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Let $\Delta ABC$ have given angle $\theta$ at vertex A and opposite side BC of given length $a$.

I would like to find the maximum area of such a triangle in terms of $\theta$ and $a$.

I think the triangle of maximum area is isosceles, but how can I show this (if this is correct)?


(I realize that this can also be worked using calculus, but I am looking for a more elementary solution.)

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There's a theorem of Euclid which can be applied to this problem:

  • Hypothesis: Given $\triangle ABC$ as in your answer, let $X$ be the circle passing through $A$, $B$, $C$. The points $B,C$ cut $X$ into two circular arcs; let $\alpha$ be the one containing $A$.
  • Conclusion: for every $A' \in \alpha$ the angles of $\triangle ABC$ and $\triangle A'BC$ at $A,A'$ are equal.

Hence, your problem is solved by choosing $A' \in \alpha$ so as to maximize the height of the triangle $A'BC$ with base $\overline{BC}$. That choice of $A'$ is unique, namely the intersection of $\alpha$ with the perpendicular bisector of $\overline{BC}$, and that does indeed give an isosceles triangle $A'BC$.

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  • $\begingroup$ Thanks for your answer. (The way I set up the notation, would I want to use the arcs cut out by the points B and C, and let $\alpha$ be the arc containing A?) $\endgroup$ – user84413 Dec 23 '15 at 18:01
  • $\begingroup$ Oops, you are right. Let me fix the notation of my answer. $\endgroup$ – Lee Mosher Dec 23 '15 at 19:40
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Because you don't really know where point A should be relative to the segment BC, let me suggest the following: Take the perpendicular bisector of BC and at C construct half of angle A on one side of BC and the complement of half of angle A at C on the other side of BC. Suppose the new sides of those two angles meet the perpendicular bisector of BC at D and E respectively. Let O be the midpoint of segment DE. Then any point on the arc BEC of the circle with center O could be the point A, because an angle at a point on a circle measures half of the arc that it "subtends", so for any point A on that arc, angle BAC measures the same as angle BEC, which is the desired angle A. (To see this, recall that segment BC is perpendicular to DE -- say they meet at F -- and because we built angle FCE to be the complement of half of angle A, angle FEC is half of angle A, so angle BEC is equal to angle A.) The choice of point A on the arc BEC that gives the greatest area is the one that is the greatest perpendicular distance away from segment BC, and that is the point E, halfway along that arc. So, yes, the greatest area comes from an isosceles triangle: the lengths EB and EC are congruent.

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