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

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All the faces of tetrahedron ABCD are acute-angled triangles. We consider
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==Problem==
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all closed polygonal paths of the form XY ZTX defined as follows: X is a
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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:
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point on edge AB distinct from A and B; similarly, Y;Z; T are interior points
 
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of edges BCCD;DA; respectively. Prove:
 
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(a) If  DAB +  BCD is not equal to  CDA +  ABC; then among the polygonal paths,
 
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there is none of minimal length.
 
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(b) If DAB + BCD =  CDA + ABC; then there are infinitely many
 
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shortest polygonal paths, their common length being 2AC sin(®=2); where
 
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® =  BAC + CAD +  DAB.
 
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(triads of three letters represent angles, except ZTX)
 
  
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{{solution}}
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(a) If <math>\angle DAB + \angle BCD \neq \angle CDA + \angle ABC</math>, then among the polygonal paths, there is none of minimal length.
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(b) If <math>\angle DAB + \angle BCD = \angle CDA + \angle ABC</math>, then there are infinitely many shortest polygonal paths, their common length being <math>2AC \sin(\alpha / 2)</math>, where <math>\alpha = \angle BAC + \angle CAD + \angle DAB</math>.
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==Solution==
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Rotate the triangle <math>BCD</math> around the edge <math>BC</math> until <math>ABCD</math> are in one plane. It is clear that in a shortest path, the point Y lies on the line connecting <math>X</math> and <math>Z</math>. Therefore, <math>XYB=ZYC</math>.
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Summing the four equations like this, we get exactly <math>\angle ABC+\angle ADC=\angle BCD+\angle BAD</math>.
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Now, draw all four faces in the plane, so that <math>BCD</math> is constructed on the exterior of the edge <math>BC</math> of <math>ABC</math> and so on with edges <math>CD</math> and <math>AD</math>.
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The final new edge <math>AB</math> (or rather <math>A'B'</math>) is parallel to the original one (because of the angle equation). Call the direction on <math>AB</math> towards <math>B</math> "right" and towards <math>A</math> "left". If we choose a vertex <math>X</math> on <math>AB</math> and connect it to the corresponding vertex <math>X'</math> on A'B'. This works for a whole interval of vertices <math>X</math> if <math>C</math> lies to the left of <math>B</math> and <math>D</math> and <math>D</math> lies to the right of <math>A</math>. It is not hard to see that these conditions correspond to the fact that various angles are acute by assumption.
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Finally, regard the sine in half the isosceles triangle <math>ACA'</math> which gives the result with the angles around <math>C</math> instead of <math>A</math>, but the role of the vertices is symmetric.
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== See Also == {{IMO box|year=1971|num-b=3|num-a=5}}
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:3D Geometry Problems]]
 
[[Category:3D Geometry Problems]]

Latest revision as of 13:02, 29 January 2021

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.

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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