We have a line d and a point A.
We pass a line perpendicular to d passing through A called D.
Let H be any point on D.
Then can we place points B and C on d such that H
is the orthocenter of triangle ABC and the angle(BAC)=60 degrees?
If so, then how?
P.S. The question appears in a chapter for vector angles,
and inscribed angles, so these may be useful in solving
perhaps I did not specify it that way in my question but:
A is any point in the plane not belonging to d, and H is any point on D.
(in the sense that we must take care of considering every case in solving the problem)
I had figured that if the distance between d and H is greater the distance between d and A then it is not possible to find the given triangle.
Though I don't know if I am right. And if I am, I don't know how to prove it.
now, supposing otherwise, the answer is a little harder to find. i haven't put much thought into it, as i am about to go to bed, but i have this idea. we can place our system on the coordinate axis. let d be the x-axis let's say. and A be an arbitrary point, say . then the line D will be the line . and then we place a point H on the line D, let H be the point, say , where . now place points on d, and . and without loss of generality, we can let so that C lies on the line d after B. and if B and C are on one side of the line that connects A to d, then it is again impossible to get the desired outcome, so we must have B on the left side of this line, and C on the right.
now, we start the hard task of constructing lines from B to A and C to A. and also, a line from AB to C so that it is perpendicular to AB and a line from AC to B such that it is perpendicular to AC. and at the same time, we have to make sure that these two lines, intersect with the line connecting A to d at the point H. this is doable, but it will be hard and frustrating, since we are given no definite information and all of this will be based on arbitrary points.
i leave it to you to hash this out. my head hurts enough right now, and i'm sleepy
I've made several sketches of the problem. The result by first sight: H is situated on D below A. (see attachment: I translated B to the left, then H is coming down)
To prove this result I used vectors. (The writing of vectors (in Germany) differs slightly from the way you are probably used to. But I hope you can get what I'm doing)
Place d on the x-axis and A on the y-axis = D. Thus
The perpendicular direction of is and the perpendicular direction of is
The orthocenter is the intersection of the heights of the triangle. The equation of the height passing through B is:
Calculating the intersection:
This will give you a system of 2 simultaneous equations:
If solved the 2nd equation for s and plugged this value into the 1rst equation. After few stepa of simplifying I got:
This value for r will give you the position vector of H when plugged into the 2nd equation:
That means the vector is pointing in the same direction as the vector and therefore
As you certainly have noticed I didn't use the property
You know the formula to calculate the angle which is enclosed by 2 vectors. I use this formula and get:
Unfortunately I haven't found a way to use this property to get a more specific value for the components of .