Question P.415: Construct a triangle knowing an angle, the perimeter, and th......
Construct a triangle knowing an angle, the perimeter, and the area (Exercises 90b, 299). Among all triangles with a given angle and given perimeter, which one has largest area?
Suppose we know the measure of angle \hat{{A}} of triangle ABC, as well as its perimeter 2s. Then, if E_{1},\ F_{1} are the points of contact of the escribed circle opposite vertex A (fig. t415), then the result of exercise 90b tells us that A F_{1}=A F_{1}=s. This allows us to construct points {{E}}_{1} and {{F}}_{1}. We can then get excenter I_{\mathrm{1}} (by erecting perpendiculars to A F_{1},\ A E_{1}{\mathrm{~at~}}F_{1},\ E_{1} respectively), and construct the escribed circle itself.
We know (254) that the inradius r of the triangle is equal to {\frac{K}{s}}, where k is its area. This circumstance allows us to construct the length r. If we then draw a parallel to A E_{1} at a distance r, it will intersect the bisector of angle \hat{{A}} at the incenter I of triangle ABC. We can then construct the incircle, and side BC is simply the common internal tangent of circles I and I_{\mathrm{I}} . This procedure accomplishes the construction required by the problem.
We next determine the triangle of largest area among those with a given angle and given perimeter. We know that rs = K, so if the perimeter is fixed, so is s (the semiperimeter), and to maximize K we must maximize r. It is not hard to see, from the figure, that this will occur when circles I and I1 are tangent; that is, when triangle ABC is isosceles.
Note. The original problem leaves ambiguous exactly how we are ‘given’ the area of the triangle. The easiest way to fill in this gap is to assume that a triangle with the same area is given. Then the construction of r involves drawing a triangle with a given base (s) equal in area to a given triangle. Students can solve this sub-problem themselves.
