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

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Since <math>\theta < \frac{\pi}{2}</math>, then <math>\angle APC + \angle BPD < \pi</math>
 
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 > \pi</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}}
 
{{alternate solutions}}
  

Revision as of 21:11, 12 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 $\angle APC + \angle BPD > \pi$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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