Find the length x such that the two distances in the triangle are the same Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to position rectangles such that they are as close as possible to a reference point but do not overlap?Showing a line is parallel to a plane and finding the distance between themAn interesting point of a triangle. (Help needed to prove a statement.)How is a vertex of a triangle moving while another vertex is moving on its angle bisector?A conjecture about an angle on a solid bodyIs it possible to compute Right Triangle's Legs starting from another Right Triangle with the same Hypotenuse?Constructing a Regular Pentagon of a Desired LengthCalculate the projected distance on an inclined planeIs the blue area greater than the red area?In an isosceles triangle $ABC$ show that $PM+PN$ does not depend on the position of the chosen point P.

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Find the length x such that the two distances in the triangle are the same



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to position rectangles such that they are as close as possible to a reference point but do not overlap?Showing a line is parallel to a plane and finding the distance between themAn interesting point of a triangle. (Help needed to prove a statement.)How is a vertex of a triangle moving while another vertex is moving on its angle bisector?A conjecture about an angle on a solid bodyIs it possible to compute Right Triangle's Legs starting from another Right Triangle with the same Hypotenuse?Constructing a Regular Pentagon of a Desired LengthCalculate the projected distance on an inclined planeIs the blue area greater than the red area?In an isosceles triangle $ABC$ show that $PM+PN$ does not depend on the position of the chosen point P.










2












$begingroup$


I have been working on the following problem



Statement




Assume you have a right angle triangle $Delta ABC$ with cateti $a$, $b$ and hypotenuse $c = sqrta^2 + b^2$. Find or construct a point $D$ on the hypothenuse such that the distance $|CD| = |DE|$, where $E$ is positioned on $AB$ in such a way that $DEparallel BC$ ($DE$ is parallel to $BC$).




enter image description here



Background



My background for wanting such a distance is that I want to create a semicircle from C onto the line $AB$. This can be made clearer in the image below



enter image description here



To be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem



Solution



Using similar triangles one arrives at the three equations



$$
beginalign*
fraccolorbluetextbluea - x & = fracba \
fraccolorredtextredx & = fracca \
colorredtextred & = colorbluetextblue
endalign*
$$



Where one easily can solve for $colorbluetextblue$, $colorredtextred$, $x$.



Question



I feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $D$ in a simpler matter?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    You still have not described what $F$ is either from the statement or from the graph.
    $endgroup$
    – Hw Chu
    10 hours ago










  • $begingroup$
    Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
    $endgroup$
    – N3buchadnezzar
    10 hours ago










  • $begingroup$
    cateti is Italian for legs
    $endgroup$
    – J. W. Tanner
    10 hours ago






  • 1




    $begingroup$
    Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
    $endgroup$
    – David K
    6 hours ago















2












$begingroup$


I have been working on the following problem



Statement




Assume you have a right angle triangle $Delta ABC$ with cateti $a$, $b$ and hypotenuse $c = sqrta^2 + b^2$. Find or construct a point $D$ on the hypothenuse such that the distance $|CD| = |DE|$, where $E$ is positioned on $AB$ in such a way that $DEparallel BC$ ($DE$ is parallel to $BC$).




enter image description here



Background



My background for wanting such a distance is that I want to create a semicircle from C onto the line $AB$. This can be made clearer in the image below



enter image description here



To be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem



Solution



Using similar triangles one arrives at the three equations



$$
beginalign*
fraccolorbluetextbluea - x & = fracba \
fraccolorredtextredx & = fracca \
colorredtextred & = colorbluetextblue
endalign*
$$



Where one easily can solve for $colorbluetextblue$, $colorredtextred$, $x$.



Question



I feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $D$ in a simpler matter?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    You still have not described what $F$ is either from the statement or from the graph.
    $endgroup$
    – Hw Chu
    10 hours ago










  • $begingroup$
    Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
    $endgroup$
    – N3buchadnezzar
    10 hours ago










  • $begingroup$
    cateti is Italian for legs
    $endgroup$
    – J. W. Tanner
    10 hours ago






  • 1




    $begingroup$
    Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
    $endgroup$
    – David K
    6 hours ago













2












2








2





$begingroup$


I have been working on the following problem



Statement




Assume you have a right angle triangle $Delta ABC$ with cateti $a$, $b$ and hypotenuse $c = sqrta^2 + b^2$. Find or construct a point $D$ on the hypothenuse such that the distance $|CD| = |DE|$, where $E$ is positioned on $AB$ in such a way that $DEparallel BC$ ($DE$ is parallel to $BC$).




enter image description here



Background



My background for wanting such a distance is that I want to create a semicircle from C onto the line $AB$. This can be made clearer in the image below



enter image description here



To be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem



Solution



Using similar triangles one arrives at the three equations



$$
beginalign*
fraccolorbluetextbluea - x & = fracba \
fraccolorredtextredx & = fracca \
colorredtextred & = colorbluetextblue
endalign*
$$



Where one easily can solve for $colorbluetextblue$, $colorredtextred$, $x$.



Question



I feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $D$ in a simpler matter?










share|cite|improve this question











$endgroup$




I have been working on the following problem



Statement




Assume you have a right angle triangle $Delta ABC$ with cateti $a$, $b$ and hypotenuse $c = sqrta^2 + b^2$. Find or construct a point $D$ on the hypothenuse such that the distance $|CD| = |DE|$, where $E$ is positioned on $AB$ in such a way that $DEparallel BC$ ($DE$ is parallel to $BC$).




enter image description here



Background



My background for wanting such a distance is that I want to create a semicircle from C onto the line $AB$. This can be made clearer in the image below



enter image description here



To be able to make sure the angles is right, I needed the red and blue line to be of same length. This lead to this problem



Solution



Using similar triangles one arrives at the three equations



$$
beginalign*
fraccolorbluetextbluea - x & = fracba \
fraccolorredtextredx & = fracca \
colorredtextred & = colorbluetextblue
endalign*
$$



Where one easily can solve for $colorbluetextblue$, $colorredtextred$, $x$.



Question



I feel my solution is quite barbaric and I feel that there is a better way to solve this problem. Is there another shorter, better, more intuitive solution. Or perhaps there exists a a way to construct the point $D$ in a simpler matter?







geometry triangles geometric-construction congruences-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 10 hours ago







N3buchadnezzar

















asked 10 hours ago









N3buchadnezzarN3buchadnezzar

6,04233475




6,04233475







  • 1




    $begingroup$
    You still have not described what $F$ is either from the statement or from the graph.
    $endgroup$
    – Hw Chu
    10 hours ago










  • $begingroup$
    Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
    $endgroup$
    – N3buchadnezzar
    10 hours ago










  • $begingroup$
    cateti is Italian for legs
    $endgroup$
    – J. W. Tanner
    10 hours ago






  • 1




    $begingroup$
    Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
    $endgroup$
    – David K
    6 hours ago












  • 1




    $begingroup$
    You still have not described what $F$ is either from the statement or from the graph.
    $endgroup$
    – Hw Chu
    10 hours ago










  • $begingroup$
    Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
    $endgroup$
    – N3buchadnezzar
    10 hours ago










  • $begingroup$
    cateti is Italian for legs
    $endgroup$
    – J. W. Tanner
    10 hours ago






  • 1




    $begingroup$
    Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
    $endgroup$
    – David K
    6 hours ago







1




1




$begingroup$
You still have not described what $F$ is either from the statement or from the graph.
$endgroup$
– Hw Chu
10 hours ago




$begingroup$
You still have not described what $F$ is either from the statement or from the graph.
$endgroup$
– Hw Chu
10 hours ago












$begingroup$
Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
$endgroup$
– N3buchadnezzar
10 hours ago




$begingroup$
Right when I said $F$ i meant $E$. I will fix it in the problem statement =)
$endgroup$
– N3buchadnezzar
10 hours ago












$begingroup$
cateti is Italian for legs
$endgroup$
– J. W. Tanner
10 hours ago




$begingroup$
cateti is Italian for legs
$endgroup$
– J. W. Tanner
10 hours ago




1




1




$begingroup$
Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
$endgroup$
– David K
6 hours ago




$begingroup$
Your system of equations very quickly and easily simplifies to $x=ab/(b+c).$ That does not seem too ugly. But the half-angle method also works. In fact, your problem is a nice way to derive at least one of the half-angle formulas for the tangent function!
$endgroup$
– David K
6 hours ago










4 Answers
4






active

oldest

votes


















6












$begingroup$

enter image description here



Let the angle bisector of $angle ACB$ intersect side $overlineAB$ at point $E$.



Let the measure of $angle ACB$ be $alpha$.
Then $m angle ACE = dfracalpha2$.



Let the line perpendicular to side $overlineAB$ at point $E$ intersect side
$overlineAC$ at point $D$.



Since $overlineED$ is parallel to $overlineBC$, then
$angle ADE cong angle ACB$.



By the exterior angle theorem, $mangle DEC = dfracalpha2$.



Hence $triangle EDC$ is isoceles.



So $CD = DE$.



(Added later). Assuming that the sides have lengths of $x$ and $y$, and that
$r = sqrtx^2+y^2$, the lengths of the segments are displayed below.



enter image description here






share|cite|improve this answer











$endgroup$












  • $begingroup$
    This is exactly what I was looking for =)
    $endgroup$
    – N3buchadnezzar
    8 hours ago


















1












$begingroup$

The curve that traces out the points that are equidistant to $C$ and the line extension of $AB$ is a parabola with focus $C$ and directrix $AB$.



Choosing a convenient coordinate system (parallel to $AB$ with $B$ as the origin), I get the equation $y = fracx^22b + fracb2$, which you want to intersect with the line $y = fracbax + b$.



Solving, I get that the point $D$ has $(x,y)$ coordinates of $x = fracb(b - c)a$ and $y = fracbc(c - b)a^2$.






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    Let $$CD=DE=y$$ then we get $$fracbc=fracyc-y$$ so $$y=fracbcb+c$$






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      @mathmandan There was an edit after my comment.
      $endgroup$
      – Michael Biro
      5 hours ago


















    0












    $begingroup$

    Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=fracxcos A$ so the equation to solve is $$(a-x)fracba=fracxsqrta^2+b^2a$$ or $$x=fracabsqrta^2+b^2+b$$



    Just another idea to construct point $E$: since $triangleDCE$ is isosceles, it's easy to find $angleACE=(90°-A)/2$






    share|cite|improve this answer











    $endgroup$













      Your Answer








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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      enter image description here



      Let the angle bisector of $angle ACB$ intersect side $overlineAB$ at point $E$.



      Let the measure of $angle ACB$ be $alpha$.
      Then $m angle ACE = dfracalpha2$.



      Let the line perpendicular to side $overlineAB$ at point $E$ intersect side
      $overlineAC$ at point $D$.



      Since $overlineED$ is parallel to $overlineBC$, then
      $angle ADE cong angle ACB$.



      By the exterior angle theorem, $mangle DEC = dfracalpha2$.



      Hence $triangle EDC$ is isoceles.



      So $CD = DE$.



      (Added later). Assuming that the sides have lengths of $x$ and $y$, and that
      $r = sqrtx^2+y^2$, the lengths of the segments are displayed below.



      enter image description here






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        This is exactly what I was looking for =)
        $endgroup$
        – N3buchadnezzar
        8 hours ago















      6












      $begingroup$

      enter image description here



      Let the angle bisector of $angle ACB$ intersect side $overlineAB$ at point $E$.



      Let the measure of $angle ACB$ be $alpha$.
      Then $m angle ACE = dfracalpha2$.



      Let the line perpendicular to side $overlineAB$ at point $E$ intersect side
      $overlineAC$ at point $D$.



      Since $overlineED$ is parallel to $overlineBC$, then
      $angle ADE cong angle ACB$.



      By the exterior angle theorem, $mangle DEC = dfracalpha2$.



      Hence $triangle EDC$ is isoceles.



      So $CD = DE$.



      (Added later). Assuming that the sides have lengths of $x$ and $y$, and that
      $r = sqrtx^2+y^2$, the lengths of the segments are displayed below.



      enter image description here






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        This is exactly what I was looking for =)
        $endgroup$
        – N3buchadnezzar
        8 hours ago













      6












      6








      6





      $begingroup$

      enter image description here



      Let the angle bisector of $angle ACB$ intersect side $overlineAB$ at point $E$.



      Let the measure of $angle ACB$ be $alpha$.
      Then $m angle ACE = dfracalpha2$.



      Let the line perpendicular to side $overlineAB$ at point $E$ intersect side
      $overlineAC$ at point $D$.



      Since $overlineED$ is parallel to $overlineBC$, then
      $angle ADE cong angle ACB$.



      By the exterior angle theorem, $mangle DEC = dfracalpha2$.



      Hence $triangle EDC$ is isoceles.



      So $CD = DE$.



      (Added later). Assuming that the sides have lengths of $x$ and $y$, and that
      $r = sqrtx^2+y^2$, the lengths of the segments are displayed below.



      enter image description here






      share|cite|improve this answer











      $endgroup$



      enter image description here



      Let the angle bisector of $angle ACB$ intersect side $overlineAB$ at point $E$.



      Let the measure of $angle ACB$ be $alpha$.
      Then $m angle ACE = dfracalpha2$.



      Let the line perpendicular to side $overlineAB$ at point $E$ intersect side
      $overlineAC$ at point $D$.



      Since $overlineED$ is parallel to $overlineBC$, then
      $angle ADE cong angle ACB$.



      By the exterior angle theorem, $mangle DEC = dfracalpha2$.



      Hence $triangle EDC$ is isoceles.



      So $CD = DE$.



      (Added later). Assuming that the sides have lengths of $x$ and $y$, and that
      $r = sqrtx^2+y^2$, the lengths of the segments are displayed below.



      enter image description here







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 1 hour ago

























      answered 9 hours ago









      steven gregorysteven gregory

      18.5k32359




      18.5k32359











      • $begingroup$
        This is exactly what I was looking for =)
        $endgroup$
        – N3buchadnezzar
        8 hours ago
















      • $begingroup$
        This is exactly what I was looking for =)
        $endgroup$
        – N3buchadnezzar
        8 hours ago















      $begingroup$
      This is exactly what I was looking for =)
      $endgroup$
      – N3buchadnezzar
      8 hours ago




      $begingroup$
      This is exactly what I was looking for =)
      $endgroup$
      – N3buchadnezzar
      8 hours ago











      1












      $begingroup$

      The curve that traces out the points that are equidistant to $C$ and the line extension of $AB$ is a parabola with focus $C$ and directrix $AB$.



      Choosing a convenient coordinate system (parallel to $AB$ with $B$ as the origin), I get the equation $y = fracx^22b + fracb2$, which you want to intersect with the line $y = fracbax + b$.



      Solving, I get that the point $D$ has $(x,y)$ coordinates of $x = fracb(b - c)a$ and $y = fracbc(c - b)a^2$.






      share|cite|improve this answer









      $endgroup$

















        1












        $begingroup$

        The curve that traces out the points that are equidistant to $C$ and the line extension of $AB$ is a parabola with focus $C$ and directrix $AB$.



        Choosing a convenient coordinate system (parallel to $AB$ with $B$ as the origin), I get the equation $y = fracx^22b + fracb2$, which you want to intersect with the line $y = fracbax + b$.



        Solving, I get that the point $D$ has $(x,y)$ coordinates of $x = fracb(b - c)a$ and $y = fracbc(c - b)a^2$.






        share|cite|improve this answer









        $endgroup$















          1












          1








          1





          $begingroup$

          The curve that traces out the points that are equidistant to $C$ and the line extension of $AB$ is a parabola with focus $C$ and directrix $AB$.



          Choosing a convenient coordinate system (parallel to $AB$ with $B$ as the origin), I get the equation $y = fracx^22b + fracb2$, which you want to intersect with the line $y = fracbax + b$.



          Solving, I get that the point $D$ has $(x,y)$ coordinates of $x = fracb(b - c)a$ and $y = fracbc(c - b)a^2$.






          share|cite|improve this answer









          $endgroup$



          The curve that traces out the points that are equidistant to $C$ and the line extension of $AB$ is a parabola with focus $C$ and directrix $AB$.



          Choosing a convenient coordinate system (parallel to $AB$ with $B$ as the origin), I get the equation $y = fracx^22b + fracb2$, which you want to intersect with the line $y = fracbax + b$.



          Solving, I get that the point $D$ has $(x,y)$ coordinates of $x = fracb(b - c)a$ and $y = fracbc(c - b)a^2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 9 hours ago









          Michael BiroMichael Biro

          11.7k21931




          11.7k21931





















              1












              $begingroup$

              Let $$CD=DE=y$$ then we get $$fracbc=fracyc-y$$ so $$y=fracbcb+c$$






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                @mathmandan There was an edit after my comment.
                $endgroup$
                – Michael Biro
                5 hours ago















              1












              $begingroup$

              Let $$CD=DE=y$$ then we get $$fracbc=fracyc-y$$ so $$y=fracbcb+c$$






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                @mathmandan There was an edit after my comment.
                $endgroup$
                – Michael Biro
                5 hours ago













              1












              1








              1





              $begingroup$

              Let $$CD=DE=y$$ then we get $$fracbc=fracyc-y$$ so $$y=fracbcb+c$$






              share|cite|improve this answer











              $endgroup$



              Let $$CD=DE=y$$ then we get $$fracbc=fracyc-y$$ so $$y=fracbcb+c$$







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 9 hours ago

























              answered 10 hours ago









              Dr. Sonnhard GraubnerDr. Sonnhard Graubner

              79.1k42867




              79.1k42867











              • $begingroup$
                @mathmandan There was an edit after my comment.
                $endgroup$
                – Michael Biro
                5 hours ago
















              • $begingroup$
                @mathmandan There was an edit after my comment.
                $endgroup$
                – Michael Biro
                5 hours ago















              $begingroup$
              @mathmandan There was an edit after my comment.
              $endgroup$
              – Michael Biro
              5 hours ago




              $begingroup$
              @mathmandan There was an edit after my comment.
              $endgroup$
              – Michael Biro
              5 hours ago











              0












              $begingroup$

              Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=fracxcos A$ so the equation to solve is $$(a-x)fracba=fracxsqrta^2+b^2a$$ or $$x=fracabsqrta^2+b^2+b$$



              Just another idea to construct point $E$: since $triangleDCE$ is isosceles, it's easy to find $angleACE=(90°-A)/2$






              share|cite|improve this answer











              $endgroup$

















                0












                $begingroup$

                Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=fracxcos A$ so the equation to solve is $$(a-x)fracba=fracxsqrta^2+b^2a$$ or $$x=fracabsqrta^2+b^2+b$$



                Just another idea to construct point $E$: since $triangleDCE$ is isosceles, it's easy to find $angleACE=(90°-A)/2$






                share|cite|improve this answer











                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=fracxcos A$ so the equation to solve is $$(a-x)fracba=fracxsqrta^2+b^2a$$ or $$x=fracabsqrta^2+b^2+b$$



                  Just another idea to construct point $E$: since $triangleDCE$ is isosceles, it's easy to find $angleACE=(90°-A)/2$






                  share|cite|improve this answer











                  $endgroup$



                  Not sure if this is less barbaric but using simple trig: $DE=(a-x)tan A$, $DC=fracxcos A$ so the equation to solve is $$(a-x)fracba=fracxsqrta^2+b^2a$$ or $$x=fracabsqrta^2+b^2+b$$



                  Just another idea to construct point $E$: since $triangleDCE$ is isosceles, it's easy to find $angleACE=(90°-A)/2$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 9 hours ago

























                  answered 9 hours ago









                  VasyaVasya

                  4,5091619




                  4,5091619



























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