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Draw the base of the triangle AC and mark its midpoint, D. Draw a perpendicular from D. Somewhere along the perpendicular will be the circumcentre, that is, the centre of the circle that circumscribes the triangle ABC.
The position of vertex B has to be such that the perpendicular bisector of AB meets the perpendicular bisector of AC at X, and the perpendicular from vertex C on to AB meets the perpendicular from vertex B on to AC at Y, such that XY is parallel to AC. This is another way of framing the question. No other construction lines are required because the missing perpendiculars are superfluous to solving the problem.
Represent the problem graphically. Plot the points A(0,0), B(p,q), C(t,0) where p and q are to be determined and t is an arbitrary constant t=AC. Midpoint of AC is N(t/2,0); midpoint of AB is M(p/2,q/2). AB is a segment of the line y=qx/p. The equation of the bisector of AB: -p/q is its gradient, so y=-px/q+c, where c is found by plugging in M: q/2=-p^2/2q+c, c=q/2+p^2/2q and the perpendicular bisector is a segment of y=q/2+p^2/2q-px/q. Therefore, the coords of X are where this line meets the perpendicular bisector of AC, which is a segment of the line x=t/2. The intersection is X(t/2,q/2+p^2/2q-pt/2q).
Now we need to find Y, which must lie on the line x=p, because this is the perpendicular from vertex B on to AC. The perpendicular from vertex C on to AB is parallel to the perpendicular bisector of AB so has the same gradient: -p/q. The equation of the perpendicular from C is y=-px/q+k, where k is found by plugging in C(t,0): 0=-pt/q+k, k=pt/q and y=-px/q+pt/q=p(t-x)/q.
The point Y is therefore Y(p,p(t-p)/q).
XY is parallel to AC, which means their y coord is the same: q/2+p^2/2q-pt/2q=p(t-p)/q; q^2+p^2-pt=2p(t-p); q^2+3p^2=3pt; q^2=3p(t-p).
Also, tanA=q/p and tanC=q/(t-p), so tanA*tanC=q^2/(pt-p^2)=3p(t-p)/(p(t-p))=3.
Note that when tanA=tanB=sqrt(3), B is (1/2,sqrt(3)/2), the triangle is equilateral and XY=0 because the circumcentre and orthocentre coincide.
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