Some secondary parts of a triangle II

We can solve this without using Heron's formula .

We have $IA \cdot IB \cdot IC = 4Rr^2$ ( see [9] ), where $R$ and $r$ are the circumradius and inradius of $\triangle ABC$ respectively. Then we have $4Rr^2 = 900R$ , $\implies r = 15$ .

We also have $\dfrac 1r = \dfrac 1{h_a} + \dfrac 1{h_b} + \dfrac 1{h_c}$ ( see [12] ) $\implies \dfrac 1{15} = \dfrac 1{48} + \dfrac 1{75} + \dfrac 1{h_c}$ $\implies h_c = \dfrac {400}{13}$ .

Since the area of $\triangle ABC$ , $[ABC] = \dfrac {ah_a}2 = \dfrac {bh_b}2 = \dfrac {ch_c}2$ $\implies a:b:c = \dfrac 1{48} : \dfrac 1{75} : \dfrac {13}{400} = 25:16:39$ . Then by cosine rule we have $39^2 = 25^2 + 16^2 - 2(25)(16) \cos C$ , $\implies \cos C = -0.8 \implies \sin C = 0.6$ .

The area of $\triangle ABC$ is also given by $[ABC] = \dfrac {ab\sin C}2 = \dfrac {ah_a}2$ , $\implies 0.6 b = 48 \implies b = 80$ $\implies a = 125$ and $c=195$ and the perimeter $a+b+c = 125+80+195 = \boxed{400}$ .

Let $r$ be the inradius and $s$ the semiperimeter of the triangle. Denote $AB$ by $c$ .

For this solution, we use some properties of the incenter one can find here and the result Mark Hennings proved in his solution to this problem.

Precisely, we need the formulae $\begin{aligned} IA\cdot IB\cdot IC=4R{{r}^{2}} \ \ \ \ \ & (1) \\ \dfrac{1}{{{h}_{a}}}+\dfrac{1}{{{h}_{b}}}+\dfrac{1}{{{h}_{c}}}=\dfrac{1}{r} \ \ \ \ \ & (2) \\ \text{Heron’s formula: } \ \ \left[ ABC \right]=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)} \ \ \ \ \ & (3) \\ \end{aligned}$ Step 1 : the inradius $\left( 1 \right)\Rightarrow 900R=4R{{r}^{2}}\Rightarrow {{r}^{2}}=225\Rightarrow r=15$ Step 2 : the third altitude: $\left( 2 \right)\Rightarrow \frac{1}{48}+\frac{1}{75}+\frac{1}{{{h}_{c}}}=\frac{1}{15}\Rightarrow {{h}_{c}}=\frac{400}{13}$ Step 3 : the triangle’s sides $\begin{aligned} \left[ ABC \right]=\frac{1}{2}{{h}_{a}}\cdot a=\frac{1}{2}{{h}_{b}}\cdot b=\frac{1}{2}{{h}_{c}}\cdot c& \Rightarrow 48a=75b=\frac{400}{13}c \Rightarrow \left\{ \begin{matrix} b=\frac{16}{25}a \\ \ \ \\ c=\frac{39}{25}a \\ \end{matrix} \right. \\ \end{aligned}$

$s=\dfrac{a+b+c}{2}=\left( 1+\frac{16}{25}+\frac{39}{25} \right)\dfrac{a}{2} \Rightarrow s=\frac{40}{25}a$

$s-a=\frac{40}{25}a-a=\frac{15}{25}a \ \ \ \ \ \ \ \ \ \ s-b=\frac{40}{25}a-\frac{16}{25}a=\frac{24}{25}a \ \ \ \ \ \ \ \ \ \ s-c=\frac{40}{25}a-\frac{39}{25}a=\frac{1}{25}a$

Now, Heron’s formula $\begin{aligned} \left( 3 \right) & \Rightarrow {{\left( \frac{1}{2}{{h}_{a}}a \right)}^{2}}=\frac{40}{25}\cdot \frac{15}{25}\cdot \frac{24}{25}\cdot \frac{1}{25}\cdot {{a}^{4}} \\ & \Rightarrow {{24}^{2}}{{a}^{2}}=\frac{14400}{{{25}^{4}}}{{a}^{4}} \\ & \Rightarrow {{a}^{2}}=\frac{{{24}^{2}}\cdot {{25}^{4}}}{{{120}^{2}}} \\ & \Rightarrow a=125 \\ \end{aligned}$ Consequently,
$b=\frac{16}{25}\cdot 125\Rightarrow b=80$ $c=\frac{39}{25}\cdot 125\Rightarrow c=195$ For the answer, the perimeter is $a+b+c=125+80+195=\boxed{400}$

By the properties of an incenter , $IA\cdot IB \cdot IC = 4Rr^2 = 900R$ , which solves to $r = 15$ .

The area of a triangle is $T = rs$ , so $T = 15s$ .

The area of a triangle is also $T = \frac{1}{2}bh$ , so $2T = 48a$ and $2T = 75b$ .

By Heron's formula the area of a triangle is also $T^2 = s(s - a)(s - b)(s - c)$ , where $2s = a + b + c$ .

These five equations solve to positive solutions of $a = 125$ , $b = 80$ , $c = 195$ , $s = 200$ , and $T = 3000$ .

Therefore, the perimeter is $2s = 2 \cdot 200 = \boxed{400}$ .