2005-07-09, 16:38 | #1 |
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts |
Cross's Theorem.
This theorem was discovered by a schoolboy-David Cross Squares have been drawn on the sides of an arbitrary triangle ABC and the free corners joined to make three more triangles. The areas of these new triangles are all (each ) equal to the area of the original triangle. WHY? Mally |
2005-07-10, 01:43 | #2 |
Aug 2003
Upstate NY, USA
2·163 Posts |
Let's say we have a triangle with sides a,b,c and opposite angles A,B,C.
We know that the area of the original triangle equals all of .5ab sinC, .5bc sinA, .5ca sinB Let's look at the triangle created from the squares off of sides a & b The triangle has two sides with length a and b, and the angle between them equals 180-C deg Thus the area of the new triangle equals .5ab sin(180-C) = .5ab sinC for C<180 The same applies for the triangles between the other pairs of squares. |
2005-07-10, 16:12 | #3 |
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts |
Cross's Theorem
Very good Tom 11784. The deduction of the angle (180 - C) is a good piece of deduction. You are very strong in Trig. I notice
Try this alternate method. The proof rests on the Theorem that triangles on equal bases and same height are equal in area. This condition can be met by simply rotating a square by 90 degrees. If you dont get it I'll show you. With the absence of a diagram its difficult to explain wthout suitable notation Cheers Mally |
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