The following problem is discussed in [SalmonConics, p. 51] by applying polar coordinates: Given base c=AB and sum of sides m = a+b=AK of a triangle ABC, erect at B the perpendicular to side a= BC intersecting the external bisector of angle C at P. To find the locus of P.
The solution proceeds by using the angle θ = π/2-B in a system of polar coordinates centered at B. The first remark is that BCP = π/2 - C/2 and from triangle PCB: a = r*tan(C/2). Then, in order to find the dependence of the radius r from θ it suffices to express a and tan(C/2) in terms of θ. For this use the cosine formula for ABC: b2 = a2 + c2 -2ac*cos(B), substituting b = m-a and cosB = sin(θ) we get, m2 - 2ma + a2 = a2 + c2 - 2ac sin(θ), and from this, a = (m2-c2)/(2(m-csin(θ)). Also tan(C/2) can be expressed in terms of the given data and θ. tan(C/2) = bsinC/(b(1+cos(C))), but bsin(C) = csin(B) = ccos(θ), and bcos(C) = a-ccos(B) = a-csin(θ), hence tan(C/2) = ccos(θ)/(m-csin(θ)). Substitution of the values found into a=r*tan(C/2) produces, (m2-c2)/(2(m-csin(θ))) = r*c*cos(θ)/(m-c*sin(θ)) => rcos(θ) = (m2-c2)/(2c). Hence the locus is a line perpendicular to AB at a distance (m2-c2)/(2c) from B.
What is the mysterious line (PQ in the figure) found by the previous calculation? The analytic solution is good and fine to locate it quickly but gives no information for its significance and geometric content. Next synthetic solution gives a much more detailed description of the relation of this locus to the triangle and its geometry.
First the bisector CP invites to reflect everything on it, thus getting K, symmetric to B, such that m=a+b is the given length of the sum of the sides. The middle H of BK describes a circle k constructible from the given data. In fact, draw from H parallels to the orthogonals KE, KI intersecting AB at G, J which are the middles of BE and BI correspondingly. Points G, J are constructible and H is viewing segment GJ under a right angle. Thus H is on the circle k with diameter GJ. The locus (line PQ) is the polar of B with respect to this circle with diameter GJ. To see this notice that the parallels to KA, KP from H intersect segments AB, BP respectively at their middles. Since AKP is a right angle so is LHN, thus circles k and BHPQ are orthogonal, consequently k and the circle k' with diameter BQ are also orthogonal. This because both k' and BHPQ belong to the circle-pencil of circles passing through {B,Q} which are simultaneously orthogonal to k and line AB. This proves the claim. The proof shows also that the locus, which is line PQ, is the inverse on k of the small circle with diameter BL.
Moral The analytic proof is easy, general applicable, powerfull and quick to find the solution. The synthetic is difficult, specialized to the particular problem, slow, requiring knowledge and experience on the special area to which pertains the problem. Its big advantage though is that it offers insight into the geometric connexions involved, something that is totally absent from the analytic methods. Remark Often a first synthetic solution can be replaced by a simpler one found later. Here, for example, a much simpler synthetic solution results by oberving that {HG,HJ} are the bisectors at H of triangle BHQ. In fact angle(BHQ) = angle(BPQ)=π/2-θ=B, angle(BHG) = angle(BKE)=angle(AKE)-C/2=(π-A)/2-C/2=B/2. Thus, it is a matter of taste, how to spend (waste?) the time, in a vita calculativa or a vita contemplativa. I am inclined towards the second.
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1. Analytic solution ![[0_0]](AnalyticVersusSynthetic_files/AnalyticVersusSynthetic_0_0_0.png)
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