Inscribed Triangle in a Parabola: othogonal projection and mirroring the triangle
Hello,
I am new to Geogebra and finding it both extraordinary and quite fiddly. I am getting confused trying to do he following on this triangle inscribed in a parabola (screenshot attached. The parabola is x^2 (can I do LaTeX or similar in this forum?). A, B and M are all points on the parabola and they will all move. My task is to determine how to maximise the area of ABM.
The first thing I wanted to do was to have an orthogonal projection of M on AB. I played around with the perpendicular line tools and I am getting all kinds of results except the orthogonal projection. I also tried a segment drawn from M to AV, but of course, the segment does not adjust when I move A, B and/or M. I know this should be simple, but I just cannot work it out.
Another way of approaching the task at hand is to create a parallelogram by mirroring ABM around AB and flipping it horizontally. I used the rotation/translation and mirroring tools but again, I just can't figure out the combination that creates the parallelogram. By mirroring ABM about AB, I get what I think is the right first step, but I don't know how to flip the new triangle obtained to get there (see the second screenshot for where I end up after this first step).
Finally, I have worked out playing around with the points A, B and C that in order for the area of the triangle to be maximised, xm has to be 1/2 (xa+xb). The above are two methods I am trying out to help me reach that conclusion with proof. But is there a way for Geogebra to work out how to maximise the area of ABC wherever A and B are located? I looked at some of the max/maximise functions but I cannot see anything that fits. I also wondered whether that could be done in the CAS perspective but I am just beginning to use Geogebra in earnest and I just don't know how to lay this out in CAS anyway. All suggestions, pointers and comments are very welcome.
Thank you for taking the time to look at this. Any partial answer will help. I am sorry that my knowledge of Geogebra is so basic that I have to ask so many questions: thank you for your patience.
Chris
0 Links
The same question 
Bonsoir,
Je ne suis pas sûr que c'est ce que tu veux mais on peut démontrer que l'aire du triangle ABM est maximale lorsque M est le point de contact de la tangente à la parabole parallèle à (AB).
Bonsoir,
Merci beaucoup de ta réponse. Je cherche tout d’abord à comprendre comment effectuer les manipulations sur le graphique; le projeté orthogonal de M sur AB et la création d'un triangle "miroir inversé" de ABM qui avec ABM forme un parallélogramme qui va s'ajuster quand je bouge A et B: je ne sais pas le faire.
Je comptais utiliser une de ces deux méthodes pour en déduire (sur papier) que l'aire d'ABM est maximisée quand xm = (xa+xb)/2, ce que je vois bien dans mon graphique. Je pense que ta méthode est plus élégante mais je ne suis pas sur de comprendre comment faire au sein même de Geogebra (ni encore sur papier d'ailleurs, mais je vais regarder bien sûr).
Merci beaucoup de prendre le temps de répondre: c"est très encourageant pour les néophytes comme moi (en maths comme en Geogebra).
J'avais oublié de mettre mon fichier Geogebra en fichier joint: je le fais plus bas.
Bien cordialement
Chris.
I dont know wether I understood exactly what you want
Calculate Area with Gauss-Trapezformula
https://www.geogebra.org/m/A5jFPkJS
M=(x,x^2), B=(a2,a^2), A=(a1,a1^2)
F(x):=1 / 2 ((a1 - x) (a1² + x²) + (a1² + a2²) (-a1 + a2) + (a2² + x²) (-a2 + x))
F'(x)=1 / 2 (-a1² + a2² + 2a1 x - 2a2 x)
F'(x)=0 ====> M_{max}:=((a1 + a2) / 2, (a1² + 2a1 a2 + a2²) / 4)
A=(-2,4), B=(1,1) ===> M=(-1/2,1/4)
Hello hawe,
Thank you very much for our response. This is not the way I did it on paper but that's really useful, thank you very much. I don't know the Gauss-Trapeze formula but I think it may be the same in the end as using the fact that the triangle is half a parallelogram. The end formula seems the same.
But although there are several approaches to solving this, what I really don't know is how to do that in Geogebra. I can do it on paper, but I don't know whether I lay it out in CAS in Geogebra, or somehow only use the graph perspective. That is why I was asking how to flip the triangle to create a parallelogram (as in my second screenshot) or (alternative method for solving) by projecting M onto AB and reasoning with vectors, which I can just about manage...except I don't know how to do an orthogonal projection in Geogebra (just tracing a line segment to AB from M but making sure that this segment stays perpendicular to AB when I move A, B and M on the graph. I can draw a perpendicular from M to AB, but that is a static drawing: when I move the points A or B, that segment does not move. How to do this was one of my two questions.
Anyway, in terms of actually solving, I will definitely look at the above and see if it is the equivalent of the parallelogramm method.
I really appreciate your answer. Thank you!
Best regards
Chris
0 Links
Hm,
you can dilate the triangle to create a parallelogram see file.
I set M to max, you may go back to bind M to f(x) again...
Comments have been locked on this page!