MHB IBV4 Quadrilateral OABC: O(0, 0), A(5, 1), B(10, 5), C(2, 7)

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wasn't sure about $\overrightarrow{AC}$
 
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Looks good to me. :D
 
MarkFL said:
Looks good to me. :D

thanks to MHB ...making some progress ... (Wasntme)
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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