The sum of the measures of all the face angles of a given complex polyhedral angle is equal to the sum of all its dihedral angles. Prove that the polyhedral angle is a trihedral angle. $\mathbf{Note:}$ A convex polyhedral angle may be formed by drawing rays from an exterior point to all points of a convex polygon.
Problem
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Tags: AMC, USA(J)MO, USAMO, geometry unsolved, geometry
adhikariprajitraj
16.03.2018 16:41
$n = 3$ is certainly possible. For example, take $\angle {APB} = \angle {APC} = \angle {BPC} = 90^{o}$ (so that the lines $PA, PB, PC$ are mutually perpendicular). Then the three planes through P are also mutually perpendicular, so the two sums are both $270{o}$.
We show that $n > 3$ is not possible.
The sum of the n angles $APB$ etc at $P$ is less than $360^o$. This is almost obvious. Take another plane which meets the lines $PA, PB, PC$ etc at $A', B', C', ...$ and so that the foot of the perpendicular from P to the plane lies inside the n-gon $A'B'C' ...$ then as we move P down the perpendicular the angles $A'PB'$ etc all increase. But when it reaches the plane their sum is $360^o$.
However, I do not immediately see how to make that rigorous. Instead, take any point O inside the n-gon ABC... . We have ∠PBA + ∠PBC > ∠ABC.
Adding the n such equations we get $\sum(180^o - APB) > \sum ABC = (n - 2) 180^o$. So, $\sum APB < 360o$.
The sum of the n angles between the planes is at least $(n - 2)*180^o$. If we take a sphere center $P$. Then the lines $PA, PB$ intersect it at $A", B", ...$ which form a spherical polygon. The angles of this polygon are the angles between the planes. We can divide the polygon into $n - 2$ triangles. The angles in a spherical triangle sum to at least $180^o$. So the angles in the spherical polygon are at least $(n - 2) 180^o$. So we have $(n - 2) 180^o < 360^o$ and hence $n < 4$.
This solution is copied from the link.Nice solution.
lakecomo224
16.03.2018 16:43
Aw...You beat my bump record
adhikariprajitraj
17.03.2018 07:47
Solution done!
adhikariprajitraj
17.03.2018 08:21
lllllllllll
adhikariprajitraj
18.03.2018 16:03
Bump!!!!!!!!!
Vrangr
18.03.2018 16:07
Calm down. https://mks.mff.cuni.cz/kalva/usa/usoln/usol814.html