At least an orthogonal projection inside a side

Question

Let $P$ a point inside a convex n-agon and let $P_1, P_2, ..., P_n$ the ortogonal projections of $P$ on the sides of the n-agon. How can I show that at least one of these projections lies inside a side of the poligon? I tried to prove that a convex n-agon is dibisible in $n$ triangle with $P$ and two vertex of the n-agone as vertex with all the angles $\leq 90°$ but I failed... any advice/solution? Thanks :)

One easy way is work with a triangle, and for general polygon $P_1P_2\dots P_n$, divide it into $n-1$ triangles. — Quang Hoang, Dec 11 '18 at 17:19
How do you divide the polygon? I didn't understand your construction — Lance, Dec 11 '18 at 18:51
Ah, never mind, it seems I was trying to solve another problem. — Quang Hoang, Dec 11 '18 at 18:55

score 2 · Accepted Answer · answered Dec 13 '18 at 22:32

2

Assume that's not the case, and each orthogonal projection is outside the polygon.

Let $A$ be the closest polygon vertex to $P$, and let $B$ be a neighboring vertex. Let's find the locus points such that $P_1 = \text{proj}_{AB}(P) \notin AB$.

There are two possibilities.

$\angle PAB > \frac{\pi}{2}$. Construct a line $g$ perpendicular to $PA$ through $A$. Then $B$ must be in the half-plane bordered by $g$ not containing $X$.
$\angle PBA > \frac{\pi}{2}$. In this case B is inside the circle with diameter PA. However, since $A$ was chosen to be the vertex closest to $P$, this is impossible.

Finally, $A$ has two neighbors. However, having both of then in the half-plane described below line $g$ violates convexity, a contradiction.

answered Dec 13 '18 at 22:32

Todor Markov

2,869

In your demonstration I see more or less a red line, but, pardon me, there is a lot of imprecision : I don't understand why angle PAB $> \pi/2$ and angle PBA $> \pi/2$ are exclusive one from the other, and why $C$ would be necessarily on the exterior side of g because it an not belong to the interior of the circle, etc... – Jean Marie Dec 15 '18 at 06:18
@JeanMarie Angle PAB and PBA can't both be obtuse, as triangle PAB can't have two obtuse angles. For the rest, the same logic that applies to B applies to C just as well. I'd be happy to clarify parts of the argument, but I'm not sure what is unclear. – Todor Markov Dec 15 '18 at 09:47
1

But why couldn't they both be acute ? – Jean Marie Dec 15 '18 at 09:54
@JeanMarie Then the orthogonal projection of X on AB would be on the side AB, contradiction the assumption. – Todor Markov Dec 15 '18 at 10:17
1

I begin to understand. Thanks ! – Jean Marie Dec 15 '18 at 10:18

At least an orthogonal projection inside a side

1 Answers1