The orthocentre of the triangle and a parabola

[phinds]

"origin" was one of the possible answers. My solution gives the origin. Why is that not correct?

EDIT: and by the way, see post #8[/QUOTE]

 

[LCKurtz]

"origin" was one of the possible answers. My solution gives the origin. Why is that not correct?

Because the question didn't ask whether it was possible for the orthocenter to be the origin. It gave three arbitrary points on the parabola and asked which answer was correct for the location of the orthocenter. As my example in post #20 shows, the orthocenter can be outside the parabola in which case none of the answers are correct. You have shown a particular collection of points having orthocenter at the origin, but there is nothing in the problem that specifies that the points must be symmetric to the axis of the parabola as yours are.

 

[phinds]

Because the question didn't ask whether it was possible for the orthocenter to be the origin. It gave three arbitrary points on the parabola and asked which answer was correct for the location of the orthocenter. As my example in post #20 shows, the orthocenter can be outside the parabola in which case none of the answers are correct. You have shown a particular collection of points having orthocenter at the origin, but there is nothing in the problem that specifies that the points must be symmetric to the axis of the parabola as yours are.

See post #8 and the one it was responding to.

 

[LCKurtz]

See post #8 and the one it was responding to.

HallsofIvy is incorrect that the orthocenter must be inside the triangle. See my post #20.
Gotta run for now.

 

[phinds]

HallsofIvy is incorrect that the orthocenter must be inside the triangle. See my post #20.
Gotta run for now.

Well, my interpretation and solution do not imply that the orthocenter MUST be inside triangle, although it certainly CAN be inside. My solution in fact has the orthocenter ON the triangle (and on the origin). The interpretation here is that they are saying there ARE three point that have one of the possible solutions as the correct answer and that's what my solution does.

 

[LCKurtz]

Well, my interpretation and solution do not imply that the orthocenter MUST be inside triangle, although it certainly CAN be inside. My solution in fact has the orthocenter ON the triangle (and on the origin). The interpretation here is that they are saying there ARE three point that have one of the possible solutions as the correct answer and that's what my solution does.

We aren't disagreeing about the mathematics, only the interpretation of the problem. With your interpretation, which I don't think is reasonable, all the answers are correct.

If the points are $(0,0),(a^2,a),(a^2,-a)$ then:
If $a=1$ the orthocenter is at the origin and vertex. I guess that is what your picture is supposed to represent.
If $a = \frac {\sqrt 5} {2}$ the orthocenter is at $(\frac 1 4, 0)$, which is the focus.
If $a=\sqrt 2$ the orthocenter is at $(1,0)$.

No reason to prefer one answer over the others then.

 

[Raghav Gupta]

If $a = \frac 5 {\sqrt 2}$ the orthocenter is at $(\frac 1 4, 0)$, which is the focus.

Should be $a = \frac{\sqrt{5}}{2}$
Anyways I knew the answer that there is no answer because answer to be chosen should be one from options in much earlier posts but that phinds guy was confusing me.
Thanks for showing the calculations.
By the way thanks to all for a discussion.

 

[phinds]

Should be $a = \frac{\sqrt{5}}{2}$
Anyways I knew the answer that there is no answer because answer to be chosen should be one from options in much earlier posts but that phinds guy was confusing me.
Thanks for showing the calculations.
By the way thanks to all for a discussion.

I wish you would stop calling me "that phinds guy". It is quite rude.

 

[Raghav Gupta]

Sorry, Mr phinds or Sir phinds. I earlier thought guy was a nice word.

In a virtual world we don't know the identity of a person actually and that avatar of you gives me some other feeling.

The problem arises for me when more then two persons are involved and we have to refer to someone in third person.

Recently looked at your profile information and you are a very experienced person.
Thanks Mr.Phinds for the discussion.

 

[LCKurtz]

Should be $a = \frac{\sqrt{5}}{2}$

Yes. Thanks for catching that typo. I will correct it.

 

[Ray Vickson]

Sorry, Mr phinds or Sir phinds. I earlier thought guy was a nice word.

In a virtual world we don't know the identity of a person actually and that avatar of you gives me some other feeling.

The problem arises for me when more then two persons are involved and we have to refer to someone in third person.

You can just say "phinds in post #xx" or something similar. In principle, you may not know if a user is actually a "guy" at all!

Recently looked at your profile information and you are a very experienced person.
Thanks Mr.Phinds for the discussion.