Euclidia isosceles triangle on tangent points 3.10












1












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I've been struggling to come up with a solution for the Euclidia problem 3.10, creating an isosceles triangle from two given points on a circle. I can do it, but not in the # of steps given in the hints, for both the L and E solutions. Web search only find solutions up to 3.8. Has anyone here figured these out? Maybe provide another hint?
Thanks






euclidean-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the number of steps given in the hint? And, for those not familiar, you should provide a link to Euclidea, note what counts as a step, and what "L and E solutions" are.
    $endgroup$
    – Blue
    Apr 15 '18 at 19:05
















1












$begingroup$


I've been struggling to come up with a solution for the Euclidia problem 3.10, creating an isosceles triangle from two given points on a circle. I can do it, but not in the # of steps given in the hints, for both the L and E solutions. Web search only find solutions up to 3.8. Has anyone here figured these out? Maybe provide another hint?
Thanks






euclidean-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is the number of steps given in the hint? And, for those not familiar, you should provide a link to Euclidea, note what counts as a step, and what "L and E solutions" are.
    $endgroup$
    – Blue
    Apr 15 '18 at 19:05














1












1








1


2



$begingroup$


I've been struggling to come up with a solution for the Euclidia problem 3.10, creating an isosceles triangle from two given points on a circle. I can do it, but not in the # of steps given in the hints, for both the L and E solutions. Web search only find solutions up to 3.8. Has anyone here figured these out? Maybe provide another hint?
Thanks






euclidean-geometry







share|cite|improve this question









$endgroup$




I've been struggling to come up with a solution for the Euclidia problem 3.10, creating an isosceles triangle from two given points on a circle. I can do it, but not in the # of steps given in the hints, for both the L and E solutions. Web search only find solutions up to 3.8. Has anyone here figured these out? Maybe provide another hint?
Thanks






euclidean-geometry




euclidean-geometry






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Apr 15 '18 at 17:22









Laser RayLaser Ray

63




63












  • $begingroup$
    What is the number of steps given in the hint? And, for those not familiar, you should provide a link to Euclidea, note what counts as a step, and what "L and E solutions" are.
    $endgroup$
    – Blue
    Apr 15 '18 at 19:05


















  • $begingroup$
    What is the number of steps given in the hint? And, for those not familiar, you should provide a link to Euclidea, note what counts as a step, and what "L and E solutions" are.
    $endgroup$
    – Blue
    Apr 15 '18 at 19:05
















$begingroup$
What is the number of steps given in the hint? And, for those not familiar, you should provide a link to Euclidea, note what counts as a step, and what "L and E solutions" are.
$endgroup$
– Blue
Apr 15 '18 at 19:05




$begingroup$
What is the number of steps given in the hint? And, for those not familiar, you should provide a link to Euclidea, note what counts as a step, and what "L and E solutions" are.
$endgroup$
– Blue
Apr 15 '18 at 19:05










1 Answer
1






active

oldest

votes


















0












$begingroup$

For L:
Draw the line that pases for the bottom point and the centre of the circunference, then draw the mediatrix between the two given points. Draw the perpendicular to the bottom point(tangent to the circunference) and where this line cross with the mediatrix, draw the line that passes from that point and from the other given point, last step is intuitive.



For E:
Lets call the centre of the circunference C, the top given point A and the bottom given point B.



1-Draw the circunfence with centre on B and radius BA.It intersects the first circunference on points A and C.



2-Circunference with centre on C and radius CA.It intersects the last circunference on A and D.



3- Draw the line CD



4-Draw the circunference with centre on A and radius AB. It intersects the original circunference on two points, lets the one on the right be E.



5-Draw the circunference with centre on B and radius AE. It intersects the one created on step 4 on points E and F.



6-Draw the segment BF. It crosses the line CD on the point G.



7-Circunference with centre on B and radius BG. Intersets with BF on G and H.



8- Draw the segment HA.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, but that is what Ive been trying. These steps give the base and one side. The other side has only one point defined (top), and seems to lack either a tangent point on the circle, or the intersection point on the base. Sorry, but I seem to be missing something. Im sure its obvious...
    $endgroup$
    – Laser Ray
    Apr 15 '18 at 17:45












  • $begingroup$
    The last side is the perpendicular to the mediatrix we created before, in the point where it intersects with the circunference.
    $endgroup$
    – Alfredo
    Apr 15 '18 at 18:14










  • $begingroup$
    OK, I see now it is obvious. I was so fixated on the base being on the bottom, I failed to see it. Thank you!
    $endgroup$
    – Laser Ray
    Apr 15 '18 at 18:26










  • $begingroup$
    I edited the E solution aswell.
    $endgroup$
    – Alfredo
    Apr 15 '18 at 18:27











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1 Answer
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votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

For L:
Draw the line that pases for the bottom point and the centre of the circunference, then draw the mediatrix between the two given points. Draw the perpendicular to the bottom point(tangent to the circunference) and where this line cross with the mediatrix, draw the line that passes from that point and from the other given point, last step is intuitive.



For E:
Lets call the centre of the circunference C, the top given point A and the bottom given point B.



1-Draw the circunfence with centre on B and radius BA.It intersects the first circunference on points A and C.



2-Circunference with centre on C and radius CA.It intersects the last circunference on A and D.



3- Draw the line CD



4-Draw the circunference with centre on A and radius AB. It intersects the original circunference on two points, lets the one on the right be E.



5-Draw the circunference with centre on B and radius AE. It intersects the one created on step 4 on points E and F.



6-Draw the segment BF. It crosses the line CD on the point G.



7-Circunference with centre on B and radius BG. Intersets with BF on G and H.



8- Draw the segment HA.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, but that is what Ive been trying. These steps give the base and one side. The other side has only one point defined (top), and seems to lack either a tangent point on the circle, or the intersection point on the base. Sorry, but I seem to be missing something. Im sure its obvious...
    $endgroup$
    – Laser Ray
    Apr 15 '18 at 17:45












  • $begingroup$
    The last side is the perpendicular to the mediatrix we created before, in the point where it intersects with the circunference.
    $endgroup$
    – Alfredo
    Apr 15 '18 at 18:14










  • $begingroup$
    OK, I see now it is obvious. I was so fixated on the base being on the bottom, I failed to see it. Thank you!
    $endgroup$
    – Laser Ray
    Apr 15 '18 at 18:26










  • $begingroup$
    I edited the E solution aswell.
    $endgroup$
    – Alfredo
    Apr 15 '18 at 18:27
















0












$begingroup$

For L:
Draw the line that pases for the bottom point and the centre of the circunference, then draw the mediatrix between the two given points. Draw the perpendicular to the bottom point(tangent to the circunference) and where this line cross with the mediatrix, draw the line that passes from that point and from the other given point, last step is intuitive.



For E:
Lets call the centre of the circunference C, the top given point A and the bottom given point B.



1-Draw the circunfence with centre on B and radius BA.It intersects the first circunference on points A and C.



2-Circunference with centre on C and radius CA.It intersects the last circunference on A and D.



3- Draw the line CD



4-Draw the circunference with centre on A and radius AB. It intersects the original circunference on two points, lets the one on the right be E.



5-Draw the circunference with centre on B and radius AE. It intersects the one created on step 4 on points E and F.



6-Draw the segment BF. It crosses the line CD on the point G.



7-Circunference with centre on B and radius BG. Intersets with BF on G and H.



8- Draw the segment HA.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks, but that is what Ive been trying. These steps give the base and one side. The other side has only one point defined (top), and seems to lack either a tangent point on the circle, or the intersection point on the base. Sorry, but I seem to be missing something. Im sure its obvious...
    $endgroup$
    – Laser Ray
    Apr 15 '18 at 17:45












  • $begingroup$
    The last side is the perpendicular to the mediatrix we created before, in the point where it intersects with the circunference.
    $endgroup$
    – Alfredo
    Apr 15 '18 at 18:14










  • $begingroup$
    OK, I see now it is obvious. I was so fixated on the base being on the bottom, I failed to see it. Thank you!
    $endgroup$
    – Laser Ray
    Apr 15 '18 at 18:26










  • $begingroup$
    I edited the E solution aswell.
    $endgroup$
    – Alfredo
    Apr 15 '18 at 18:27














0












0








0





$begingroup$

For L:
Draw the line that pases for the bottom point and the centre of the circunference, then draw the mediatrix between the two given points. Draw the perpendicular to the bottom point(tangent to the circunference) and where this line cross with the mediatrix, draw the line that passes from that point and from the other given point, last step is intuitive.



For E:
Lets call the centre of the circunference C, the top given point A and the bottom given point B.



1-Draw the circunfence with centre on B and radius BA.It intersects the first circunference on points A and C.



2-Circunference with centre on C and radius CA.It intersects the last circunference on A and D.



3- Draw the line CD



4-Draw the circunference with centre on A and radius AB. It intersects the original circunference on two points, lets the one on the right be E.



5-Draw the circunference with centre on B and radius AE. It intersects the one created on step 4 on points E and F.



6-Draw the segment BF. It crosses the line CD on the point G.



7-Circunference with centre on B and radius BG. Intersets with BF on G and H.



8- Draw the segment HA.






share|cite|improve this answer











$endgroup$



For L:
Draw the line that pases for the bottom point and the centre of the circunference, then draw the mediatrix between the two given points. Draw the perpendicular to the bottom point(tangent to the circunference) and where this line cross with the mediatrix, draw the line that passes from that point and from the other given point, last step is intuitive.



For E:
Lets call the centre of the circunference C, the top given point A and the bottom given point B.



1-Draw the circunfence with centre on B and radius BA.It intersects the first circunference on points A and C.



2-Circunference with centre on C and radius CA.It intersects the last circunference on A and D.



3- Draw the line CD



4-Draw the circunference with centre on A and radius AB. It intersects the original circunference on two points, lets the one on the right be E.



5-Draw the circunference with centre on B and radius AE. It intersects the one created on step 4 on points E and F.



6-Draw the segment BF. It crosses the line CD on the point G.



7-Circunference with centre on B and radius BG. Intersets with BF on G and H.



8- Draw the segment HA.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 15 '18 at 18:26

























answered Apr 15 '18 at 17:30









AlfredoAlfredo

895




895












  • $begingroup$
    Thanks, but that is what Ive been trying. These steps give the base and one side. The other side has only one point defined (top), and seems to lack either a tangent point on the circle, or the intersection point on the base. Sorry, but I seem to be missing something. Im sure its obvious...
    $endgroup$
    – Laser Ray
    Apr 15 '18 at 17:45












  • $begingroup$
    The last side is the perpendicular to the mediatrix we created before, in the point where it intersects with the circunference.
    $endgroup$
    – Alfredo
    Apr 15 '18 at 18:14










  • $begingroup$
    OK, I see now it is obvious. I was so fixated on the base being on the bottom, I failed to see it. Thank you!
    $endgroup$
    – Laser Ray
    Apr 15 '18 at 18:26










  • $begingroup$
    I edited the E solution aswell.
    $endgroup$
    – Alfredo
    Apr 15 '18 at 18:27


















  • $begingroup$
    Thanks, but that is what Ive been trying. These steps give the base and one side. The other side has only one point defined (top), and seems to lack either a tangent point on the circle, or the intersection point on the base. Sorry, but I seem to be missing something. Im sure its obvious...
    $endgroup$
    – Laser Ray
    Apr 15 '18 at 17:45












  • $begingroup$
    The last side is the perpendicular to the mediatrix we created before, in the point where it intersects with the circunference.
    $endgroup$
    – Alfredo
    Apr 15 '18 at 18:14










  • $begingroup$
    OK, I see now it is obvious. I was so fixated on the base being on the bottom, I failed to see it. Thank you!
    $endgroup$
    – Laser Ray
    Apr 15 '18 at 18:26










  • $begingroup$
    I edited the E solution aswell.
    $endgroup$
    – Alfredo
    Apr 15 '18 at 18:27
















$begingroup$
Thanks, but that is what Ive been trying. These steps give the base and one side. The other side has only one point defined (top), and seems to lack either a tangent point on the circle, or the intersection point on the base. Sorry, but I seem to be missing something. Im sure its obvious...
$endgroup$
– Laser Ray
Apr 15 '18 at 17:45






$begingroup$
Thanks, but that is what Ive been trying. These steps give the base and one side. The other side has only one point defined (top), and seems to lack either a tangent point on the circle, or the intersection point on the base. Sorry, but I seem to be missing something. Im sure its obvious...
$endgroup$
– Laser Ray
Apr 15 '18 at 17:45














$begingroup$
The last side is the perpendicular to the mediatrix we created before, in the point where it intersects with the circunference.
$endgroup$
– Alfredo
Apr 15 '18 at 18:14




$begingroup$
The last side is the perpendicular to the mediatrix we created before, in the point where it intersects with the circunference.
$endgroup$
– Alfredo
Apr 15 '18 at 18:14












$begingroup$
OK, I see now it is obvious. I was so fixated on the base being on the bottom, I failed to see it. Thank you!
$endgroup$
– Laser Ray
Apr 15 '18 at 18:26




$begingroup$
OK, I see now it is obvious. I was so fixated on the base being on the bottom, I failed to see it. Thank you!
$endgroup$
– Laser Ray
Apr 15 '18 at 18:26












$begingroup$
I edited the E solution aswell.
$endgroup$
– Alfredo
Apr 15 '18 at 18:27




$begingroup$
I edited the E solution aswell.
$endgroup$
– Alfredo
Apr 15 '18 at 18:27


















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