At least an ortogonal projection inside a side
$begingroup$
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 :)
geometry
asked Dec 11 '18 at 16:58
LanceLance
10112
$endgroup$
add a comment |
$begingroup$
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 :)
geometry
asked Dec 11 '18 at 16:58
LanceLance
10112
$endgroup$
$begingroup$
One easy way is work with a triangle, and for general polygon $P_1P_2dots P_n$, divide it into $n-1$ triangles.
$endgroup$
– Quang Hoang
Dec 11 '18 at 17:19
$begingroup$
How do you divide the polygon? I didn't understand your construction
$endgroup$
– Lance
Dec 11 '18 at 18:51
$begingroup$
Ah, never mind, it seems I was trying to solve another problem.
$endgroup$
– Quang Hoang
Dec 11 '18 at 18:55
add a comment |
$begingroup$
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 :)
geometry
asked Dec 11 '18 at 16:58
LanceLance
10112
$endgroup$
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 :)
geometry
geometry
asked Dec 11 '18 at 16:58
LanceLance
10112
asked Dec 11 '18 at 16:58
LanceLance
10112
asked Dec 11 '18 at 16:58
LanceLance
10112
asked Dec 11 '18 at 16:58
LanceLance
10112
asked Dec 11 '18 at 16:58
LanceLance
10112
10112
$begingroup$
One easy way is work with a triangle, and for general polygon $P_1P_2dots P_n$, divide it into $n-1$ triangles.
$endgroup$
– Quang Hoang
Dec 11 '18 at 17:19
$begingroup$
How do you divide the polygon? I didn't understand your construction
$endgroup$
– Lance
Dec 11 '18 at 18:51
$begingroup$
Ah, never mind, it seems I was trying to solve another problem.
$endgroup$
– Quang Hoang
Dec 11 '18 at 18:55
add a comment |
$begingroup$
One easy way is work with a triangle, and for general polygon $P_1P_2dots P_n$, divide it into $n-1$ triangles.
$endgroup$
– Quang Hoang
Dec 11 '18 at 17:19
$begingroup$
How do you divide the polygon? I didn't understand your construction
$endgroup$
– Lance
Dec 11 '18 at 18:51
$begingroup$
Ah, never mind, it seems I was trying to solve another problem.
$endgroup$
– Quang Hoang
Dec 11 '18 at 18:55
$begingroup$
One easy way is work with a triangle, and for general polygon $P_1P_2dots P_n$, divide it into $n-1$ triangles.
$endgroup$
– Quang Hoang
Dec 11 '18 at 17:19
$begingroup$
One easy way is work with a triangle, and for general polygon $P_1P_2dots P_n$, divide it into $n-1$ triangles.
$endgroup$
– Quang Hoang
Dec 11 '18 at 17:19
$begingroup$
How do you divide the polygon? I didn't understand your construction
$endgroup$
– Lance
Dec 11 '18 at 18:51
$begingroup$
How do you divide the polygon? I didn't understand your construction
$endgroup$
– Lance
Dec 11 '18 at 18:51
$begingroup$
Ah, never mind, it seems I was trying to solve another problem.
$endgroup$
– Quang Hoang
Dec 11 '18 at 18:55
$begingroup$
Ah, never mind, it seems I was trying to solve another problem.
$endgroup$
– Quang Hoang
Dec 11 '18 at 18:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 MarkovTodor Markov
2,420412
$endgroup$
$begingroup$
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...
$endgroup$
– Jean Marie
Dec 15 '18 at 6:18
$begingroup$
@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.
$endgroup$
– Todor Markov
Dec 15 '18 at 9:47
1
$begingroup$
But why couldn't they both be acute ?
$endgroup$
– Jean Marie
Dec 15 '18 at 9:54
$begingroup$
@JeanMarie Then the orthogonal projection of X on AB would be on the side AB, contradiction the assumption.
$endgroup$
– Todor Markov
Dec 15 '18 at 10:17
1
$begingroup$
I begin to understand. Thanks !
$endgroup$
– Jean Marie
Dec 15 '18 at 10:18
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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 MarkovTodor Markov
2,420412
$endgroup$
$begingroup$
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...
$endgroup$
– Jean Marie
Dec 15 '18 at 6:18
$begingroup$
@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.
$endgroup$
– Todor Markov
Dec 15 '18 at 9:47
1
$begingroup$
But why couldn't they both be acute ?
$endgroup$
– Jean Marie
Dec 15 '18 at 9:54
$begingroup$
@JeanMarie Then the orthogonal projection of X on AB would be on the side AB, contradiction the assumption.
$endgroup$
– Todor Markov
Dec 15 '18 at 10:17
1
$begingroup$
I begin to understand. Thanks !
$endgroup$
– Jean Marie
Dec 15 '18 at 10:18
add a comment |
$begingroup$
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 MarkovTodor Markov
2,420412
$endgroup$
$begingroup$
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...
$endgroup$
– Jean Marie
Dec 15 '18 at 6:18
$begingroup$
@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.
$endgroup$
– Todor Markov
Dec 15 '18 at 9:47
1
$begingroup$
But why couldn't they both be acute ?
$endgroup$
– Jean Marie
Dec 15 '18 at 9:54
$begingroup$
@JeanMarie Then the orthogonal projection of X on AB would be on the side AB, contradiction the assumption.
$endgroup$
– Todor Markov
Dec 15 '18 at 10:17
1
$begingroup$
I begin to understand. Thanks !
$endgroup$
– Jean Marie
Dec 15 '18 at 10:18
add a comment |
$begingroup$
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 MarkovTodor Markov
2,420412
$endgroup$
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 MarkovTodor Markov
2,420412
answered Dec 13 '18 at 22:32
Todor MarkovTodor Markov
2,420412
answered Dec 13 '18 at 22:32
Todor MarkovTodor Markov
2,420412
answered Dec 13 '18 at 22:32
Todor MarkovTodor Markov
2,420412
2,420412
$begingroup$
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...
$endgroup$
– Jean Marie
Dec 15 '18 at 6:18
$begingroup$
@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.
$endgroup$
– Todor Markov
Dec 15 '18 at 9:47
1
$begingroup$
But why couldn't they both be acute ?
$endgroup$
– Jean Marie
Dec 15 '18 at 9:54
$begingroup$
@JeanMarie Then the orthogonal projection of X on AB would be on the side AB, contradiction the assumption.
$endgroup$
– Todor Markov
Dec 15 '18 at 10:17
1
$begingroup$
I begin to understand. Thanks !
$endgroup$
– Jean Marie
Dec 15 '18 at 10:18
add a comment |
$begingroup$
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...
$endgroup$
– Jean Marie
Dec 15 '18 at 6:18
$begingroup$
@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.
$endgroup$
– Todor Markov
Dec 15 '18 at 9:47
1
$begingroup$
But why couldn't they both be acute ?
$endgroup$
– Jean Marie
Dec 15 '18 at 9:54
$begingroup$
@JeanMarie Then the orthogonal projection of X on AB would be on the side AB, contradiction the assumption.
$endgroup$
– Todor Markov
Dec 15 '18 at 10:17
1
$begingroup$
I begin to understand. Thanks !
$endgroup$
– Jean Marie
Dec 15 '18 at 10:18
$begingroup$
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...
$endgroup$
– Jean Marie
Dec 15 '18 at 6:18
$begingroup$
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...
$endgroup$
– Jean Marie
Dec 15 '18 at 6:18
$begingroup$
@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.
$endgroup$
– Todor Markov
Dec 15 '18 at 9:47
$begingroup$
@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.
$endgroup$
– Todor Markov
Dec 15 '18 at 9:47
1
1
$begingroup$
But why couldn't they both be acute ?
$endgroup$
– Jean Marie
Dec 15 '18 at 9:54
$begingroup$
But why couldn't they both be acute ?
$endgroup$
– Jean Marie
Dec 15 '18 at 9:54
$begingroup$
@JeanMarie Then the orthogonal projection of X on AB would be on the side AB, contradiction the assumption.
$endgroup$
– Todor Markov
Dec 15 '18 at 10:17
$begingroup$
@JeanMarie Then the orthogonal projection of X on AB would be on the side AB, contradiction the assumption.
$endgroup$
– Todor Markov
Dec 15 '18 at 10:17
1
1
$begingroup$
I begin to understand. Thanks !
$endgroup$
– Jean Marie
Dec 15 '18 at 10:18
$begingroup$
I begin to understand. Thanks !
$endgroup$
– Jean Marie
Dec 15 '18 at 10:18
add a comment |
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$begingroup$
One easy way is work with a triangle, and for general polygon $P_1P_2dots P_n$, divide it into $n-1$ triangles.
$endgroup$
– Quang Hoang
Dec 11 '18 at 17:19
$begingroup$
How do you divide the polygon? I didn't understand your construction
$endgroup$
– Lance
Dec 11 '18 at 18:51
$begingroup$
Ah, never mind, it seems I was trying to solve another problem.
$endgroup$
– Quang Hoang
Dec 11 '18 at 18:55