Given a point P inside an acute angle XAY, show how to construct a line through P meeting the line AX at B and the line AY at C such that the area of the triangle ABC is AP2.
Solution
Solution by Milan Novakovic
Take D on AX such that PD is parallel to AY. Let AP = p, AD = d, DB = x and let the perpendicular distance from P to AX be h. The triangles ABC and DBP are similar, so (x + d)2/x2 = area ABC/area DBP = p2/xh. Hence x2 - 2bx + d2 = 0 where b = p2/h - d.
To find b, we can take a line parallel to AX and a distance p from it (on the same side as P). Suppose it meets the line AP at Z. Then AP/h = AZ/p, so AZ = p2/h. Then if the circle center A radius AD meets AZ at W, we have WZ = b.
Now take a circle radius b and take a point T on the circle a distance d from the diameter UV. Let the foot of the perpendicular from T to UV be S. Then SU and SV are the two possible values of x.
© John Scholes
jscholes@kalva.demon.co.uk
18 Dec 2002
Last corrected/updated 18 Dec 2002