The altitude lengths are 12, 15 and 20. I would like a process rather than just a single solution.


Let $A$ be the area of our triangle (not yet known), and let the altitudes be $p$, $q$, and $r$ (known). Let the sides (not yet known) be, in order, $x$, $y$, and $z$. Then $$px=qy=rz=2A,$$ or equivalently $$x=\frac{2A}{p},\qquad y=\frac{2A}{q}, \qquad z=\frac{2A}{r}.$$

By Heron's Formula, $$A=\sqrt{s(s-x)(s-y)(s-z)}$$ where $s=(x+y+z)/2$.

Substitute our expressions for $x$, $y$, and $z$ in terms of $A$, $p$, $q$, and $r$ into Heron's Formula.

We obtain a messy-looking but fundamentally simple equation for $A$ (after cancellation, it is linear). Solve for $A$. Now that we know $A$, we can find the sides, and hence the perimeter.

  • 1
    $\begingroup$ It may be $A=\dfrac{1}{\sqrt{\frac{2}{{q}^{2}\,{r}^{2}}+\frac{2}{{p}^{2}\,{r}^{2}}+\frac{2}{{p}^{2}\,{q}^{2}}-\frac{1}{{p}^{4}}-\frac{1}{{q}^{4}}-\frac{1}{{r}^{4}}}}$ $\endgroup$ – Henry Aug 3 '11 at 22:59
  • $\begingroup$ @Henry: Undoubtedly you are right. The terms in Heron's Formula combine nicely. In the proof that I (barely) remember, they are obtained by factoring. $\endgroup$ – André Nicolas Aug 3 '11 at 23:10
  • $\begingroup$ @Henry: that's the formula I get, as well. $\endgroup$ – robjohn Aug 4 '11 at 14:50

Consider the triangle

enter image description here

Triangle with angles $A,B,C$ and opposite sides $a,b,c$

Its area is given e.g. by $$S=\dfrac{bc\sin A}{2}.\tag{1}$$ From the following trigonometric relations valid for a triangle

$$\sin A=2\sin \frac{A}{2}\cos \frac{A}{2}\tag{2}$$


$$\sin \frac{A}{2}=\sqrt{\dfrac{(p-b)(p-c)}{bc}}\tag{3}$$

$$\cos \frac{A}{2}=\sqrt{\dfrac{p(p-a)}{bc}},\tag{4}$$

we obtain the Heron's formula

$$S=\sqrt{p\left( p-a\right) \left( p-b\right) \left( p-c\right) },\tag{5}$$

where $$2p=a+b+c\tag{6}$$ is the perimeter. A geometric proof of $(5)$ can be found in this post, in Portuguese. In this case $S=\dfrac{12a}{2}=\dfrac{15b}{2}=\dfrac{20c}{2}$. Then $a=\dfrac{1}{6}S,b=\dfrac{2}{15}S$ and $c=\dfrac{1}{10}S$. Hence

$$2p=a+b+c=\frac{1}{6}S+\frac{2}{15}S+\frac{1}{10}S=\frac{2}{5}S,\qquad p=\frac{1}{5}\tag{7}S$$


$$S=\sqrt{\frac{1}{5}S\left(\frac{1}{5}S-\frac{1}{6}S\right)\left(\frac{1}{5}S-\frac{2}{15}S\right)\left(% \frac{1}{5}S-\frac{1}{10}S\right)}=\frac{1}{150}S^{2}.\tag{8}$$

Solving for $S$, we get $S=150$. So the perimeter is $$\boxed{2p=\frac{2}{5}150=60.}\tag{9}$$

Note: the sides are $a=\dfrac{1}{6}150=25,b=\dfrac{2}{15}150=20,c=\dfrac{1}{10}150=15$.


enter image description here

Triangle with angles $A,B,C$, opposite sides $a,b,c$ and altitudes $h_1,h_2,h_3$

For the general case of a triangle with altitudes $h_{1},h_{2},h_{3}$ perpendicular respectively to sides $a,b,c$ its area is $S=\dfrac{ah_{1}}{2}=% \dfrac{bh_{2}}{2}=\dfrac{ch_{3}}{2}$. Consequently $a=\dfrac{2S}{h_{1}}$, $b=% \dfrac{2S}{h_{2}}$, $c=\dfrac{2S}{h_{3}}$. So the perimeter $2p$ and semi-perimeter $p$ of the triangle are given by

$$\begin{eqnarray} 2p &=&a+b+c=\frac{2S}{h}, \\ p &=&\frac{S}{h}, \end{eqnarray}\tag{A}$$

where $h$ is such that

$$\begin{equation} \frac{1}{h}=\frac{1}{h_{1}}+\frac{1}{h_{2}}+\frac{1}{h_{3}} \end{equation}\Leftrightarrow h=\frac{h_{1}h_{2}h_{3}}{h_{2}h_{3}+h_{1}h_{3}+h_{1}h_{2}}.\tag{B}$$


$$\begin{eqnarray*} S &=&\sqrt{\frac{S}{h}\left( \frac{S}{h}-\frac{2S}{h_{1}}\right) \left( \frac{S}{h}-\frac{2S}{h_{2}}\right) \left( \frac{S}{h}-\frac{2S}{h_{3}}% \right) } \\ &=&\frac{S^{2}}{h^{2}}\sqrt{\frac{\left( h_{1}-2h\right) \left( h_{2}-2h\right) \left( h_{3}-2h\right) }{h_{1}h_{2}h_{3}}} \end{eqnarray*}\tag{C}$$

Solving for $S$, we now get

$$\begin{equation} S=\dfrac{h^{2}}{\sqrt{\dfrac{\left( h_{1}-2h\right) \left( h_{2}-2h\right) \left( h_{3}-2h\right) }{h_{1}h_{2}h_{3}}}} \end{equation}\tag{D}$$

and finally the perimeter as a function of $h,h_1,h_2$ and $h_3$, with $h$ being a function of $h_1,h_2,h_3$ given by $(\text{B})$:

$$\boxed{\begin{equation} 2p=\dfrac{2h}{\sqrt{\dfrac{\left( h_{1}-2h\right) \left( h_{2}-2h\right) \left( h_{3}-2h\right) }{h_{1}h_{2}h_{3}}}}. \end{equation}}\tag{E}$$


The sides are proportional to 1/(altitudes). Find the constant of proportionality so that the resulting triangle has exactly the given altitudes and not some multiple thereof.

You could, for example, construct one scale model of the triangle (in the given example it is similar to a 3-4-5 right triangle) and see how much it has to be enlarged to match the given heights. Or use the area formula as in the other solutions. No matter the approach, the shape of the triangle is known immediately and only the exact scale is left to be determined.


$$ t=area, x=ha, y=hb, z=hc $$ $$ t=\frac{x^2*y^2*z^2}{\sqrt{(xy+yz+zx)(-xy+yz+zx)(xy-yz+zx)(xy+yz-zx)}} $$ or $$ t=\frac{1}{\sqrt{\frac{2}{x^2*y^2}+\frac{2}{y^2*z^2}+\frac{2}{z^2*x^2}-\frac{1}{x^4}-\frac{1}{y^4}-\frac{1}{z^4}}} $$ before each side = area * 2 / height

finaly perimeter = a + b + c

Use TrianCal.esy.es (Triangle Calculator) Example: http://triancal.esy.es/?lang=en&tip=2&x=45&y=60&z=36


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