Among all the isosceles triangles whose circumferences are smaller than 24 centimeters, what fraction of them has their apex greater than pi/3? [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)What are the odds of drawing a isosceles triangle whose apex is greater than pi/3 and has its circumference smaller than 24?Isosceles triangles with perimeter less than 24 that have apex angle at least $tfracpi3$Rectangle divided into three triangles with two lines. One angle is given, what are all the others?What is the flaw in this proof that all triangles are isosceles?Usage of law of sinesIsosceles triangle has the least perimeter among triangles on the same base with same area?What is the probability that $triangle ABP$ has a greater area than each of $triangle ACP$ and $triangle BCP$?How show with at least one of the triangles $PAB, PBC, PAC$ has a smaller perimeter than ABC?Let D denote a point on base AB, and let E denote a point on leg BC of an isosceles triangle ABC.Finding the Lengths of Two unknown squaresShow that no matter how $12$ points are put on a plane, there are $3$ among them forming an angle not greater than $18^o$.What are the odds of drawing a isosceles triangle whose apex is greater than pi/3 and has its circumference smaller than 24?
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Among all the isosceles triangles whose circumferences are smaller than 24 centimeters, what fraction of them has their apex greater than pi/3? [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)What are the odds of drawing a isosceles triangle whose apex is greater than pi/3 and has its circumference smaller than 24?Isosceles triangles with perimeter less than 24 that have apex angle at least $tfracpi3$Rectangle divided into three triangles with two lines. One angle is given, what are all the others?What is the flaw in this proof that all triangles are isosceles?Usage of law of sinesIsosceles triangle has the least perimeter among triangles on the same base with same area?What is the probability that $triangle ABP$ has a greater area than each of $triangle ACP$ and $triangle BCP$?How show with at least one of the triangles $PAB, PBC, PAC$ has a smaller perimeter than ABC?Let D denote a point on base AB, and let E denote a point on leg BC of an isosceles triangle ABC.Finding the Lengths of Two unknown squaresShow that no matter how $12$ points are put on a plane, there are $3$ among them forming an angle not greater than $18^o$.What are the odds of drawing a isosceles triangle whose apex is greater than pi/3 and has its circumference smaller than 24?
$begingroup$
It's is known and certain that they are all isosceles triangles and their circumferences are smaller than 24 centimeters.
geometry trigonometry
asked Apr 2 at 18:26
buhesapsahtebuhesapsahte
11
$endgroup$
closed as unclear what you're asking by InterstellarProbe, Leucippus, Eevee Trainer, John Omielan, dantopa Apr 2 at 21:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 1 more comment
$begingroup$
It's is known and certain that they are all isosceles triangles and their circumferences are smaller than 24 centimeters.
geometry trigonometry
asked Apr 2 at 18:26
buhesapsahtebuhesapsahte
11
$endgroup$
closed as unclear what you're asking by InterstellarProbe, Leucippus, Eevee Trainer, John Omielan, dantopa Apr 2 at 21:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
It is recommended that, when asking a question on Stack Exchange, one give both context and some sort of effort in how to solve (ie, how you would approach the problem)
$endgroup$
– Moko19
Apr 2 at 18:28
5
$begingroup$
There are an infinite number of isosceles triangles with circumference smaller than 24 centimeters. There are an infinite number of such triangles with angle at their apex greater than $dfracpi3$. How are you deciding the distribution? Is it for a fixed circumference? Is there a probability mass function you are using to determine the dimensions (and angles) of said triangles? Based on how you are wording your question, any answer between 0% and 100% inclusive could easily be justified. This question is far too open.
$endgroup$
– InterstellarProbe
Apr 2 at 18:32
$begingroup$
Since I am not counting them one by one, it can be calculated. You are not making any sense.
$endgroup$
– buhesapsahte
Apr 2 at 18:38
$begingroup$
I actually find this problem kinda interesting. If you want to improve upon this problem and have an answer that can actually be provided, I have some suggestions. How about consider the set of points: $$chi = left(x,y)in mathbbR^2: 0<x<dfrac144-y^224, 0<y<12right$$ This will give you the set of all pairs such that you can make an isosceles triangle with the points: (-x,0), (x,0), and (0,y). Since this set of pairs covers all triangles with circumference at most 24 units, you can take the Lebesgue measure: $$int_chi f$$ where...
$endgroup$
– InterstellarProbe
Apr 4 at 20:00
$begingroup$
$$f(x,y)=begincases1, & x>dfracysqrt3 \ 0, & textotherwiseendcases$$ And then you divide by $$int_chi 1$$ as this will give you the relative amount of data points that yield triangles with the given properties out of all possible triangles in a very uniform way.
$endgroup$
– InterstellarProbe
Apr 4 at 20:01
|
show 1 more comment
$begingroup$
It's is known and certain that they are all isosceles triangles and their circumferences are smaller than 24 centimeters.
geometry trigonometry
asked Apr 2 at 18:26
buhesapsahtebuhesapsahte
11
$endgroup$
It's is known and certain that they are all isosceles triangles and their circumferences are smaller than 24 centimeters.
geometry trigonometry
geometry trigonometry
asked Apr 2 at 18:26
buhesapsahtebuhesapsahte
11
asked Apr 2 at 18:26
buhesapsahtebuhesapsahte
11
asked Apr 2 at 18:26
buhesapsahtebuhesapsahte
11
asked Apr 2 at 18:26
buhesapsahtebuhesapsahte
11
asked Apr 2 at 18:26
buhesapsahtebuhesapsahte
11
11
closed as unclear what you're asking by InterstellarProbe, Leucippus, Eevee Trainer, John Omielan, dantopa Apr 2 at 21:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by InterstellarProbe, Leucippus, Eevee Trainer, John Omielan, dantopa Apr 2 at 21:06
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
It is recommended that, when asking a question on Stack Exchange, one give both context and some sort of effort in how to solve (ie, how you would approach the problem)
$endgroup$
– Moko19
Apr 2 at 18:28
5
$begingroup$
There are an infinite number of isosceles triangles with circumference smaller than 24 centimeters. There are an infinite number of such triangles with angle at their apex greater than $dfracpi3$. How are you deciding the distribution? Is it for a fixed circumference? Is there a probability mass function you are using to determine the dimensions (and angles) of said triangles? Based on how you are wording your question, any answer between 0% and 100% inclusive could easily be justified. This question is far too open.
$endgroup$
– InterstellarProbe
Apr 2 at 18:32
$begingroup$
Since I am not counting them one by one, it can be calculated. You are not making any sense.
$endgroup$
– buhesapsahte
Apr 2 at 18:38
$begingroup$
I actually find this problem kinda interesting. If you want to improve upon this problem and have an answer that can actually be provided, I have some suggestions. How about consider the set of points: $$chi = left(x,y)in mathbbR^2: 0<x<dfrac144-y^224, 0<y<12right$$ This will give you the set of all pairs such that you can make an isosceles triangle with the points: (-x,0), (x,0), and (0,y). Since this set of pairs covers all triangles with circumference at most 24 units, you can take the Lebesgue measure: $$int_chi f$$ where...
$endgroup$
– InterstellarProbe
Apr 4 at 20:00
$begingroup$
$$f(x,y)=begincases1, & x>dfracysqrt3 \ 0, & textotherwiseendcases$$ And then you divide by $$int_chi 1$$ as this will give you the relative amount of data points that yield triangles with the given properties out of all possible triangles in a very uniform way.
$endgroup$
– InterstellarProbe
Apr 4 at 20:01
|
show 1 more comment
$begingroup$
It is recommended that, when asking a question on Stack Exchange, one give both context and some sort of effort in how to solve (ie, how you would approach the problem)
$endgroup$
– Moko19
Apr 2 at 18:28
5
$begingroup$
There are an infinite number of isosceles triangles with circumference smaller than 24 centimeters. There are an infinite number of such triangles with angle at their apex greater than $dfracpi3$. How are you deciding the distribution? Is it for a fixed circumference? Is there a probability mass function you are using to determine the dimensions (and angles) of said triangles? Based on how you are wording your question, any answer between 0% and 100% inclusive could easily be justified. This question is far too open.
$endgroup$
– InterstellarProbe
Apr 2 at 18:32
$begingroup$
Since I am not counting them one by one, it can be calculated. You are not making any sense.
$endgroup$
– buhesapsahte
Apr 2 at 18:38
$begingroup$
I actually find this problem kinda interesting. If you want to improve upon this problem and have an answer that can actually be provided, I have some suggestions. How about consider the set of points: $$chi = left(x,y)in mathbbR^2: 0<x<dfrac144-y^224, 0<y<12right$$ This will give you the set of all pairs such that you can make an isosceles triangle with the points: (-x,0), (x,0), and (0,y). Since this set of pairs covers all triangles with circumference at most 24 units, you can take the Lebesgue measure: $$int_chi f$$ where...
$endgroup$
– InterstellarProbe
Apr 4 at 20:00
$begingroup$
$$f(x,y)=begincases1, & x>dfracysqrt3 \ 0, & textotherwiseendcases$$ And then you divide by $$int_chi 1$$ as this will give you the relative amount of data points that yield triangles with the given properties out of all possible triangles in a very uniform way.
$endgroup$
– InterstellarProbe
Apr 4 at 20:01
$begingroup$
It is recommended that, when asking a question on Stack Exchange, one give both context and some sort of effort in how to solve (ie, how you would approach the problem)
$endgroup$
– Moko19
Apr 2 at 18:28
$begingroup$
It is recommended that, when asking a question on Stack Exchange, one give both context and some sort of effort in how to solve (ie, how you would approach the problem)
$endgroup$
– Moko19
Apr 2 at 18:28
5
5
$begingroup$
There are an infinite number of isosceles triangles with circumference smaller than 24 centimeters. There are an infinite number of such triangles with angle at their apex greater than $dfracpi3$. How are you deciding the distribution? Is it for a fixed circumference? Is there a probability mass function you are using to determine the dimensions (and angles) of said triangles? Based on how you are wording your question, any answer between 0% and 100% inclusive could easily be justified. This question is far too open.
$endgroup$
– InterstellarProbe
Apr 2 at 18:32
$begingroup$
There are an infinite number of isosceles triangles with circumference smaller than 24 centimeters. There are an infinite number of such triangles with angle at their apex greater than $dfracpi3$. How are you deciding the distribution? Is it for a fixed circumference? Is there a probability mass function you are using to determine the dimensions (and angles) of said triangles? Based on how you are wording your question, any answer between 0% and 100% inclusive could easily be justified. This question is far too open.
$endgroup$
– InterstellarProbe
Apr 2 at 18:32
$begingroup$
Since I am not counting them one by one, it can be calculated. You are not making any sense.
$endgroup$
– buhesapsahte
Apr 2 at 18:38
$begingroup$
Since I am not counting them one by one, it can be calculated. You are not making any sense.
$endgroup$
– buhesapsahte
Apr 2 at 18:38
$begingroup$
I actually find this problem kinda interesting. If you want to improve upon this problem and have an answer that can actually be provided, I have some suggestions. How about consider the set of points: $$chi = left(x,y)in mathbbR^2: 0<x<dfrac144-y^224, 0<y<12right$$ This will give you the set of all pairs such that you can make an isosceles triangle with the points: (-x,0), (x,0), and (0,y). Since this set of pairs covers all triangles with circumference at most 24 units, you can take the Lebesgue measure: $$int_chi f$$ where...
$endgroup$
– InterstellarProbe
Apr 4 at 20:00
$begingroup$
I actually find this problem kinda interesting. If you want to improve upon this problem and have an answer that can actually be provided, I have some suggestions. How about consider the set of points: $$chi = left(x,y)in mathbbR^2: 0<x<dfrac144-y^224, 0<y<12right$$ This will give you the set of all pairs such that you can make an isosceles triangle with the points: (-x,0), (x,0), and (0,y). Since this set of pairs covers all triangles with circumference at most 24 units, you can take the Lebesgue measure: $$int_chi f$$ where...
$endgroup$
– InterstellarProbe
Apr 4 at 20:00
$begingroup$
$$f(x,y)=begincases1, & x>dfracysqrt3 \ 0, & textotherwiseendcases$$ And then you divide by $$int_chi 1$$ as this will give you the relative amount of data points that yield triangles with the given properties out of all possible triangles in a very uniform way.
$endgroup$
– InterstellarProbe
Apr 4 at 20:01
$begingroup$
$$f(x,y)=begincases1, & x>dfracysqrt3 \ 0, & textotherwiseendcases$$ And then you divide by $$int_chi 1$$ as this will give you the relative amount of data points that yield triangles with the given properties out of all possible triangles in a very uniform way.
$endgroup$
– InterstellarProbe
Apr 4 at 20:01
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The circumference doesn't matter.
If the apex angle is $x$,
then the other angles are
$(pi-x)/2$.
If the apex angles
are distributed uniformly,
they can be from
$0$ to $pi$,
so the chance of
them being less than
$pi/3$
is
$dfracpi/3pi
=dfrac13
$.
answered Apr 2 at 18:32
marty cohenmarty cohen
76k549130
$endgroup$
$begingroup$
Correct me if I am wrong. Its angles being a, a, b; I want b being greater than $fracpi3$. So it is $fracpi2 > fracb2 > fracpi3$. That way, shouldn't it be $frac23$?
$endgroup$
– buhesapsahte
Apr 2 at 18:46
add a comment |
$begingroup$
Assuming the triangles are uniformly distributed by the length of their legs, and for a fixed leg length, they are uniformly distributed by the length of their base, we can calculate the probability using integration. When the base is longer than the legs, it will have an apex angle greater than $dfracpi3$. The base can be no longer than double the length of each leg (this would be an apex angle of $pi$). Each leg can be no longer than 12cm (if it were longer, then the circumference would be greater than 24cm).
So, for triangles with legs of length no greater than 6cm, half of all triangles will have a base of length at least as large as the length of the leg (because we decided that for a fixed leg length, the length of the base is uniformly distributed). Since half of all triangle have legs of length 0cm to 6cm, that means (with our chosen distributions) $dfrac12cdot dfrac12 = dfrac14$ of all triangles have legs of length no greater than 6cm and an apex angle of at least $dfracpi3$.
Next, we consider triangles with legs of length between 6cm and 8cm. That is $dfrac16$ of all triangles (because the leg lengths are uniformly distributed).
We can integrate to determine the fraction of these triangles that have an apex angle greater than $dfracpi3$. For a given leg length of $x$, when the base has length between $x$ and $24-2x$, the triangle will have an apex angle of at least $dfracpi3$. So, $|(24-2x)-x| = 24-3x$. This gives the integral:
$$int_6^8 dfrac24-3x24-2xdx = 3-ln left(dfrac72964right)$$
And as discussed, this accounts for $dfrac16$ of all triangles.
For triangles with legs of length between 8cm and 12cm, it is not possible for the base to be longer than one of the legs and still have a circumference less than 24. This yields the total probability:
$$dfrac12cdot dfrac12 + dfrac16left(3-lnleft(dfrac72964right)right) approx 34.4535%$$
answered Apr 2 at 20:08
InterstellarProbeInterstellarProbe
3,249730
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The circumference doesn't matter.
If the apex angle is $x$,
then the other angles are
$(pi-x)/2$.
If the apex angles
are distributed uniformly,
they can be from
$0$ to $pi$,
so the chance of
them being less than
$pi/3$
is
$dfracpi/3pi
=dfrac13
$.
answered Apr 2 at 18:32
marty cohenmarty cohen
76k549130
$endgroup$
$begingroup$
Correct me if I am wrong. Its angles being a, a, b; I want b being greater than $fracpi3$. So it is $fracpi2 > fracb2 > fracpi3$. That way, shouldn't it be $frac23$?
$endgroup$
– buhesapsahte
Apr 2 at 18:46
add a comment |
$begingroup$
The circumference doesn't matter.
If the apex angle is $x$,
then the other angles are
$(pi-x)/2$.
If the apex angles
are distributed uniformly,
they can be from
$0$ to $pi$,
so the chance of
them being less than
$pi/3$
is
$dfracpi/3pi
=dfrac13
$.
answered Apr 2 at 18:32
marty cohenmarty cohen
76k549130
$endgroup$
$begingroup$
Correct me if I am wrong. Its angles being a, a, b; I want b being greater than $fracpi3$. So it is $fracpi2 > fracb2 > fracpi3$. That way, shouldn't it be $frac23$?
$endgroup$
– buhesapsahte
Apr 2 at 18:46
add a comment |
$begingroup$
The circumference doesn't matter.
If the apex angle is $x$,
then the other angles are
$(pi-x)/2$.
If the apex angles
are distributed uniformly,
they can be from
$0$ to $pi$,
so the chance of
them being less than
$pi/3$
is
$dfracpi/3pi
=dfrac13
$.
answered Apr 2 at 18:32
marty cohenmarty cohen
76k549130
$endgroup$
The circumference doesn't matter.
If the apex angle is $x$,
then the other angles are
$(pi-x)/2$.
If the apex angles
are distributed uniformly,
they can be from
$0$ to $pi$,
so the chance of
them being less than
$pi/3$
is
$dfracpi/3pi
=dfrac13
$.
answered Apr 2 at 18:32
marty cohenmarty cohen
76k549130
answered Apr 2 at 18:32
marty cohenmarty cohen
76k549130
answered Apr 2 at 18:32
marty cohenmarty cohen
76k549130
answered Apr 2 at 18:32
marty cohenmarty cohen
76k549130
76k549130
$begingroup$
Correct me if I am wrong. Its angles being a, a, b; I want b being greater than $fracpi3$. So it is $fracpi2 > fracb2 > fracpi3$. That way, shouldn't it be $frac23$?
$endgroup$
– buhesapsahte
Apr 2 at 18:46
add a comment |
$begingroup$
Correct me if I am wrong. Its angles being a, a, b; I want b being greater than $fracpi3$. So it is $fracpi2 > fracb2 > fracpi3$. That way, shouldn't it be $frac23$?
$endgroup$
– buhesapsahte
Apr 2 at 18:46
$begingroup$
Correct me if I am wrong. Its angles being a, a, b; I want b being greater than $fracpi3$. So it is $fracpi2 > fracb2 > fracpi3$. That way, shouldn't it be $frac23$?
$endgroup$
– buhesapsahte
Apr 2 at 18:46
$begingroup$
Correct me if I am wrong. Its angles being a, a, b; I want b being greater than $fracpi3$. So it is $fracpi2 > fracb2 > fracpi3$. That way, shouldn't it be $frac23$?
$endgroup$
– buhesapsahte
Apr 2 at 18:46
add a comment |
$begingroup$
Assuming the triangles are uniformly distributed by the length of their legs, and for a fixed leg length, they are uniformly distributed by the length of their base, we can calculate the probability using integration. When the base is longer than the legs, it will have an apex angle greater than $dfracpi3$. The base can be no longer than double the length of each leg (this would be an apex angle of $pi$). Each leg can be no longer than 12cm (if it were longer, then the circumference would be greater than 24cm).
So, for triangles with legs of length no greater than 6cm, half of all triangles will have a base of length at least as large as the length of the leg (because we decided that for a fixed leg length, the length of the base is uniformly distributed). Since half of all triangle have legs of length 0cm to 6cm, that means (with our chosen distributions) $dfrac12cdot dfrac12 = dfrac14$ of all triangles have legs of length no greater than 6cm and an apex angle of at least $dfracpi3$.
Next, we consider triangles with legs of length between 6cm and 8cm. That is $dfrac16$ of all triangles (because the leg lengths are uniformly distributed).
We can integrate to determine the fraction of these triangles that have an apex angle greater than $dfracpi3$. For a given leg length of $x$, when the base has length between $x$ and $24-2x$, the triangle will have an apex angle of at least $dfracpi3$. So, $|(24-2x)-x| = 24-3x$. This gives the integral:
$$int_6^8 dfrac24-3x24-2xdx = 3-ln left(dfrac72964right)$$
And as discussed, this accounts for $dfrac16$ of all triangles.
For triangles with legs of length between 8cm and 12cm, it is not possible for the base to be longer than one of the legs and still have a circumference less than 24. This yields the total probability:
$$dfrac12cdot dfrac12 + dfrac16left(3-lnleft(dfrac72964right)right) approx 34.4535%$$
answered Apr 2 at 20:08
InterstellarProbeInterstellarProbe
3,249730
$endgroup$
add a comment |
$begingroup$
Assuming the triangles are uniformly distributed by the length of their legs, and for a fixed leg length, they are uniformly distributed by the length of their base, we can calculate the probability using integration. When the base is longer than the legs, it will have an apex angle greater than $dfracpi3$. The base can be no longer than double the length of each leg (this would be an apex angle of $pi$). Each leg can be no longer than 12cm (if it were longer, then the circumference would be greater than 24cm).
So, for triangles with legs of length no greater than 6cm, half of all triangles will have a base of length at least as large as the length of the leg (because we decided that for a fixed leg length, the length of the base is uniformly distributed). Since half of all triangle have legs of length 0cm to 6cm, that means (with our chosen distributions) $dfrac12cdot dfrac12 = dfrac14$ of all triangles have legs of length no greater than 6cm and an apex angle of at least $dfracpi3$.
Next, we consider triangles with legs of length between 6cm and 8cm. That is $dfrac16$ of all triangles (because the leg lengths are uniformly distributed).
We can integrate to determine the fraction of these triangles that have an apex angle greater than $dfracpi3$. For a given leg length of $x$, when the base has length between $x$ and $24-2x$, the triangle will have an apex angle of at least $dfracpi3$. So, $|(24-2x)-x| = 24-3x$. This gives the integral:
$$int_6^8 dfrac24-3x24-2xdx = 3-ln left(dfrac72964right)$$
And as discussed, this accounts for $dfrac16$ of all triangles.
For triangles with legs of length between 8cm and 12cm, it is not possible for the base to be longer than one of the legs and still have a circumference less than 24. This yields the total probability:
$$dfrac12cdot dfrac12 + dfrac16left(3-lnleft(dfrac72964right)right) approx 34.4535%$$
answered Apr 2 at 20:08
InterstellarProbeInterstellarProbe
3,249730
$endgroup$
add a comment |
$begingroup$
Assuming the triangles are uniformly distributed by the length of their legs, and for a fixed leg length, they are uniformly distributed by the length of their base, we can calculate the probability using integration. When the base is longer than the legs, it will have an apex angle greater than $dfracpi3$. The base can be no longer than double the length of each leg (this would be an apex angle of $pi$). Each leg can be no longer than 12cm (if it were longer, then the circumference would be greater than 24cm).
So, for triangles with legs of length no greater than 6cm, half of all triangles will have a base of length at least as large as the length of the leg (because we decided that for a fixed leg length, the length of the base is uniformly distributed). Since half of all triangle have legs of length 0cm to 6cm, that means (with our chosen distributions) $dfrac12cdot dfrac12 = dfrac14$ of all triangles have legs of length no greater than 6cm and an apex angle of at least $dfracpi3$.
Next, we consider triangles with legs of length between 6cm and 8cm. That is $dfrac16$ of all triangles (because the leg lengths are uniformly distributed).
We can integrate to determine the fraction of these triangles that have an apex angle greater than $dfracpi3$. For a given leg length of $x$, when the base has length between $x$ and $24-2x$, the triangle will have an apex angle of at least $dfracpi3$. So, $|(24-2x)-x| = 24-3x$. This gives the integral:
$$int_6^8 dfrac24-3x24-2xdx = 3-ln left(dfrac72964right)$$
And as discussed, this accounts for $dfrac16$ of all triangles.
For triangles with legs of length between 8cm and 12cm, it is not possible for the base to be longer than one of the legs and still have a circumference less than 24. This yields the total probability:
$$dfrac12cdot dfrac12 + dfrac16left(3-lnleft(dfrac72964right)right) approx 34.4535%$$
answered Apr 2 at 20:08
InterstellarProbeInterstellarProbe
3,249730
$endgroup$
Assuming the triangles are uniformly distributed by the length of their legs, and for a fixed leg length, they are uniformly distributed by the length of their base, we can calculate the probability using integration. When the base is longer than the legs, it will have an apex angle greater than $dfracpi3$. The base can be no longer than double the length of each leg (this would be an apex angle of $pi$). Each leg can be no longer than 12cm (if it were longer, then the circumference would be greater than 24cm).
So, for triangles with legs of length no greater than 6cm, half of all triangles will have a base of length at least as large as the length of the leg (because we decided that for a fixed leg length, the length of the base is uniformly distributed). Since half of all triangle have legs of length 0cm to 6cm, that means (with our chosen distributions) $dfrac12cdot dfrac12 = dfrac14$ of all triangles have legs of length no greater than 6cm and an apex angle of at least $dfracpi3$.
Next, we consider triangles with legs of length between 6cm and 8cm. That is $dfrac16$ of all triangles (because the leg lengths are uniformly distributed).
We can integrate to determine the fraction of these triangles that have an apex angle greater than $dfracpi3$. For a given leg length of $x$, when the base has length between $x$ and $24-2x$, the triangle will have an apex angle of at least $dfracpi3$. So, $|(24-2x)-x| = 24-3x$. This gives the integral:
$$int_6^8 dfrac24-3x24-2xdx = 3-ln left(dfrac72964right)$$
And as discussed, this accounts for $dfrac16$ of all triangles.
For triangles with legs of length between 8cm and 12cm, it is not possible for the base to be longer than one of the legs and still have a circumference less than 24. This yields the total probability:
$$dfrac12cdot dfrac12 + dfrac16left(3-lnleft(dfrac72964right)right) approx 34.4535%$$
answered Apr 2 at 20:08
InterstellarProbeInterstellarProbe
3,249730
edited Apr 2 at 20:20
answered Apr 2 at 20:08
InterstellarProbeInterstellarProbe
3,249730
answered Apr 2 at 20:08
InterstellarProbeInterstellarProbe
3,249730
answered Apr 2 at 20:08
InterstellarProbeInterstellarProbe
3,249730
3,249730
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It is recommended that, when asking a question on Stack Exchange, one give both context and some sort of effort in how to solve (ie, how you would approach the problem)
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– Moko19
Apr 2 at 18:28
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There are an infinite number of isosceles triangles with circumference smaller than 24 centimeters. There are an infinite number of such triangles with angle at their apex greater than $dfracpi3$. How are you deciding the distribution? Is it for a fixed circumference? Is there a probability mass function you are using to determine the dimensions (and angles) of said triangles? Based on how you are wording your question, any answer between 0% and 100% inclusive could easily be justified. This question is far too open.
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– InterstellarProbe
Apr 2 at 18:32
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Since I am not counting them one by one, it can be calculated. You are not making any sense.
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– buhesapsahte
Apr 2 at 18:38
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I actually find this problem kinda interesting. If you want to improve upon this problem and have an answer that can actually be provided, I have some suggestions. How about consider the set of points: $$chi = left(x,y)in mathbbR^2: 0<x<dfrac144-y^224, 0<y<12right$$ This will give you the set of all pairs such that you can make an isosceles triangle with the points: (-x,0), (x,0), and (0,y). Since this set of pairs covers all triangles with circumference at most 24 units, you can take the Lebesgue measure: $$int_chi f$$ where...
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– InterstellarProbe
Apr 4 at 20:00
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$$f(x,y)=begincases1, & x>dfracysqrt3 \ 0, & textotherwiseendcases$$ And then you divide by $$int_chi 1$$ as this will give you the relative amount of data points that yield triangles with the given properties out of all possible triangles in a very uniform way.
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– InterstellarProbe
Apr 4 at 20:01