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# 2017 NT 7

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2 Answer: sqrt(42)-5 or equivalent

Apr 8, 2019

#1
+2

$$\text{on the line }y=x, \text{ oddly enough for every point }y=x\\ \text{thus we have}\\ \dfrac{x^2-x}{2x-x+2} = a\\ x^2-x=a x + 2a\\ x^2 -(1+a)x -2a=0\\ x = \dfrac{(1+a)\pm \sqrt{(1+a)^2+8a}}{2}$$

$$\text{You then have two points of the form }(x_k,x_k),~k=1,2\\ \text{Where the }x_k \text{ are the roots above}\\ \text{Find the distance between these two points. It's }\\ d=\sqrt{2} \sqrt{\left | a^2+10 a+1\right| }\\ \text{Solve for }d=6 \text{ and choose the value of }a \text{ that is positive}\\ \text{and leads to real points }x,y\\ a=\sqrt{42}-5 \text{ is the only one}$$

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Apr 9, 2019

#1
+2

$$\text{on the line }y=x, \text{ oddly enough for every point }y=x\\ \text{thus we have}\\ \dfrac{x^2-x}{2x-x+2} = a\\ x^2-x=a x + 2a\\ x^2 -(1+a)x -2a=0\\ x = \dfrac{(1+a)\pm \sqrt{(1+a)^2+8a}}{2}$$

$$\text{You then have two points of the form }(x_k,x_k),~k=1,2\\ \text{Where the }x_k \text{ are the roots above}\\ \text{Find the distance between these two points. It's }\\ d=\sqrt{2} \sqrt{\left | a^2+10 a+1\right| }\\ \text{Solve for }d=6 \text{ and choose the value of }a \text{ that is positive}\\ \text{and leads to real points }x,y\\ a=\sqrt{42}-5 \text{ is the only one}$$

Rom Apr 9, 2019
#2
+1

That's a nice solution, Rom....!!!   CPhill  Apr 9, 2019