Expected area of triangle inscribed in a circle












0












$begingroup$


On a unit circle , BC is a chord with length 4/π . Point A is picked randomly at its circumference .
Find the expected area of △ ABC .






geometric-probability







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    On a unit circle , BC is a chord with length 4/π . Point A is picked randomly at its circumference .
    Find the expected area of △ ABC .






    geometric-probability







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      On a unit circle , BC is a chord with length 4/π . Point A is picked randomly at its circumference .
      Find the expected area of △ ABC .






      geometric-probability







      share|cite|improve this question









      $endgroup$




      On a unit circle , BC is a chord with length 4/π . Point A is picked randomly at its circumference .
      Find the expected area of △ ABC .






      geometric-probability




      geometric-probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 10 at 6:56









      mrwongmrwong

      5915




      5915






















          1 Answer
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          1












          $begingroup$

          For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2phi$ at the centre of the circle symmetric to the y axis.



          Then the perpendicular distance of the chord from the centre is $D = sqrt{(1 - (frac{2}{pi})^2}$ and $sinphi = frac{2}{pi}$



          Let the line joining A and the centre makes an angle $theta$ with the x axis, then



          For $theta$ between $0$ and $frac{pi}{2}$,



          $$Area = frac{1}{2}frac{4}{pi}(sintheta + D)$$



          For $theta$ between $0$ and $frac{pi}{2} - phi$



          $$Area = frac{1}{2}frac{4}{pi}(D - sintheta)$$



          For $theta$ between $frac{pi}{2} - phi$ and $frac{pi}{2}$



          $$Area = frac{1}{2}frac{4}{pi}(sintheta - D)$$



          Hence the average area is



          $$frac{4}{2pi^2}(int_0^{frac{pi}{2}}(sintheta + D)dtheta + int_0^{frac{pi}{2} - phi}(D - sintheta)dtheta + int_{frac{pi}{2} - phi}^{frac{pi}{2}}(sintheta - D)dtheta)$$



          $$= frac{2}{pi^2}(frac{2}{pi} + Dpi - 2Dphi + sinphi) = 0.533$$



          I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.



          Also if the area is $frac{3}{2pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
            $endgroup$
            – mrwong
            Jan 12 at 4:04










          • $begingroup$
            We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
            $endgroup$
            – KY Tang
            Jan 13 at 1:37










          • $begingroup$
            :Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
            $endgroup$
            – mrwong
            Jan 13 at 5:08










          • $begingroup$
            In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
            $endgroup$
            – KY Tang
            Jan 13 at 17:16










          • $begingroup$
            I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
            $endgroup$
            – mrwong
            Jan 14 at 4:40











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          1 Answer
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          1












          $begingroup$

          For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2phi$ at the centre of the circle symmetric to the y axis.



          Then the perpendicular distance of the chord from the centre is $D = sqrt{(1 - (frac{2}{pi})^2}$ and $sinphi = frac{2}{pi}$



          Let the line joining A and the centre makes an angle $theta$ with the x axis, then



          For $theta$ between $0$ and $frac{pi}{2}$,



          $$Area = frac{1}{2}frac{4}{pi}(sintheta + D)$$



          For $theta$ between $0$ and $frac{pi}{2} - phi$



          $$Area = frac{1}{2}frac{4}{pi}(D - sintheta)$$



          For $theta$ between $frac{pi}{2} - phi$ and $frac{pi}{2}$



          $$Area = frac{1}{2}frac{4}{pi}(sintheta - D)$$



          Hence the average area is



          $$frac{4}{2pi^2}(int_0^{frac{pi}{2}}(sintheta + D)dtheta + int_0^{frac{pi}{2} - phi}(D - sintheta)dtheta + int_{frac{pi}{2} - phi}^{frac{pi}{2}}(sintheta - D)dtheta)$$



          $$= frac{2}{pi^2}(frac{2}{pi} + Dpi - 2Dphi + sinphi) = 0.533$$



          I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.



          Also if the area is $frac{3}{2pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
            $endgroup$
            – mrwong
            Jan 12 at 4:04










          • $begingroup$
            We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
            $endgroup$
            – KY Tang
            Jan 13 at 1:37










          • $begingroup$
            :Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
            $endgroup$
            – mrwong
            Jan 13 at 5:08










          • $begingroup$
            In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
            $endgroup$
            – KY Tang
            Jan 13 at 17:16










          • $begingroup$
            I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
            $endgroup$
            – mrwong
            Jan 14 at 4:40
















          1












          $begingroup$

          For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2phi$ at the centre of the circle symmetric to the y axis.



          Then the perpendicular distance of the chord from the centre is $D = sqrt{(1 - (frac{2}{pi})^2}$ and $sinphi = frac{2}{pi}$



          Let the line joining A and the centre makes an angle $theta$ with the x axis, then



          For $theta$ between $0$ and $frac{pi}{2}$,



          $$Area = frac{1}{2}frac{4}{pi}(sintheta + D)$$



          For $theta$ between $0$ and $frac{pi}{2} - phi$



          $$Area = frac{1}{2}frac{4}{pi}(D - sintheta)$$



          For $theta$ between $frac{pi}{2} - phi$ and $frac{pi}{2}$



          $$Area = frac{1}{2}frac{4}{pi}(sintheta - D)$$



          Hence the average area is



          $$frac{4}{2pi^2}(int_0^{frac{pi}{2}}(sintheta + D)dtheta + int_0^{frac{pi}{2} - phi}(D - sintheta)dtheta + int_{frac{pi}{2} - phi}^{frac{pi}{2}}(sintheta - D)dtheta)$$



          $$= frac{2}{pi^2}(frac{2}{pi} + Dpi - 2Dphi + sinphi) = 0.533$$



          I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.



          Also if the area is $frac{3}{2pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
            $endgroup$
            – mrwong
            Jan 12 at 4:04










          • $begingroup$
            We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
            $endgroup$
            – KY Tang
            Jan 13 at 1:37










          • $begingroup$
            :Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
            $endgroup$
            – mrwong
            Jan 13 at 5:08










          • $begingroup$
            In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
            $endgroup$
            – KY Tang
            Jan 13 at 17:16










          • $begingroup$
            I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
            $endgroup$
            – mrwong
            Jan 14 at 4:40














          1












          1








          1





          $begingroup$

          For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2phi$ at the centre of the circle symmetric to the y axis.



          Then the perpendicular distance of the chord from the centre is $D = sqrt{(1 - (frac{2}{pi})^2}$ and $sinphi = frac{2}{pi}$



          Let the line joining A and the centre makes an angle $theta$ with the x axis, then



          For $theta$ between $0$ and $frac{pi}{2}$,



          $$Area = frac{1}{2}frac{4}{pi}(sintheta + D)$$



          For $theta$ between $0$ and $frac{pi}{2} - phi$



          $$Area = frac{1}{2}frac{4}{pi}(D - sintheta)$$



          For $theta$ between $frac{pi}{2} - phi$ and $frac{pi}{2}$



          $$Area = frac{1}{2}frac{4}{pi}(sintheta - D)$$



          Hence the average area is



          $$frac{4}{2pi^2}(int_0^{frac{pi}{2}}(sintheta + D)dtheta + int_0^{frac{pi}{2} - phi}(D - sintheta)dtheta + int_{frac{pi}{2} - phi}^{frac{pi}{2}}(sintheta - D)dtheta)$$



          $$= frac{2}{pi^2}(frac{2}{pi} + Dpi - 2Dphi + sinphi) = 0.533$$



          I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.



          Also if the area is $frac{3}{2pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.






          share|cite|improve this answer











          $endgroup$



          For easy reference let chord BC sits in the lower portion of the unit circle. It lies horizontally making angle $2phi$ at the centre of the circle symmetric to the y axis.



          Then the perpendicular distance of the chord from the centre is $D = sqrt{(1 - (frac{2}{pi})^2}$ and $sinphi = frac{2}{pi}$



          Let the line joining A and the centre makes an angle $theta$ with the x axis, then



          For $theta$ between $0$ and $frac{pi}{2}$,



          $$Area = frac{1}{2}frac{4}{pi}(sintheta + D)$$



          For $theta$ between $0$ and $frac{pi}{2} - phi$



          $$Area = frac{1}{2}frac{4}{pi}(D - sintheta)$$



          For $theta$ between $frac{pi}{2} - phi$ and $frac{pi}{2}$



          $$Area = frac{1}{2}frac{4}{pi}(sintheta - D)$$



          Hence the average area is



          $$frac{4}{2pi^2}(int_0^{frac{pi}{2}}(sintheta + D)dtheta + int_0^{frac{pi}{2} - phi}(D - sintheta)dtheta + int_{frac{pi}{2} - phi}^{frac{pi}{2}}(sintheta - D)dtheta)$$



          $$= frac{2}{pi^2}(frac{2}{pi} + Dpi - 2Dphi + sinphi) = 0.533$$



          I managed to borrow a PC and use 1000 random points and the result is around 5,1 which is very close to the calculation.



          Also if the area is $frac{3}{2pi}$, I have used trial and error to find x to be about 1.54 which is your expected answer.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 15 at 1:27

























          answered Jan 11 at 1:50









          KY TangKY Tang

          1245




          1245












          • $begingroup$
            Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
            $endgroup$
            – mrwong
            Jan 12 at 4:04










          • $begingroup$
            We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
            $endgroup$
            – KY Tang
            Jan 13 at 1:37










          • $begingroup$
            :Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
            $endgroup$
            – mrwong
            Jan 13 at 5:08










          • $begingroup$
            In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
            $endgroup$
            – KY Tang
            Jan 13 at 17:16










          • $begingroup$
            I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
            $endgroup$
            – mrwong
            Jan 14 at 4:40


















          • $begingroup$
            Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
            $endgroup$
            – mrwong
            Jan 12 at 4:04










          • $begingroup$
            We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
            $endgroup$
            – KY Tang
            Jan 13 at 1:37










          • $begingroup$
            :Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
            $endgroup$
            – mrwong
            Jan 13 at 5:08










          • $begingroup$
            In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
            $endgroup$
            – KY Tang
            Jan 13 at 17:16










          • $begingroup$
            I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
            $endgroup$
            – mrwong
            Jan 14 at 4:40
















          $begingroup$
          Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
          $endgroup$
          – mrwong
          Jan 12 at 4:04




          $begingroup$
          Thanks for your reply .You got a result of about 0.7 sq.unit . Will you please check the result with simulation . If they match then I will accept the result .
          $endgroup$
          – mrwong
          Jan 12 at 4:04












          $begingroup$
          We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
          $endgroup$
          – KY Tang
          Jan 13 at 1:37




          $begingroup$
          We can estimate the area for those triangles above the chord. As D is 0.77 and the chord length is 1.27. The simple average height of these triangles is (1+0.77)/2 = 0.885. Therefore the area is 1/2. x 0.885 x 1.27 = 0.562. Similarly the simple area average is 1/2 x 0.115 x 1.27 = 0.07 for those triangles below the chord. Since the % of height over 0.885 is higher, the answer of 0.7 is reasonable.
          $endgroup$
          – KY Tang
          Jan 13 at 1:37












          $begingroup$
          :Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
          $endgroup$
          – mrwong
          Jan 13 at 5:08




          $begingroup$
          :Thanks again . If the question is changed to : The length of chord BC = x units , while A is picked randomly . If the expected area of △ ABC = 3/ 2π (about 0.477 sq.un) , then what is the value of x ?
          $endgroup$
          – mrwong
          Jan 13 at 5:08












          $begingroup$
          In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
          $endgroup$
          – KY Tang
          Jan 13 at 17:16




          $begingroup$
          In fact the formula in terms of D and phi is 1/(pi^2) x (4sinphi + 2piD). Now D = sqrt(1-x^2/4) and sin phi = x/2. Then put this into the area formula we have x = 1.99. That means the chord is almost the diameter. Please see if this is what you expect.
          $endgroup$
          – KY Tang
          Jan 13 at 17:16












          $begingroup$
          I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
          $endgroup$
          – mrwong
          Jan 14 at 4:40




          $begingroup$
          I got an answer of about 1.53 elsewhere . Is it useful to use simulation to clarify ?
          $endgroup$
          – mrwong
          Jan 14 at 4:40


















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