finding triangle for given orthocenter

**tombrownington** · October 13th 2007, 11:29 PM

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
the problem.

**Jhevon** · October 13th 2007, 11:54 PM

Originally Posted by tombrownington

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
the problem.

it certainly is possible, and there are perhaps several ways to get the desired outcome (and several ways so that it is impossible to get it). your question is vague, the fact that H can be "any point" on D, and the fact that A is some point that we don't even know how far from d it is, makes being specific as to how to get the desired outcome very hard. do we have the freedom of choosing where we want A and H to be?

**tombrownington** · October 14th 2007, 12:17 AM

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.

**Jhevon** · October 14th 2007, 12:31 AM

Originally Posted by tombrownington

...
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.

well, that is true. the proof is simple. if the distance between d and H is greater than the distance between d and A, then H will lie outside ANY triangle ABC, and therefore it cannot be the orthocenter. it is also impossible if H lies right on top of A, for nearly the same reason.

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 $(x_1,y_1)$ . then the line D will be the line $x = x_1$ . and then we place a point H on the line D, let H be the point, say $(x_2,y_1)$ , where $x_2 < x_1$ . now place points on d, $B(x_3, 0)$ and $C(x_4,0)$ . and without loss of generality, we can let $x_4 > x_3$ 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

**earboth** · October 14th 2007, 10:36 AM

Originally Posted by tombrownington

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
the problem.

Hello,

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 $d \perp D$

$\overrightarrow{OB}={-b\choose 0}$ , $\overrightarrow{OC}={c\choose 0}$ , $\overrightarrow{OA}={0\choose a}$

Then $\overrightarrow{AB}={-b\choose -a}$ and $\overrightarrow{AC}={c \choose -a}$

The perpendicular direction of $\overrightarrow{AB}$ is $perp(\overrightarrow{AB})={-a\choose b}$ and the perpendicular direction of $\overrightarrow{AC}$ is $perp(\overrightarrow{AC})= {a \choose c}$

The orthocenter is the intersection of the heights of the triangle. The equation of the height passing through B is:

$h_B: \vec x={-b \choose 0} + s \cdot {a \choose c}$ and

$h_C: \vec x={c \choose 0} + r \cdot {-a \choose b}$

Calculating the intersection:

$h_B \cap h_C: {-b \choose 0} + s \cdot {a \choose c} = {c \choose 0} + r \cdot {-a \choose b}$ This will give you a system of 2 simultaneous equations:

$\begin{array}{c}-b+as = c-ar\\cs = br\end{array}$ If solved the 2nd equation for s and plugged this value into the 1rst equation. After few stepa of simplifying I got: $r=\frac ca$

This value for r will give you the position vector of H when plugged into the 2nd equation:

$\vec h = {c \choose 0} + \frac ca \cdot {-a \choose b} = {0 \choose \frac{bc}{a}}$

That means the vector $\vec h$ is pointing in the same direction as the vector $\overrightarrow{OA}$ and therefore $H \in D$

As you certainly have noticed I didn't use the property $\angle(BAC) = 60^\circ$

You know the formula to calculate the angle which is enclosed by 2 vectors. I use this formula and get:

$cos(\angle(BAC)=\frac{{-b \choose -a} \cdot {c \choose -a}}{\left|{-b \choose -a}\right| \cdot \left|{c \choose -a} \right|}=\frac{a^2-bc}{\sqrt{(a^2+b^2)(a^2+c^2)}}=\frac12$

Unfortunately I haven't found a way to use this property to get a more specific value for the components of $\vec h$ .

**earboth** · October 14th 2007, 10:39 AM

Originally Posted by Jhevon

... 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....

Hi,

I don't want to pick at you but I've detected something funny. (see attachment)

**Jhevon** · October 14th 2007, 10:57 AM

Originally Posted by earboth

Hi,

I don't want to pick at you but I've detected something funny. (see attachment)

i'm sorry, earboth, i don't get it. H is not the orthocenter of triangle ABC here? why is this funny in regards to what i said?

**earboth** · October 14th 2007, 11:40 AM

Originally Posted by Jhevon

i'm sorry, earboth, i don't get it. H is not the orthocenter of triangle ABC here? why is this funny in regards to what i said?

Hi,

if the triangle ABC is obtuse (and that's the case here) the orthocenter is outside the triangle. That's possible with this problem if B and C are on the same side of D.

If ABC is a right triangle (in this case C is placed on the intersection of d and D) then H = C.

**Jhevon** · October 14th 2007, 01:09 PM

Originally Posted by earboth

Hi,

if the triangle ABC is obtuse (and that's the case here) the orthocenter is outside the triangle. That's possible with this problem if B and C are on the same side of D.

If ABC is a right triangle (in this case C is placed on the intersection of d and D) then H = C.

o ok. sorry. i didn't realize the orthocenter could be outside the triangle. after looking up the definition in more detail, i realize you are correct

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