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Joined 2 years ago
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Cake day: June 18th, 2023

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  • @TauZero@mander.xyz It is a Geogebra drawing I did to reason with the problem, I took a screenshot of the drawing to attached it.

    • In the drawing, the labels are different from the problem, but I just made a sphere whose diameter is [AP] (here point P has label A, while A has label A’),

    • then constructed the plane using A and two other points of the sphere (C and D in the picture),
      I thought like “if that property from 2D geometry holds in 3D then any point in the intersection of plane and the sphere will satisfy the perpendicularity, and thus two of them will do for a counterexample”.

    • And It is exactly what happened: Using Geogebra’s tool of measuring angles it shows that the two points, C and D, that I picked up both satisfy the orthogonality condition (in the picture angle(A,C,A’)=90°=angle(A,D,A’), but they can’t be both the projection of P, right ? Counterexample! (the hypothesis was that a point on a plane that satisfy that condition is immediately THE projection of the point that isn’t on the plane)

      Yeah It is not the best thing but I wanted to attach something, and the drawing that I used was the best thing at hand.