Some secondary parts of a triangle

By symmetry, $\triangle ABC$ is an isosceles triangle.

Let $AD = h_1 = 40$ , $BE = h_2 = 48$ , and $CF = h_3 = 48$ . Also, let $x = BD = DC$ , $y = BF = CE$ , and $z = FA = EA$ .

By the Pythagorean Theorem on $\triangle ADB$ , $x^2 + 40^2 = (y + z)^2$ .

By the Pythagorean Theorem on $\triangle BFC$ , $y^2 + 48^2 = (2x)^2$ .

By the Pythagorean Theorem on $\triangle AFC$ , $z^2 + 48^2 = (y + z)^2$ .

The three equations have positive solutions of $x = 30$ , $y = 36$ , and $z = 14$ , so the perimeter of $\triangle ABC$ is $P = 2x + 2y + 2z = \boxed{160}$ .

Assuming the perpendicular from $A$ to $BC$ be $h_1$ , then $h_2$ and $h_3$ are the altitude from $B$ and $C$ . Since $h_2=h_3$ , then $b=c$ . Then the area of $\triangle ABC$ , $[ABC] = \dfrac {ah_1}2 = \dfrac {bh_2}2 \implies 20a = 24 b \implies b = \dfrac 56a$ . We note that

$\begin{aligned} b^2 - \left(\dfrac a2 \right)^2 & = h_1^2 \\ \frac {25}{36}a^2 + \frac {a^2}4 & = 40^2 \\ \frac {16}{36}a^2 & = 40^2 \\ \frac 46 a & = 40 \\ \implies a & = 60 \\ b & = \frac 56 a = 50 \end{aligned}$

The perimeter of $\triangle ABC = a + 2b = 60 + 2\times 50 = \boxed{160}$ .

For any triangle with area $\Delta$ , the sides of the triangle can be expressed in terms of the altitudes and $\Delta$ by $a \; = \; 2\Delta h_1^{-1} \hspace{1cm} b \; =\; 2\Delta h_2^{-1} \hspace{1cm} c \; = \; 2\Delta h_3^{-1}$ and hence we have the semiperimeter $s = \Delta(h_1^{-1} + h_2^{-1} + h_3^{-1})$ and $s-a = \Delta(-h_1^{-1} + h_2^{-1} + h_3^{-1}) \hspace{1cm} s-b = \Delta(h_1^{-1} - h_2^{-1} + h_3^{-1}) \hspace{1cm} s-c = \Delta(h_1^{-1} + h_2^{-1} - h_3^{-1})$ Using Heron's Formula, we deduce that $\Delta \; = \; \frac{1}{\sqrt{(h_1^{-1} + h_2^{-1} + h_3^{-1})(-h_1^{-1} + h_2^{-1} + h_3^{-1})(h_1^{-1} - h_2^{-1} + h_3^{-1})(h_1^{-1} + h_2^{-1} - h_3^{-1})}}$ Thus the perimeter is $2s \; = \; 2\sqrt{\frac{h_1^{-1} + h_2^{-1} + h_3^{-1}}{(-h_1^{-1} + h_2^{-1} + h_3^{-1})(h_1^{-1} - h_2^{-1} + h_3^{-1})(h_1^{-1} + h_2^{-1} - h_3^{-1})}}$ With $h_1=48$ and $h_2=h_3=40$ we find the perimeter to be $\boxed{160}$ . This method will work for all triangles, not just isosceles ones.

The second equation is not needed. The equality of two altitudes imply that the triangle is isosceles. Let the sides of the triangle have lengths $a,b$ and $b$ . Then

$b^2=\dfrac {a^2}{4}+1600$

$\dfrac 12\times a\times 40=\dfrac 12\times b\times 48$

These two equations yield

$a=\dfrac {6b}{5}\implies b^2=\dfrac {36b^2}{100}+1600$

$\implies b=50,a=60,a+2b=160$

Hence the perimeter of the triangle is $\boxed {160}$ .