Vladimir Nikolin

Definition 1: Two lines a and b meet at a point O, beyond the drawing surface. O is the inaccessible point.

- You can't use the point O in your constructions.
- Lines a and b are, of course, accessible.
- Let (O) be the notation for the inaccessible point O.

The line p passes through the points P and (O)

Basic problem: Construct a line p through the given point P and the inaccessible point (0), defined by the pair of lines a and b.

Famous solution : Point P as orthocenter of the ΔA(O)B

Solution 1: Let Ha \in a, PHa \perp a, and let A be the intersection of the lines b and PHa. Similarly, let's define points Hb (as a point of line b,PHb \perp b ) and B (as a point of line a and B, P and Hb are collinear).
Now, point P is the ortocenter of the ΔA(O)B. As we know, altitudes of the triangle are concurrent , so the line p, which is perpendicular to side AB and passes through the point P, is the third altitude of the ΔA(O)B. Hence, (O) \in p and P \in p.
Discussion: This sounds fine, but what if lines a and b are perpendicular? Then construction is not possible at all, because lines b and PHaare parallel, there is no point A. Second, what if angle between those lines are obtuse? Then construction is not possible as well, from the other reason. In this case, orthocenter of the obtuse triangle is higher than the vertex (O), beyond the drawing surface too.
In conclusion: Solution 1 may be correct only if angle between lines a and b is acute and the drawing surface is enough wide.

Better and more formal definition of an inaccessible point would be:
Definition 2: Two lines a and b meet at a point O and this point is not allowed in the construction. We call it inaccessible point.
Solution 2: Let Ha \in a, PHa \perp a and Hb \in b, PHb \perp b. If the point R is intersection of perpendicular bisectors of the PHa and PHb, then points P, R and (O) are collinear i.e. (O) \in PR.

Proof: The quadrilateral (O)HaPHb is a cyclic, point R is its circumcenter and P(O) is a diameter.

Solution 3: Let SP (a) = a', SP (b) = b' where SP is a half turn - reflection in point P. Let M = a \cap b' , N = b \cap a' and O' = a' \cap b'. Solution is line PO'.

Proof: In this case quadrilateral (O)MO'N is a parallelogram (a||a', b||b') where point P is intersection of diagonals, so (O)\in PO'.

Solution 4: Let Q \in a, R \in b and PQR is a triangle. Let Q1 \in a, R1 \in b and triangles PQR and P1Q1R1 have parallel sides. Then solution is line PP1.

Proof:. Pairs of points P,P1, Q,Q1 and R,R1 determine a homothety, where point (O) is the center of homothety, so lines PP1, QQ1 and RR1 are concurrent.