle C'D'P = 0</math>, which is true if and only if
+<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 <math>ZZ'</math> of the same size (see Figure 2 below).
+[[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>, which is the formula we wanted.
-[TO BE CONTINUED.  SAVING MID WAY SO I DON"T LOOS WORK DONE SO FAR.]
+[Solution by pf02, December 2024]

1971 IMO (Problems) • Resources
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