Quadratic Straightedge and Compass Solver
Given integer values of , , and , use a straightedge and compass to construct rectangle such that and , and then construct rectangle such that . Then construct the diagonals of rectangle and call their intersection , and finally construct a circle with center and radius . Let the intersection points of the circle and be and .
Then the lengths of and are solutions to which of the following quadratic equations?
(inspired by the book Number: The Language of Science by Tobias Dantzig)
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1 solution
Let P be the origin. Then Q has coordinates ( 0 , 1 ) and X has coordinates ( 2 a b , 2 a c + a ) .
Using the distance formula on X Q , X Q 2 = ( 2 a c − a ) 2 + ( 2 a b ) 2 .
Then the circle has an equation of ( x − 2 a b ) 2 + ( y − 2 a c + a ) 2 = X Q 2 , or ( x − 2 a b ) 2 + ( y − 2 a c + a ) 2 = ( 2 a c − a ) 2 + ( 2 a b ) 2 .
Since P is the origin, P U is the x -axis, and W and V are the x -intercepts of the circle when y = 0 , so ( x − 2 a b ) 2 + ( 0 − 2 a c + a ) 2 = ( 2 a c − a ) 2 + ( 2 a b ) 2 , which simplifies to x = 2 a b ± b 2 − 4 a c . Substituting − b ′ for b gives x = 2 a − b ′ ± b ′ 2 − 4 a c , which is the quadratic equation that would solve a x 2 + b ′ x + c = 0 , and since b ′ = − b , this straightedge and compass method solves a x 2 − b x + c = 0 .