For All Of U Who Are Good At Math..

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22crazy

Senior Member
whats up. i have a question that been pissing me off and i was wondering if any of you could solve it. the question: if you construct equalateral triangles of the sides of a quaderlateral, alternately inward and outward, when you connect the vertex of each triangle, the result is a parallelogram. Prove it (the parallelogram that is formed). ill see if i can make the picture.
 
Yeah looks to me like some seriously hard proof that must be written to prove that one. I am just in mathanalysis in sophomore year of high school so I am little under that problem for now.
 
well simply u just take the second derivative of the intergral of A to B, assuming that the distance from A to C is equal.... u then muliply that by the sides of the parrallogram, which gives u = j/p... im taking calculus for my senior year right now, and i dont have ur answer.
im sure some of the older college grad guys can answer that though...
 
well can anyone help me, its been bugging me alot and i need to complete it. thanx for the help though. i no it has to do something with the 60 degrees in the equalateral triangles and also, u have to construct a parallelogram within the quaderlateral: that is done by connecting the midpoints.
 
is this supposed to be a 3D object?
If it is i sorta see what you are going for but a little lost at the moment


as for the calculas answer vpspoon you just reminded me that i need to study for my AP calculas test, damn you j/k
 
what math class is it for?
there are probably several ways of doing it..

and which part am i suposed to prove.. that the red lines are a parallelogra?M
 
man I can't remember the last time I did a geometric proof. I'd have to bust out my books, but I've got finals this week so that ain't happening.
 
If you just have to prove that a parallelogram would exist, I would look at the relation between the sides of the quad. and the parallel. to see if they correspond ?
 
s3eb676b246cde.jpg

You need to prove that Angles 1 & 6 are equal, and anges 3 & 4 are equal. This proves that angles 2&5 are equal because the sums of 1, 2, & 3 and 4, 5, & 6 are both 60 degrees. Then all you have to do is prove that each pair of opposing sides are equal length. Then you have a parallelogram.
 
I had to sit and think for a while. It has been forever since I did a Geometric proof. And even then (freshman year of HS) I wasn't very good at them.

You could also simply prove that the opposite inside angles of the parallelogram are equal instead of showing the opposite sides equal.
 
Originally posted by lsvtec@May 5 2003, 08:17 AM
You could also simply prove that the opposite inside angles of the parallelogram are equal instead of showing the opposite sides equal.
Click to expand...

Exactly what I was going to say.

*pushes non-existent glasses up on nose, goes for high-five with lsvtec, misses horribly, gives nerdy, snorty laugh*
 
actually those angles 4 3 and 1 6 are not equal, but yet 5 and 2 are....i was able to prove it using geometer sketch pad, but i dont no how to do it paper and pencil method
 
i didn't even read the question, i just looked at the picture and it made my head hurt :blink:
 
What really matters in the angle-2 side proof is that the sum of 1 and 3 and the sum of 4 and 6 are equal (1 + 3 == 4 + 6 ), and that the larger triangle are isosolese (sp?). If the triangles are isosolese the the sum of the three angles is 60. If you can prove that 1 + 3 == 4 + 6 you have proven that angle 2 == angle 5.
 
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