Looking through the recent, and excellent, 67th Carnival of MathematicsI came across a link to a theorem at "Cut the Knot" that I had never known.
It begins with the simple fact that for a triangle ABC, Cos2(A) + Cos2(B) + Cos2(C) = 1 IFF (if and only if) the triangle is a right triangle..that is, one of Cos(A), Cos(B) or Cos(C) = 0.
But the part I loved is the more general extension... that for ANY triangle, Cos2(A) + Cos2(B) + Cos2(C)+2 Cos(A)Cos(B)Cos(C) = 1.
Since only one of the angles can be obtuse (and hence the quantity 2 Cos(A)Cos(B)Cos(C) would be negative only in the obtuse case), we can use Cos2(A) + Cos2(B) + Cos2(C) as a determinant for triangles. When the sum is > 1 the triangle is obtuse. If it is equal to one, the triangle is a right triangle; and if the sum is< 1, the triangle is acute. Not sure how I got so old without knowing that.
Can anyone tell me who/when this general identity was first discovered?
Showing posts with label Pythagorean-like relations. Show all posts
Showing posts with label Pythagorean-like relations. Show all posts
Sunday, 4 July 2010
Wednesday, 13 January 2010
Almost Pythagoras
For some reason I really like little simple geometric relations that remind me of the Pythagorean Theorem. I still remember the first time I saw 32 + 42=52 set next to 33+43+53=63and wondered.... Could it be????... Is it possible???? Oh go on, you know you are going to check... how could you resist?
Just reminded of some other "almost Pythagorean" relations in geometry from an old (1932) Mathematics Teacher article...(Thank you, again, Dave Renfro)
The Two Triangles formed when the median is drawn to any side are almost Pythagorean
Treat the two sides not cut by the median as if they were the hypotenii (hypotenuses?) of a right triangle. Both are wrong, but in sum, they are right..
That is, while neither of the following are true, AM2+MC2= AC2
AM2+MB2=AB2
Adding the two equations produces a true relation.
AM2+MC2+AM2+MB2= AC2+AB2
The proof is pretty easy using the Law of Cosines.
A second, and about as easy to prove.... Let G be the centroid of Triangle ABC, then AB2+BC2+CA2 = 3 (GA2+GB2+GC2)
A third, which is, I believe, both necessary and sufficient to prove a parallelogram is that the sum of the squares of the diagonals is equal to the sum of the squares of the sides.
And if it is not a parallelogram, but is a trapezoid, then the sum of the squares of the diagonals is equal to the sum of the squares of the non-parallel sides, plus twice the product of the parallel bases (sort of a law-of-cosines look-a-like).
Recently saw the old Wizard of Oz mis-spoken statement of the Theorem by Ray Bolger as the Scarecrow posted at 360, so I thought this was a good time to include it for my students (and anyone else who has never seen it, or just wants to chuckle one more time). [] Play it for your students and see if they can catch all the mistakes...
Just reminded of some other "almost Pythagorean" relations in geometry from an old (1932) Mathematics Teacher article...(Thank you, again, Dave Renfro)
The Two Triangles formed when the median is drawn to any side are almost Pythagorean
Treat the two sides not cut by the median as if they were the hypotenii (hypotenuses?) of a right triangle. Both are wrong, but in sum, they are right..
That is, while neither of the following are true, AM2+MC2= AC2
AM2+MB2=AB2
Adding the two equations produces a true relation.
AM2+MC2+AM2+MB2= AC2+AB2
The proof is pretty easy using the Law of Cosines.
A second, and about as easy to prove.... Let G be the centroid of Triangle ABC, then AB2+BC2+CA2 = 3 (GA2+GB2+GC2)
A third, which is, I believe, both necessary and sufficient to prove a parallelogram is that the sum of the squares of the diagonals is equal to the sum of the squares of the sides.
And if it is not a parallelogram, but is a trapezoid, then the sum of the squares of the diagonals is equal to the sum of the squares of the non-parallel sides, plus twice the product of the parallel bases (sort of a law-of-cosines look-a-like).
Recently saw the old Wizard of Oz mis-spoken statement of the Theorem by Ray Bolger as the Scarecrow posted at 360, so I thought this was a good time to include it for my students (and anyone else who has never seen it, or just wants to chuckle one more time). [] Play it for your students and see if they can catch all the mistakes...
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