How many maximum triangles can be made?












7












$begingroup$


There are $8$ points on a plane no three are colinear how many maximum triangles can be made s.t no two triangles have more than one point in common.



Now I can choose $3$ points from $8$ points in $^8C_3$ ways and two triangles can have two points in common if I choose $5$ points make $2$ triangles out of it and that can be made in $^8C_6 times ^6C_3 times frac 12$ ways. So the answer should be $^8C_3-(^8C_5 times ^5C_3times frac 12)$.



Am I double counting anything?



After seeing one comment and thinking a bit I feel that method of complementation will be harder here and I am thinking about how many ways to draw a triangle instead of maximum how many triangles,



So another approach: I can choose three points from $8$ points and draw the 1st triangle then the second triangle can be drawn taking one point from the first(because we are maximizing) and $2$ others from the remaining $5$ points. So we have used $5$ points and drew $2$ triangles. Then we can draw atmost one more triangle. So $3$ is the answer.






combinatorics discrete-mathematics proof-verification graph-theory contest-math







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Seeing as your answer is a negative number, probably not.
    $endgroup$
    – bof
    Jan 4 at 5:08










  • $begingroup$
    I am double counting something then...
    $endgroup$
    – Gimgim
    Jan 4 at 5:09










  • $begingroup$
    Ah - what's the question? How many ways to do it (whatever it is)? The maximum number of triangles we can fit in under these rules? If its' the latter, the answer is certainly more than three - it only takes six points for four triangles, based on alternating faces of an octahedron.
    $endgroup$
    – jmerry
    Jan 4 at 5:40










  • $begingroup$
    @jmerry And $7$ points for $7$ triangles, and $9$ points for $12$ triangles, Steiner triple systems.
    $endgroup$
    – bof
    Jan 4 at 5:49










  • $begingroup$
    What does "can be made" mean?
    $endgroup$
    – DanielV
    Jan 4 at 10:05
















7












$begingroup$


There are $8$ points on a plane no three are colinear how many maximum triangles can be made s.t no two triangles have more than one point in common.



Now I can choose $3$ points from $8$ points in $^8C_3$ ways and two triangles can have two points in common if I choose $5$ points make $2$ triangles out of it and that can be made in $^8C_6 times ^6C_3 times frac 12$ ways. So the answer should be $^8C_3-(^8C_5 times ^5C_3times frac 12)$.



Am I double counting anything?



After seeing one comment and thinking a bit I feel that method of complementation will be harder here and I am thinking about how many ways to draw a triangle instead of maximum how many triangles,



So another approach: I can choose three points from $8$ points and draw the 1st triangle then the second triangle can be drawn taking one point from the first(because we are maximizing) and $2$ others from the remaining $5$ points. So we have used $5$ points and drew $2$ triangles. Then we can draw atmost one more triangle. So $3$ is the answer.






combinatorics discrete-mathematics proof-verification graph-theory contest-math







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Seeing as your answer is a negative number, probably not.
    $endgroup$
    – bof
    Jan 4 at 5:08










  • $begingroup$
    I am double counting something then...
    $endgroup$
    – Gimgim
    Jan 4 at 5:09










  • $begingroup$
    Ah - what's the question? How many ways to do it (whatever it is)? The maximum number of triangles we can fit in under these rules? If its' the latter, the answer is certainly more than three - it only takes six points for four triangles, based on alternating faces of an octahedron.
    $endgroup$
    – jmerry
    Jan 4 at 5:40










  • $begingroup$
    @jmerry And $7$ points for $7$ triangles, and $9$ points for $12$ triangles, Steiner triple systems.
    $endgroup$
    – bof
    Jan 4 at 5:49










  • $begingroup$
    What does "can be made" mean?
    $endgroup$
    – DanielV
    Jan 4 at 10:05














7












7








7





$begingroup$


There are $8$ points on a plane no three are colinear how many maximum triangles can be made s.t no two triangles have more than one point in common.



Now I can choose $3$ points from $8$ points in $^8C_3$ ways and two triangles can have two points in common if I choose $5$ points make $2$ triangles out of it and that can be made in $^8C_6 times ^6C_3 times frac 12$ ways. So the answer should be $^8C_3-(^8C_5 times ^5C_3times frac 12)$.



Am I double counting anything?



After seeing one comment and thinking a bit I feel that method of complementation will be harder here and I am thinking about how many ways to draw a triangle instead of maximum how many triangles,



So another approach: I can choose three points from $8$ points and draw the 1st triangle then the second triangle can be drawn taking one point from the first(because we are maximizing) and $2$ others from the remaining $5$ points. So we have used $5$ points and drew $2$ triangles. Then we can draw atmost one more triangle. So $3$ is the answer.






combinatorics discrete-mathematics proof-verification graph-theory contest-math







share|cite|improve this question











$endgroup$




There are $8$ points on a plane no three are colinear how many maximum triangles can be made s.t no two triangles have more than one point in common.



Now I can choose $3$ points from $8$ points in $^8C_3$ ways and two triangles can have two points in common if I choose $5$ points make $2$ triangles out of it and that can be made in $^8C_6 times ^6C_3 times frac 12$ ways. So the answer should be $^8C_3-(^8C_5 times ^5C_3times frac 12)$.



Am I double counting anything?



After seeing one comment and thinking a bit I feel that method of complementation will be harder here and I am thinking about how many ways to draw a triangle instead of maximum how many triangles,



So another approach: I can choose three points from $8$ points and draw the 1st triangle then the second triangle can be drawn taking one point from the first(because we are maximizing) and $2$ others from the remaining $5$ points. So we have used $5$ points and drew $2$ triangles. Then we can draw atmost one more triangle. So $3$ is the answer.






combinatorics discrete-mathematics proof-verification graph-theory contest-math




combinatorics discrete-mathematics proof-verification graph-theory contest-math






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 6:03







Gimgim

















asked Jan 4 at 5:03









GimgimGimgim

28313




28313








  • 1




    $begingroup$
    Seeing as your answer is a negative number, probably not.
    $endgroup$
    – bof
    Jan 4 at 5:08










  • $begingroup$
    I am double counting something then...
    $endgroup$
    – Gimgim
    Jan 4 at 5:09










  • $begingroup$
    Ah - what's the question? How many ways to do it (whatever it is)? The maximum number of triangles we can fit in under these rules? If its' the latter, the answer is certainly more than three - it only takes six points for four triangles, based on alternating faces of an octahedron.
    $endgroup$
    – jmerry
    Jan 4 at 5:40










  • $begingroup$
    @jmerry And $7$ points for $7$ triangles, and $9$ points for $12$ triangles, Steiner triple systems.
    $endgroup$
    – bof
    Jan 4 at 5:49










  • $begingroup$
    What does "can be made" mean?
    $endgroup$
    – DanielV
    Jan 4 at 10:05














  • 1




    $begingroup$
    Seeing as your answer is a negative number, probably not.
    $endgroup$
    – bof
    Jan 4 at 5:08










  • $begingroup$
    I am double counting something then...
    $endgroup$
    – Gimgim
    Jan 4 at 5:09










  • $begingroup$
    Ah - what's the question? How many ways to do it (whatever it is)? The maximum number of triangles we can fit in under these rules? If its' the latter, the answer is certainly more than three - it only takes six points for four triangles, based on alternating faces of an octahedron.
    $endgroup$
    – jmerry
    Jan 4 at 5:40










  • $begingroup$
    @jmerry And $7$ points for $7$ triangles, and $9$ points for $12$ triangles, Steiner triple systems.
    $endgroup$
    – bof
    Jan 4 at 5:49










  • $begingroup$
    What does "can be made" mean?
    $endgroup$
    – DanielV
    Jan 4 at 10:05








1




1




$begingroup$
Seeing as your answer is a negative number, probably not.
$endgroup$
– bof
Jan 4 at 5:08




$begingroup$
Seeing as your answer is a negative number, probably not.
$endgroup$
– bof
Jan 4 at 5:08












$begingroup$
I am double counting something then...
$endgroup$
– Gimgim
Jan 4 at 5:09




$begingroup$
I am double counting something then...
$endgroup$
– Gimgim
Jan 4 at 5:09












$begingroup$
Ah - what's the question? How many ways to do it (whatever it is)? The maximum number of triangles we can fit in under these rules? If its' the latter, the answer is certainly more than three - it only takes six points for four triangles, based on alternating faces of an octahedron.
$endgroup$
– jmerry
Jan 4 at 5:40




$begingroup$
Ah - what's the question? How many ways to do it (whatever it is)? The maximum number of triangles we can fit in under these rules? If its' the latter, the answer is certainly more than three - it only takes six points for four triangles, based on alternating faces of an octahedron.
$endgroup$
– jmerry
Jan 4 at 5:40












$begingroup$
@jmerry And $7$ points for $7$ triangles, and $9$ points for $12$ triangles, Steiner triple systems.
$endgroup$
– bof
Jan 4 at 5:49




$begingroup$
@jmerry And $7$ points for $7$ triangles, and $9$ points for $12$ triangles, Steiner triple systems.
$endgroup$
– bof
Jan 4 at 5:49












$begingroup$
What does "can be made" mean?
$endgroup$
– DanielV
Jan 4 at 10:05




$begingroup$
What does "can be made" mean?
$endgroup$
– DanielV
Jan 4 at 10:05










2 Answers
2






active

oldest

votes


















7












$begingroup$

I have only a partial answer to your question: the maximum number of triangles is either $8$ or $9$.



You can't have more than $9$ triangles, because there are only $^8C_2=28$ edges determined by the $8$ points, each triangle needs $3$ edges, and no edge may be used by more than one triangle. So the number of triangles is at most $lfloor28/3rfloor=9$.



I don't see how to construct a set of $9$ triangles satisfying your conditions, but I can get $8$. Namely, if we label the points $A,B,C,D,E,F,G,H$, the following $8$ triangles work:
$$ABC, ADG, AFH, BEH, BFG, CDH, CEG, DEH$$



P.S. In fact, $8$ is the maximum. Let $p$ be the number of points (so $p=8$), let $t$ be the number of triangles, and let $n$ be the number of pairs $(P,T)$ where $T$ is a triangle and $P$ is a vertex of $T$. Then $n=3t$ (since each triangle has $3$ vertices), and $nle3p$ (since at most $3$ triangles can contain a given point), so $tle p=8$.



P.P.S. In case you're wondering how I found that set of $8$ triangles, I started with the well-known example of $12$ triangles on $9$ points (Steiner triple system) and deleted one point. Namely, I wrote the letters A to I in a $3times3$ square array, took the $6$ rows and columns and the $6$ "diagonals", and then deleted the ones containing the letter I.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ohhh! yes!! I didn't think in that way
    $endgroup$
    – Gimgim
    Jan 4 at 5:54



















6












$begingroup$

bof gives a great justification of why it is eight or nine with an example of $8$.



Here is why it can't be nine. If there were nine triangles, they would use $27$ points, so one point would have to be used at least $4$ times. Each of these four triangles creates two edges containing this point, so we have at least $8$ edges containing this point. But there are only $7$ other points there must be a duplicate edge.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Right. I didn't notice that you posted this answer while I was typing the P.S. to my answer.
    $endgroup$
    – bof
    Jan 4 at 6:04











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

I have only a partial answer to your question: the maximum number of triangles is either $8$ or $9$.



You can't have more than $9$ triangles, because there are only $^8C_2=28$ edges determined by the $8$ points, each triangle needs $3$ edges, and no edge may be used by more than one triangle. So the number of triangles is at most $lfloor28/3rfloor=9$.



I don't see how to construct a set of $9$ triangles satisfying your conditions, but I can get $8$. Namely, if we label the points $A,B,C,D,E,F,G,H$, the following $8$ triangles work:
$$ABC, ADG, AFH, BEH, BFG, CDH, CEG, DEH$$



P.S. In fact, $8$ is the maximum. Let $p$ be the number of points (so $p=8$), let $t$ be the number of triangles, and let $n$ be the number of pairs $(P,T)$ where $T$ is a triangle and $P$ is a vertex of $T$. Then $n=3t$ (since each triangle has $3$ vertices), and $nle3p$ (since at most $3$ triangles can contain a given point), so $tle p=8$.



P.P.S. In case you're wondering how I found that set of $8$ triangles, I started with the well-known example of $12$ triangles on $9$ points (Steiner triple system) and deleted one point. Namely, I wrote the letters A to I in a $3times3$ square array, took the $6$ rows and columns and the $6$ "diagonals", and then deleted the ones containing the letter I.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ohhh! yes!! I didn't think in that way
    $endgroup$
    – Gimgim
    Jan 4 at 5:54
















7












$begingroup$

I have only a partial answer to your question: the maximum number of triangles is either $8$ or $9$.



You can't have more than $9$ triangles, because there are only $^8C_2=28$ edges determined by the $8$ points, each triangle needs $3$ edges, and no edge may be used by more than one triangle. So the number of triangles is at most $lfloor28/3rfloor=9$.



I don't see how to construct a set of $9$ triangles satisfying your conditions, but I can get $8$. Namely, if we label the points $A,B,C,D,E,F,G,H$, the following $8$ triangles work:
$$ABC, ADG, AFH, BEH, BFG, CDH, CEG, DEH$$



P.S. In fact, $8$ is the maximum. Let $p$ be the number of points (so $p=8$), let $t$ be the number of triangles, and let $n$ be the number of pairs $(P,T)$ where $T$ is a triangle and $P$ is a vertex of $T$. Then $n=3t$ (since each triangle has $3$ vertices), and $nle3p$ (since at most $3$ triangles can contain a given point), so $tle p=8$.



P.P.S. In case you're wondering how I found that set of $8$ triangles, I started with the well-known example of $12$ triangles on $9$ points (Steiner triple system) and deleted one point. Namely, I wrote the letters A to I in a $3times3$ square array, took the $6$ rows and columns and the $6$ "diagonals", and then deleted the ones containing the letter I.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ohhh! yes!! I didn't think in that way
    $endgroup$
    – Gimgim
    Jan 4 at 5:54














7












7








7





$begingroup$

I have only a partial answer to your question: the maximum number of triangles is either $8$ or $9$.



You can't have more than $9$ triangles, because there are only $^8C_2=28$ edges determined by the $8$ points, each triangle needs $3$ edges, and no edge may be used by more than one triangle. So the number of triangles is at most $lfloor28/3rfloor=9$.



I don't see how to construct a set of $9$ triangles satisfying your conditions, but I can get $8$. Namely, if we label the points $A,B,C,D,E,F,G,H$, the following $8$ triangles work:
$$ABC, ADG, AFH, BEH, BFG, CDH, CEG, DEH$$



P.S. In fact, $8$ is the maximum. Let $p$ be the number of points (so $p=8$), let $t$ be the number of triangles, and let $n$ be the number of pairs $(P,T)$ where $T$ is a triangle and $P$ is a vertex of $T$. Then $n=3t$ (since each triangle has $3$ vertices), and $nle3p$ (since at most $3$ triangles can contain a given point), so $tle p=8$.



P.P.S. In case you're wondering how I found that set of $8$ triangles, I started with the well-known example of $12$ triangles on $9$ points (Steiner triple system) and deleted one point. Namely, I wrote the letters A to I in a $3times3$ square array, took the $6$ rows and columns and the $6$ "diagonals", and then deleted the ones containing the letter I.






share|cite|improve this answer











$endgroup$



I have only a partial answer to your question: the maximum number of triangles is either $8$ or $9$.



You can't have more than $9$ triangles, because there are only $^8C_2=28$ edges determined by the $8$ points, each triangle needs $3$ edges, and no edge may be used by more than one triangle. So the number of triangles is at most $lfloor28/3rfloor=9$.



I don't see how to construct a set of $9$ triangles satisfying your conditions, but I can get $8$. Namely, if we label the points $A,B,C,D,E,F,G,H$, the following $8$ triangles work:
$$ABC, ADG, AFH, BEH, BFG, CDH, CEG, DEH$$



P.S. In fact, $8$ is the maximum. Let $p$ be the number of points (so $p=8$), let $t$ be the number of triangles, and let $n$ be the number of pairs $(P,T)$ where $T$ is a triangle and $P$ is a vertex of $T$. Then $n=3t$ (since each triangle has $3$ vertices), and $nle3p$ (since at most $3$ triangles can contain a given point), so $tle p=8$.



P.P.S. In case you're wondering how I found that set of $8$ triangles, I started with the well-known example of $12$ triangles on $9$ points (Steiner triple system) and deleted one point. Namely, I wrote the letters A to I in a $3times3$ square array, took the $6$ rows and columns and the $6$ "diagonals", and then deleted the ones containing the letter I.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 10:02

























answered Jan 4 at 5:45









bofbof

52.4k558121




52.4k558121












  • $begingroup$
    Ohhh! yes!! I didn't think in that way
    $endgroup$
    – Gimgim
    Jan 4 at 5:54


















  • $begingroup$
    Ohhh! yes!! I didn't think in that way
    $endgroup$
    – Gimgim
    Jan 4 at 5:54
















$begingroup$
Ohhh! yes!! I didn't think in that way
$endgroup$
– Gimgim
Jan 4 at 5:54




$begingroup$
Ohhh! yes!! I didn't think in that way
$endgroup$
– Gimgim
Jan 4 at 5:54











6












$begingroup$

bof gives a great justification of why it is eight or nine with an example of $8$.



Here is why it can't be nine. If there were nine triangles, they would use $27$ points, so one point would have to be used at least $4$ times. Each of these four triangles creates two edges containing this point, so we have at least $8$ edges containing this point. But there are only $7$ other points there must be a duplicate edge.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Right. I didn't notice that you posted this answer while I was typing the P.S. to my answer.
    $endgroup$
    – bof
    Jan 4 at 6:04
















6












$begingroup$

bof gives a great justification of why it is eight or nine with an example of $8$.



Here is why it can't be nine. If there were nine triangles, they would use $27$ points, so one point would have to be used at least $4$ times. Each of these four triangles creates two edges containing this point, so we have at least $8$ edges containing this point. But there are only $7$ other points there must be a duplicate edge.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Right. I didn't notice that you posted this answer while I was typing the P.S. to my answer.
    $endgroup$
    – bof
    Jan 4 at 6:04














6












6








6





$begingroup$

bof gives a great justification of why it is eight or nine with an example of $8$.



Here is why it can't be nine. If there were nine triangles, they would use $27$ points, so one point would have to be used at least $4$ times. Each of these four triangles creates two edges containing this point, so we have at least $8$ edges containing this point. But there are only $7$ other points there must be a duplicate edge.






share|cite|improve this answer











$endgroup$



bof gives a great justification of why it is eight or nine with an example of $8$.



Here is why it can't be nine. If there were nine triangles, they would use $27$ points, so one point would have to be used at least $4$ times. Each of these four triangles creates two edges containing this point, so we have at least $8$ edges containing this point. But there are only $7$ other points there must be a duplicate edge.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 6:06









Gimgim

28313




28313










answered Jan 4 at 6:00









Erik ParkinsonErik Parkinson

1,17519




1,17519












  • $begingroup$
    Right. I didn't notice that you posted this answer while I was typing the P.S. to my answer.
    $endgroup$
    – bof
    Jan 4 at 6:04


















  • $begingroup$
    Right. I didn't notice that you posted this answer while I was typing the P.S. to my answer.
    $endgroup$
    – bof
    Jan 4 at 6:04
















$begingroup$
Right. I didn't notice that you posted this answer while I was typing the P.S. to my answer.
$endgroup$
– bof
Jan 4 at 6:04




$begingroup$
Right. I didn't notice that you posted this answer while I was typing the P.S. to my answer.
$endgroup$
– bof
Jan 4 at 6:04


















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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Gyllenstierna

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