- #1

Hannisch

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## Homework Statement

Show that a triangle's medians after suitable parallel movement (I'm directly translating from Swedish here, excuse any weird sentences) can form a new triangle. Also show that the quotation between the median triangle area and the original triangle area always are the same.

## Homework Equations

## The Attempt at a Solution

Okay, so I have a triangle that I've labled ABC. D is the midpoint on AB, E is the midpoint on AC and F is the midpoint on BC and then I draw the medians. I show the first problem by expressing AF, BE and CD through the other vectors and since ABC is a triangle, AB+BC+CA equals zero and I get AF+BE+CD = AB+BC+CA + ½(AB+BC+CA) = 0.

The problem comes with the second part. I know I can express the area of the original triangle in a few different ways;

Area = ½|AB×AC| = P

Area = ½|BA×BC| = P

Area = ½|CA×CB| = P

For the median triangle I can say that FA×FE is the median area, but since FE = BE, I can use that instead.

FA is the vector from the midpoint on BC to A, which can be expressed as AB + ½BC.

BE is the vector from B to the midpoint on AC, which can be expressed as BC + ½CA.

½|FA×BE| = ½|(-AB - ½BC)×(BC + ½CA)|

Since the cross product is distributive over addition and BC×BC = 0:

= ½|-AB×BC - ½BC×BC - AB×½CA - ½BC×½CA| = ½|-AB×BC -½AB×CA - (1/4)BC×CA|

So, AB = -BA. CA = -AC. BC = -CB. Thus:

= ½|-(-BA)×BC - ½AB×(-AC) - (1/4)(-CB)×CA| = ½|BA×BC + ½AB×AC + (1/4)CB×CA|

All of those three equals the area, P:

=½|2P+ P + (1/2)P| = ½|(7/2)P| = (7/4)P.

My question is: does this make sense? It seems odd that the median triangle area should be bigger than the original area.. But then again, it's not impossible - just strange.

Anyway, a yes or a no+explanation where I went wrong would be very helpful!

(should I care that the last expression is CB×CA - because when I expressed the areas in the beginning, I said CA×CB - but in this case (since it's the value we're talking about) it's equal.. even though CA×CB = - CB×CA..)