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In triangle inscribed circle with radius $r = 1$ and one of it sides $a=3$. Find the minimum area of triangle? Ans = 5.4

My reasonings:

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$BC = a$, $AC = b$, $AB = c$

$AD=AF=x$

$FC=CE=y$

$BD=BE=z$

$a=z+y$, $b=x+y$, $c=x+z$

The radius of the incircle is $$r =\frac{A_{ABC}}{s}$$ where $s = \frac{a+b+c}{2} = x + y+ z$. By condition $z+y=3$ so $s=x+3$

By Heron's formula, the area of the triangle is $A=\sqrt{s(s-a)(s-b)(s-c)}$

In other side $A=sr = x+3$.

What is next? I think that I should get a fucntion for which I will can find a minimum, but I don't know how.

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If you can prove that the least area occurs when side $a$ is the base of an isosceles triangle then you are done. –  RicardoCruz Mar 25 at 3:43
    
@RicardoCruz Isnt that case for the maximum area? –  Sawarnik Mar 26 at 9:43

2 Answers 2

up vote 3 down vote accepted

You were on the right track.
Using your notation, $A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{(x + 3)xyz}$, but also $A = x + 3$. So $$\begin{align} \sqrt{(x + 3)xyz} &= x + 3, \\ xyz &= x + 3, \\ x &= \frac{3}{yz - 1}. \end{align}$$

Substituting that into $A$ we get $$A = \sqrt{\left(\frac{3}{yz - 1} + 3\right)\frac{3}{yz - 1}yz} = \sqrt{\frac{3yz}{yz - 1}\cdot\frac{3}{yz - 1}yz} = \frac{3yz}{yz - 1} = \frac{3}{yz - 1} + 3.$$

Finally, by AM-GM we get lower bound: $$A = \frac{3}{yz - 1} + 3 \geqslant \frac{3}{\frac{(y + z)^2}{4} - 1} + 3 = \frac{3}{\frac{9}{4} - 1} + 3 = \frac{12}{5} + 3 = \frac{27}{5} = 5.4$$

Because it is AM-GM, equality is reached when $y = z = 1.5$


P.S. You may wonder, why for some values of $y$ and $z$ expression $\frac{3}{yz - 1}$ may become infinite or negative. Isn't it strange? Besides, I myself silently assumed that it is positive. And there's a reason for that.
Positive values of $\frac{3}{yz - 1}$ correspond to the situation you describe: a circle inscribe in a triangle.
When it becomes infinite two sides $AB$ and $AC$ become parallel.
And finally, when it's negative, your circle is no longer an incircle, it becomes excircle.

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Thank you. But can you explain last part yours solving? How do you get $\frac{(y+z)^2}{4} - 1$ in right side in denominator? –  Alexander Side Mar 27 at 12:50
1  
@AlexanderSide AM-GM looks like $\sqrt{yz} \leqslant \frac{y+z}{2}$. Starting from there we get $$\begin{align} yz &\leqslant \left(\frac{y+z}{2}\right)^2 = \frac{(y+z)^2}{4}, \\ yz-1 &\leqslant \frac{(y+z)^2}{4} -1, \\ \frac{3}{yz-1} &\geqslant \frac{3}{\frac{(y+z)^2}{4} - 1}. \end{align}$$ –  ElThor Mar 27 at 13:36

Trigonometric approach:

In your notation,

$AD=AF=x$, $FC=CE=y$, $BD=BE=z$,

denote in addition

$\alpha=\angle OAF=\angle OAD$, $~~~\beta=\angle OCF=\angle OCE$, $~~~\gamma=\angle OBE=\angle OBD$.

So, if $r=1$, then $x=\dfrac{1}{\tan\alpha}$, $~~~y=\dfrac{1}{\tan\beta}$, $~~~z=\dfrac{1}{\tan\gamma}$.

Then, as you said, $$ A=r(x+y+z)=r(x+3)=x+3. $$

$$ x = \dfrac{1}{\tan(90^\circ - \beta-\gamma)} = \tan(\beta+\gamma) = \dfrac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma}. $$ Dividing numerator and denominator by $(\tan\beta\tan\gamma)$, we get: $$ x=\frac{z+y}{yz-1}=\frac{3}{yz-1}. $$

$$ A=x+3=\frac{3}{yz-1}+3. $$

Further thoughts - as in answer of ElThor.

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