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I proved that the isosceles triangle has bigger area, using Heron's formula. But, I couldn't prove this fact:

Given two triangles ABC and ABC' with same perimeter, if $|\overline{AC}-\overline{BC}|<|\overline{AC'}-\overline{BC'}|$, then the area of ABC is bigger than ABC'.

Sketch of the solution:

I tried to show that $(p-b)(p-c)>(p-d)(p-e)$, where $b=\overline{BC}, c=\overline{AC}$ and $d=\overline{BC'}, e=\overline{AC'}$, to use Heron's formula, but I can't get anywhere from that. I also used the fact that $b+c=d+e$, since the two triangles have the same perimeter and the same side AB.

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  • $\begingroup$ Since you have already solved the original problem, why do you need a solution to the second? Is it just curiosity? What have you tried to prove it? Welcome to Math Stack Exchange, I hope we can be of help in your studies. Thanks for using MathJax and telling us what you have done on this problem already. These are good habits to get into on this site. $\endgroup$ – Stephen Meskin Nov 17 '17 at 21:37
  • $\begingroup$ Thanks very much Stephen, I read the "How to Ask" guide (heheh). In fact, this is a second part of the problem. I'll write what I have already done to solve it. $\endgroup$ – Irlexi Nov 17 '17 at 21:58
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The maximum area is attained by the isosceles triangle. A proof without words:

enter image description here

For the second part, exploit the convexity/concavity of hyperbolas (loci such that a difference of distances is constant) and ellipses (loci such that a sum of distances is constant).

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As it was already noted, the third vertex, $C$ is located on an ellipse.

enter image description here

Let $|BC|=a,\ |AC|=b,\ |AB|=c$ be the sides of $\triangle ABC$, and let $|OP|=u,\ |OQ|=v$ be the semi-major and semi-minor axis of the ellipse, respectively.

Due to the symmetry, we can consider only the I quadrant of the ellipse.

Following the cosine rule,

\begin{align} a^2&=b^2+c^2-2bc\cos\alpha ,\\ b^2&=a^2+c^2-2ac\cos\beta ,\\ 2(b^2-a^2)&=2bc\cos\alpha-2ac\cos\beta ,\\ b-a&=\frac{c}{a+b}(b\cos\alpha-a\cos\beta) \\ &= \frac{c}{a+b}(|AH|-|BH|) = \frac{2\,c}{a+b}|OH| = \frac{2\,c}{a+b}x . \\ \text{Since } \frac{2\,c}{a+b}&=\mathrm{const}, \quad b-a\sim x . \end{align}

Also, the area of $\triangle ABC$ is directly proportional to its height, that is, $[ABC]\sim y$.

Expression for the point $C=(x,y)$ on the elliptic arc $PQ$ in terms of parameter $\theta\in(0,\tfrac\pi2)$ gives \begin{align} x(\theta)&=u\cos\theta ,\\ y(\theta)&=v\sin\theta ,\\ x'(\theta)&=-u\sin\theta<0\quad \forall\theta\in(0,\tfrac\pi2) ,\\ y'(\theta)&=v\cos\theta>0\quad \forall\theta\in(0,\tfrac\pi2) , \end{align}

so, smaller $x$ is equivalent to bigger $y$ and the smaller the side difference, the bigger the area.

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  • $\begingroup$ I think it resolves my problem. I had not thought in this problem in terms of a ellipse, I was trying to use the Heron's formula directly, but... $\endgroup$ – Irlexi Nov 18 '17 at 21:09
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Well, I don't know if this is appropriate answer, but here it goes. $C$ and $C'$ must lie on a the same ellipse since we have: $$AC+BC = AC'+BC'$$ Imagine this ellipse is with focuses $A$ and $B$ on x-axis. Since $|\overline{AC}-\overline{BC}|<|\overline{AC'}-\overline{BC'}|$ the triangle is ''more'' isosceles then $ABC'$. So $ABC$ has bigger altitude on $AB$ then $ABC'$ does and thus the conclusion.

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  • $\begingroup$ It's a great idea. But how can I prove that $ABC$ das bigger altitude? How can I show, in general, that $ABC$ is more isosceles than $ABC'$? It seems to me that it needs a little bit more argumentation. $\endgroup$ – Irlexi Nov 17 '17 at 22:11
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Using the Heron formula we obtain that the areas of $ABC$ and $ABC'$are respectively $S$ and $S'$. Suppose $H$ and $H'$are the heights relative to the $AB$ side. Write

$$S= \frac{AB * H}{2} \,\,\,\ \textrm{and}\,\,\,\ S'= \frac{AB * H'}{2}.$$

Use the hypothesis and the fact that $AC'+BC'=AC+BC$

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  • $\begingroup$ Yeah, but how can I assure that by the inequality $|\overline{AC}-\overline{BC}|<|\overline{AC'}-\overline{BC'}|$, $ABC$ has bigger area? In this case, the triangle $ABC$ can not even be isosceles. $\endgroup$ – Irlexi Nov 17 '17 at 23:28

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