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As title,

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I could only figure out that

$$a^2 + r^2 = 9$$ $$b^2 + r^2 = 16$$ $$c^2 + r^2 = 25$$ $$r^2 = \frac{abc}{a+b+c}$$

But couldn't get how to derive $2(a+b+c)$.

athos
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    I wonder if we can do some kind of rotation or reflection of the incenter, given the nice lengths might form a right triangle – mpnm Dec 27 '20 at 08:35
  • When we try to get an equation with just $a,b,c$ we get an equation with a bunch of symmetric polynomials: $(a+b+c)^3-2(ab+bc+ca)(a+b+c)-50(a+b+c)+3abc=0.$ Can we do so kind of manipulation with this, possibly with Vieta's? – mpnm Dec 27 '20 at 17:25

3 Answers3

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Not sure if there is a more straightforward solution but in meantime,

$3 \sin\frac{A}{2} = 4 \sin\frac{B}{2} = 5 \sin\frac{C}{2} = r$

$\frac{C}{2} = 90^0 - \frac{A+B}{2}$

So, $\cos \frac{A+B}{2} = \frac{r}{5}$

Expanding and writing values in terms of $r$ and squaring both sides, we simplify it to a cubic

$\frac{\sqrt {9-r^2} \times \sqrt {16-r^2}}{12} - \frac{r^2}{12} = \frac{r}{5}$

$25(9-r^2)(16-r^2) = (5r^2 + 12r)^2$

$120r^3 + 769r^2 - 3600 = 0$.

As $0 \lt r \lt 3$, we get $r \approx 1.9$ (with help from WolframAlpha)

From here on, we get sides as $\approx 5.842, 6.947, 8.145$.

Math Lover
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I doubt if it can be solved by hand. By Mathematica, we have $$r=\frac{1}{360} \left(\sqrt[3]{245083391+21600 i \sqrt{314509827}}+\frac{591361}{\sqrt[3]{245083391+21600 i \sqrt{314509827}}}-769\right)$$

Computing its minimal polynomial, it is a root of $120 r^3+769 r^2-3600=0$

And thus it is very complicated to represent the result in radicals.

The numerical solutions is $a= 2.32148,b= 3.51984,c= 4.62485,r= 1.9002$.

FFjet
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In a triangle $ABC$ with sides $a,b,c$ opposite to $A,B,C$, respectively, and with incenter $I$, one has $$AI^2 = bc\cdot \frac{b+c-a}{a+b+c}, \quad BI^2 = ca\cdot \frac{c+a-b}{a+b+c}, \quad CI^2 = ab\cdot \frac{a+b-c}{a+b+c}.$$ The conditions $AI=3$, $BI=4$ and $CI=5$ can be rewritten as $$\lambda := \frac{2abc}{a+b+c} = bc-9 = ca-16 = ab - 25.$$ One gets $$a^2=\frac{ab\cdot ac}{bc} = \frac{(\lambda+16)(\lambda+25)}{\lambda+9}$$ and similarly $$b^2=\frac{(\lambda+9)(\lambda+25)}{\lambda+16}, \quad c^2=\frac{(\lambda+9)(\lambda+16)}{\lambda+25}.$$ Substituting this into $\lambda = \dfrac{2abc}{a+b+c}$ one obtains $$\lambda = \frac{2(\lambda+9)(\lambda+16)(\lambda+25)}{(\lambda+9)(\lambda+16)+(\lambda+9)(\lambda+25)+(\lambda+16)(\lambda+25)}$$ which, according to wolframalpha, is equivalent to $$\lambda^3-769\lambda-7200=0,$$ hence the answer is $$\sqrt{\frac{(\lambda+16)(\lambda+25)}{\lambda+9}} + \sqrt{\frac{(\lambda+9)(\lambda+25)}{\lambda+16}} + \sqrt{\frac{(\lambda+9)(\lambda+16)}{\lambda+25}}$$ where $$\lambda = \frac{\sqrt[3]{32400 + i \sqrt{314509827}}}{3^{2/3}} + \frac{769}{\sqrt[3]{3(32400 + i \sqrt{314509827})}} \approx 31.5756803606316$$ is the positive root of $\lambda^3-769\lambda-7200=0$.

timon92
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