Difference between revisions of "1984 USAMO Problems/Problem 3"

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== Solution ==
 
== Solution ==
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Greatest value is achieved when all the points are as close as possible to all being on a plane.
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Since <math>\theta < \frac{\pi}{2}</math>, then <math>\angle APC + \angle BPD < \pi</math>
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Smallest value is achieved when point P is above and the remaining points are as close as possible to colinear when <math>\theta > 0</math>, then <math>\angle APC + \angle BPD > 0</math>
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and the inequality for this problem is:
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<math>0 < \angle APC + \angle BPD < \pi</math>
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~Tomas Diaz. orders@tomasdiaz.com
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{{alternate solutions}}
  
 
== See Also ==
 
== See Also ==
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[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
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[[Category:3D Geometry Problems]]

Latest revision as of 02:26, 21 November 2023

Problem

$P$, $A$, $B$, $C$, and $D$ are five distinct points in space such that $\angle APB = \angle BPC = \angle CPD = \angle DPA = \theta$, where $\theta$ is a given acute angle. Determine the greatest and least values of $\angle APC + \angle BPD$.

Solution

Greatest value is achieved when all the points are as close as possible to all being on a plane.

Since $\theta < \frac{\pi}{2}$, then $\angle APC + \angle BPD < \pi$

Smallest value is achieved when point P is above and the remaining points are as close as possible to colinear when $\theta > 0$, then $\angle APC + \angle BPD > 0$

and the inequality for this problem is:

$0 < \angle APC + \angle BPD < \pi$

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1984 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions

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