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We call a triangle "nice if all angles are between $45$ and $90$ degrees (including $90$ and $45$ itself) and all sides are between $1$ and $2$ (including $1$ and $2$ itself). What is the greatest area difference between two "nice triangles"?

My attempt: Because we have side and angle limits the best way to finding area is using the formula $S=bc\cos{A}$. We should find the greatest and lowest area. But here I got stuck and I don't know how to have both limits with each other. First I thought that the maximum area is $A=90^\circ $ and $b=c=2$. But then I saw that then we have$a=2\sqrt{2}>2$. Could you please give a way?

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    It would also be nice if you have spaces between some of the words. :( – 2017-02-03
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    Set $b=c=2$, we have $$a^2=2^2+2^2-2(2)(2)\cos A=8-8\cos A=8(1-\cos A)$$ thus $a^2=16\sin^2\frac A2$, as a result $a=4\sin\frac A2$. On the other hand, we have $a\le 2$ therefore $4\sin\frac A2\le 2$. , So $45^\circ\le A\le 60^\circ$. $$ S_{\text{max}}=\frac {1}{2} \times 2\times 2 \times \frac {\sqrt{3}}{2}=\sqrt{3}$$ – 2017-02-03
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    @BehrouzMaleki How do you know the maximum area happens when $b=c=2$? – 2017-02-03
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    $y=\sin x$ is an increasing function on $[45^\circ,90^\circ] $ thus $$S_{\max}=\frac12 b_{\max}c_{\max}\sin A_{\max}$$ – 2017-02-03
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    But maybe for some $b$ the increase of $A$ will effect more. – 2017-02-03
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    Note $A\in[45^\circ,60^\circ]$ – 2017-02-03
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    @BehrouzMaleki The $3^{rd}$ side must be $\le2$ as well, which your $S_{max}$ doesn't observe. P.S. (to the latest comment) There is no obvious reason why $A \le 60^\circ$. – 2017-02-03
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    yes , $a=4\sin\frac A2\le 2$ thus $A\in[45^\circ,60^\circ]$ – 2017-02-03
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    @BehrouzMaleki Sorry, I don't follow. This appears to be based on your premise that $b=c=2\,$, but it's not obvious why that should hold, from just what you wrote. – 2017-02-03
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    $S=\frac 12 bc \sin A$ and $45^\circ\le A\le 90^\circ$. and $y=\sin x$ is an increasing function in this region . Thus $S_{max}=\frac 12 (2)(2) \sin 60^\circ$ – 2017-02-03
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    @BehrouzMaleki Again you are *assuming* that $b=c=2$ but you don't justify that assumption. Note that $b,c,A$ are *not* independent variables, but they are constrained by the law of cosines to $1 \le b^2+c^2 - 2bc \cos A \le 2\,$. Also note that the question is not about $S_{max}$, but about the $\max(S_{max}-S_{min})$. – 2017-02-03

2 Answers 2

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The Maximal triangle has two sides of length $2.$

for any triangle that had only 2 sides of lenght 2, we could extend the second longest side until it was length 2, and create a larger triangle.

$Area = \frac 12 BC\sin a\\ B,C = 2$

maximize $2\sin a$

constrained by: $4 \sin \frac {a}{2} \le 2$

$\sin \frac {a}{2}\le \frac 12\\ \frac {a}{2}\le 30\\ a\le 60$

Since $\sin a$ is strictly increasing between 0 and 90.

$a = 60\\ Area = 2\sin 60 = \sqrt 3$

then minimize $\frac 12 \sin a$

constrained by: $4 \sin \frac {a}{2} \ge 1$

and again $a = 60$

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    Why "The Maximal triangle has two sides of length 2"? – 2017-02-03
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    By the rule no triangle has sides longer than 2. Suppose you have a triangle that is (2,1.9,1.7) I say that a triangle with sides (2,2,1.7) will always be larger. In fact go through this argument, I could have dispensed with the trig altogether. – 2017-02-03
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By the isoperimetric inequality the largest area of a nice triangle is $\sqrt{3}$, i.e. the area of a equilateral triangle with side length $2$. By the formula $2\Delta = ab\sin\gamma$ the smallest area of a nice triangle is $\frac{1}{2}\cdot 1\cdot 1\cdot\sin 60^\circ = \frac{1}{4}\sqrt{3}$.

It follows that the largest area difference between two nice triangles is $\color{red}{\frac{3}{4}\sqrt{3}}$.

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    Wait, but $(1,1,45^\circ)$ makes the third side shorter than 1. – 2017-02-03
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    @IvanNeretin: you are clearly right, fixed. – 2017-02-03
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    Can you elaborate on how the isoperimetric inequality is applied here? I know only the one which states that $4 \pi A \le L^2$ (A=area, L=length of circumference). – 2017-02-03
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    @MartinR: the isoperimetric inequality for triangles, of course - $$A\leq \frac{1}{12\sqrt{3}}(a+b+c)^2$$ – 2017-02-03
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    OK, thanks .... – 2017-02-03