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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Martoman on March 03, 2010, 09:30:49 pm

Title: Prove that the altitudes of a triangle are concurrent
Post by: Martoman on March 03, 2010, 09:30:49 pm: So i'm bored and want to determine if the altitudes of a triangle are indeed concurrent by vector methods.

I have http://img685.imageshack.us/i/prooconf.jpg/

Half of the problem was getting it into this form and declaring those vectors.

Now from that diagram i know that
$b$ is perpendicular to AC.
$\therefore b\cdot(c-a)=0$
$b\cdot c - a\cdot b = 0$
Hence $b\cdot c = a\cdot b$

Similarly for the other side BC $a\cdot b = a \cdot c$

Hrmmm, now I am wondering where to go from here.

I was thinking along the lines of $|a| \cdot |b| cos(\theta) = |a| \cdot |c| cos(\theta)$

Then $|b| = |c|$
and similarly $|a| = |c|$

Then they all intersect at the same point, but I think this reasoning is fallacious as it assumes the angles to be the same.... correct?

If this doesn't work, then where to go?
Title: Re: Prove that the altitudes of a triangle are concurrent
Post by: kamil9876 on March 04, 2010, 12:31:37 am: correct: you can't assume the $\theta$ are equal.

A good way to finish it is to stretch out the line segment $c$ so that it hits $AB$ . This is vector is basically $\lambda c$ there $\lambda$ is a non-zero real number (scale factor).

Now you wish to show $\lambda c$ is perpendicular to $AB$ , so try to do a dot product again.

Great start on the proof btw, I remember comming up with something similair for this one:

"Prove that if the midpoint of each line segment is connected to the point on the opposite side, then these three line segments are concurrent"
Title: Re: Prove that the altitudes of a triangle are concurrent
Post by: Martoman on March 04, 2010, 07:51:10 am: Yes that was where I started as well with that proof "prove that the medians of a triangle are concurrent". That was bitchy, but a bit easier than this one.
Title: Re: Prove that the altitudes of a triangle are concurrent
Post by: Martoman on March 06, 2010, 01:06:23 am: Quote from: kamil9876 on March 04, 2010, 12:31:37 am
correct: you can't assume the $\theta$ are equal.

A good way to finish it is to stretch out the line segment $c$ so that it hits $AB$ . This is vector is basically $\lambda c$ there $\lambda$ is a non-zero real number (scale factor).

Now you wish to show $\lambda c$ is perpendicular to $AB$ , so try to do a dot product again.

Great start on the proof btw, I remember comming up with something similair for this one:

"Prove that if the midpoint of each line segment is connected to the point on the opposite side, then these three line segments are concurrent"

I may be being silly here.... given its 1 in the freaking morning... but if $\lambda c$ perp to AB means $\lambda c \cdot (-a + b) = 0$ Then the lambda just gets canceled out.
Title: Re: Prove that the altitudes of a triangle are concurrent
Post by: kamil9876 on March 06, 2010, 01:09:40 am: yeah true, just makes the expression easier. I pointed out $\lambda$ is non-zero, and that is why you can 'cancel' it.
Title: Re: Prove that the altitudes of a triangle are concurrent
Post by: Martoman on March 06, 2010, 01:18:17 am: Oh. Disgusting. Interestingly, i just thought if you inscribe it in a circle then instantly |a|=|b|=|c| but it isn't a vector proof. Fuck my creativity.