Question 65

Two vertices of a triangle $$ABC$$ are $$A(3, -1)$$ and $$B(-2, 3)$$, and its orthocentre is $$P(1, 1)$$. If the coordinates of the point $$C$$ are $$(\alpha, \beta)$$ and the centre of the circle circumscribing the triangle $$PAB$$ is $$(h, k)$$, then the value of $$(\alpha + \beta) + 2(h + k)$$ equals

Orthocenter property:

$$(AP\perp BC),(BP\perp AC)$$

Find (C):

$$Slope(AP=-1\Rightarrow BC)slope(=1\Rightarrow y=x+5)$$

$$Slope(BP=-\frac{2}{3}\Rightarrow AC)slope(=\frac{3}{2}\Rightarrow y=\frac{3}{2}x-\frac{11}{2})$$

Intersection ⇒ (C(21,26))
$$\alpha+\beta=47$$

Circumcenter of (PAB):

Solve perpendicular bisectors ⇒ (h + k = -21)

Final:
$$(\alpha+\beta)+2(h+k)=47+2(-21)=5$$

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