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Geometry Difficulty 7.6 National Olympiad, round 2 Find the answer

PP , AA , BB , CC , and DD are five distinct points in space such that ∠APB=∠BPC=∠CPD=∠DPA=θ\angle APB = \angle BPC = \angle CPD = \angle DPA = \theta , where θ\theta is a given acute angle. Determine the greatest and least values of ∠APC+∠BPD\angle APC + \angle BPD .

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solutions — 2

Solution 1

Greatest value is achieved when all the points are as close as possible to all being on a plane.
Since θ<π2\theta < \frac{\pi}{2} , then ∠APC+∠BPD<π\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 θ>0\theta > 0 , then ∠APC+∠BPD>0\angle APC + \angle BPD > 0
and the inequality for this problem is:
0<∠APC+∠BPD<π0 < \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.

Solution 2

Consider five distinct points P,A,B,C, P, A, B, C, and D D in space where the angles formed at P P satisfy ∠APB=∠BPC=∠CPD=∠DPA=θ \angle APB = \angle BPC = \angle CPD = \angle DPA = \theta . We are tasked with finding the greatest and least possible values of the sum of angles ∠APC+∠BPD \angle APC + \angle BPD .

### Analyzing the Geometry

Since each angle ∠APB,∠BPC,∠CPD,∠DPA\angle APB, \angle BPC, \angle CPD, \angle DPA equals θ\theta, it suggests some symmetrical arrangement around point P P . One way to visualize this is by considering a circular arrangement with equal angles subtended at the center by these points A,B,C, A, B, C, and D D .

1. **Sum Around Point P P :**
The total sum of angles around point P P should be 360∘ 360^\circ . Therefore, if the points A,B,C, A, B, C, and D D symmetrically divide the plane or sphere around P P , these angles ensure all points maintain the respective θ \theta .

2. **Possible Values of ∠APC+∠BPD \angle APC + \angle BPD :**
To determine ∠APC+∠BPD \angle APC + \angle BPD , consider:

- The configuration can be manipulated by changing the relative position of A,B,C, A, B, C, and D D . These could shift as vectors with fixed directions but different initial points or offsets around P P , all still maintaining the equal angles with neighboring vectors.

- The least configuration for ∠APC+∠BPD\angle APC + \angle BPD is when APC APC and BPD BPD form an overlapping or complementary pair within the same plane, logically resulting in their sum being near zero. Hence, the minimum is 0∘ 0^\circ .

- The greatest sum occurs if the paths from P P create no overlap with 180∘ 180^\circ other paths, enclosing an entire circular path without swaps or overlaps between them. Ideally, they may be manipulated within separate contiguous quadrants or arrangements that maximize separation, reflecting back upon each other to ensure 360∘ 360^\circ .

### Conclusion

Through spatial manipulation respecting the given θ\theta, ∠APC+∠BPD \angle APC + \angle BPD can range from a state where sum of zero superposition exists (collapsing the enclosing angle) to a fully rotational backtrack forming the maximum cycle without intersection points. Thus, the least and greatest values of ∠APC+∠BPD\angle APC + \angle BPD are:

0∘ and 360∘. \boxed{0^\circ} \text{ and } \boxed{360^\circ}.

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Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.