45. a) What values can the following quantities take:
1) angle $A$ of triangle $A B C$;
2) angle $B$
3) angle $C$,

if $A \leqslant B \leqslant C$?

b) The area of a triangle with sides $a, b$, and $c$, where $a \leqslant b \leqslant c$, is 1. What values can

1) side $a$ of the triangle;
2) side $b$
3) side $c$ 

take?
45. a) It is clear that
1) $0<A \leqslant 60^{\circ}$,

where $A$ is the smallest angle of triangle $ABC$ ($\angle A \leqslant 60^{\circ}$, because $A+B+C=180^{\circ}=3 \cdot 60^{\circ}$ and $A \leqslant B \leqslant C$; $\angle A=60^{\circ}$, if triangle $ABC$ is equilateral);
2) $0<B<90^{\circ}$,
where $B$ is the middle-sized angle of triangle $ABC$ ($\angle B$ can be arbitrarily small if triangle $ABC$ is isosceles with a very large angle $C$ at the vertex (Fig. $118, a$); $\angle B$ can be arbitrarily close to $90^{\circ}$ if the triangle is isosceles with a very small angle $A$ at the vertex (Fig. 118, b); $\angle B<90^{\circ}$, because $\angle B \leqslant \angle C$ and $\left.\angle B+\angle C<180^{\circ}\right\rceil$)
3) $60^{\circ} \leqslant C<180^{\circ}$

where $C$ is the largest angle of triangle $ABC$ ($\angle C \geqslant 60^{\circ}$, because $A+B+C=180^{\circ}=3 \cdot 60^{\circ}$ and $C \geqslant B \geqslant A$; $\angle C=60^{\circ}$, if triangle $ABC$ is equilateral).

b) 1) It is clear that the smallest side $a$ of triangle $ABC$ with an area of 1 can be arbitrarily small: if the legs of a right triangle are $\varepsilon$ (where $\varepsilon$ can be arbitrarily small!) and $2 / \varepsilon$, then the area of the triangle is 1. The side $a$ can also be arbitrarily large: if two sides of the triangle are $M$ and $b \geqslant M$ (where $M$ can be arbitrarily large!), and the angle $C$ between these sides varies, starting from $90^{\circ}$ and approaching $180^{\circ}$, then the third side $c$ is always the largest, and the area of the triangle, starting from a large value $\frac{1}{2} b M$, decreases without bound, and thus, passing through all values between $\frac{1}{2} b M$ and 0, it also passes through the value 1.

2) Since the smallest side $a$ of triangle $ABC$ with an area of 1 can be arbitrarily large, its middle side $AC=b$ can also be arbitrarily large. Furthermore, if the triangle has a (fixed) side $AC=b$ and some (arbitrary) side $BC=a \leqslant b$, then the vertex $B$ of the triangle belongs to the (shaded in Fig. $119, a$) circle $K$ of radius $b$ centered at point $C$. Since every point of this circle is at a distance $d \leqslant b$ from the line $AC$, the height $BD$ of triangle $ABC$ does not exceed $b$, and its area cannot be greater than $b^{2} / 2$ (it equals $b^{2} / 2$ if $BC=b$ and $BC \perp AC$). Therefore, the middle side $b$ of triangle $ABC$ with an area of 1 cannot be less than $\sqrt{2}$ (otherwise, it would be $b^{2} / 2<1$).
3) It is clear that the largest side $c$ of a triangle with an area of 1 can be arbitrarily large, but it cannot be arbitrarily small. If the side $AB=c$ is fixed and $AC=b \leqslant c, BC=a \leqslant c$, then the vertex $C$ of the triangle belongs to the intersection of two circles $K_{1}$ and $K_{2}$ of radius $c$ centered at $B$ and $A$; therefore, the height $CE$ of the triangle, and thus its area, will be the largest when $C$ coincides with the vertex of the shaded curvilinear triangle in Fig. 119, b, i.e., when $ABC$ is an equilateral triangle with side $c$ and area $\frac{c^{2} \sqrt{3}}{4}$. From this, it follows that if the area of triangle $ABC$ is 1, then its largest side $c$ cannot be less than $\frac{2}{\sqrt[4]{3}}=\frac{2 \sqrt[4]{27}}{3}$.

Thus, $0<a<\infty ; \sqrt{2} \leqslant b<\infty ; \frac{2 \sqrt[4]{27}}{3}<c<\infty$.

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-146.jpg?height=413&width=452&top_left_y=424&top_left_x=224)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-146.jpg?height=387&width=272&top_left_y=452&top_left_x=764)
b)

![](https://cdn.mathpix.com/cropped/2024_05_21_15011fcc7946713185f0g-146.jpg?height=597&width=456&top_left_y=992&top_left_x=387)

Fig. 119.