I've been scratching my head with this problem: "Draw a triangle given one of its sides, the height of that side and the inradius."

Now, I can calculate the area and obtain the semiperimeter. From that I can find the sum of the two other sides (let's say, $b+c$). But I can't find neither $b$ nor $c$. Whatever I try I always end up with complicated formulas that go nowhere.

Any ideas?


  • $\begingroup$ what do you mean by the height of that side? $\endgroup$ – user31280 Nov 19 '12 at 22:10
  • $\begingroup$ I mean the height relative to that side i.e. think of the side as the base of the triangle. $\endgroup$ – Luís Pureza Nov 19 '12 at 22:15

Let's denote the given side and the equivalent height by $a$ and $h$ respectively, then the area of the triangle $A$ is

  • $A=\cfrac 12 a\cdot h=\cfrac 12 b\cdot c \cdot \sin Â$

Using law of cosine

  • $a^2=b^2+c^2-2b\cdot c\cdot \cos Â$

You already have the sum of the two sides

  • $b+c=k$

These three equations should help in finding $b$, $c$ and $Â$.

$$b\cdot c = \cfrac {2A} {\sin Â}\\(b+c)^2 = b^2 +c^2 +2bc = k^2 \implies b^2+c^2=k^2-2bc\\ a^2=k^2-2bc(1+\cosÂ)=k^2- 4A \cfrac {1+\cos Â} {\sin Â}=k^2- 4A \cdot\cot \cfrac Â2 \implies \tan \cfrac Â2=\cfrac {4A} {k^2-a^2}$$

We found $Â$!

The rest should be easy.

  • $\begingroup$ Algebraically, this should be straightforward: three equations and three unknowns. However, when I try to mix these equations together, I always end up with two unknowns: $b$ (or $c$) and $Â$. Somehow, I can't get rid of the angle... $\endgroup$ – Luís Pureza Nov 19 '12 at 23:21
  • $\begingroup$ @LuísPureza I think finding the angle first is the trick here, if one can simplify $\cfrac {1+\cos Â} {\sin Â}=\cot \cfrac Â2$ then the work is almost done. $\endgroup$ – user31280 Nov 19 '12 at 23:27
  • 1
    $\begingroup$ There's a much faster way to $\angle A$. Denote the semi-perimeter $s=x+y+z$ where $a=x+y$. Then we know $z = s - a$ where $z$ is the tangential semi-perimeter segments opposite of $a$. We then get $$\tan\left(\frac{A}{2}\right) = \frac{r}{z} = \frac{r}{s-a}$$ $\endgroup$ – EuYu Nov 19 '12 at 23:27
  • $\begingroup$ @EuYu I'm just seeing the Law of Cotangent for the first time in my life. Had I known, I would have applied it. It certainly would have made my work easier. $\endgroup$ – user31280 Nov 19 '12 at 23:30
  • $\begingroup$ Thanks, that clears it up! $\endgroup$ – Luís Pureza Nov 19 '12 at 23:31

The following uses plain coordinate geometry.

Let the known side be $a$, and let the height be $h$. As you point out, one can find (and construct) the sum $k$ of the lengths of the other two sides.

Let one end of the known side be $(-a/2,0)$, and another end be $(a/2,0)$. Let the coordinates of the third vertex of the triangle be $(x,h)$.

We can use the usual distance formula to find the sum of the distances from $(x,h)$ to the points $(-a/2,0)$ and $(a/2,0)$. This yields the equation $$\sqrt{(x+a/2)^2+h^2}+\sqrt{(x-a/2)^2+h^2}=k.\tag{$1$}$$ Now comes a cute little trick. Multiply top and (virtual) bottom of the left-hand side of $(1)$ by $\sqrt{(x+a/2)^2+h^2}-\sqrt{(x-a/2)^2+h^2}$. We get after some simplification $$\frac{2ax}{\sqrt{(x+a/2)^2+h^2}-\sqrt{(x-a/2)^2+h^2}}=k.$$ Flip both sides over, and simplify a little. We get $$\sqrt{(x+a/2)^2+h^2}-\sqrt{(x-a/2)^2+h^2}=\frac{2ax}{k}.\tag{$2$}$$ "Add" Equations $(1)$ and $(2)$. We get $$2\sqrt{(x+a/2)^2+h^2}=k+\frac{2ax}{k}.$$ Now it is safe to square both sides and not get a mess. We get a quadratic equation in $x$. This can be solved algebraically as usual, or by compass and straightedge.


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