A friend gave me recently the following interesting problem and I would like to share a couple of solutions. Any additional contributions are welcome.

A triangle $\vartriangle ABC$ is given and we know the radius $r$ of its incircle $(O, r)$. Let $D$ be the tangential point of the incircle on $AC$ which partitions $AC$ into $AD=\alpha$ and $DC=\beta$. Determine the area of $\vartriangle ABC$ as a function of $\alpha$, $\beta$ and $r$.

Figure 1: Triangle ABC and its incircle centered at O.

  • $\begingroup$ Have you got $B$ and $C$ swapped around in your diagram? $\endgroup$ – Rob Arthan Jul 31 '15 at 14:54
  • $\begingroup$ @DavidQuinn The area of $ABC$ is $S=r\tau$ where $\tau$ is the semiperimeter of $ABC$, i.e., $\tau=(AB+BC+CA)/2$. I updated my answer with the complete formula of $S$. $\endgroup$ – Pantelis Sopasakis Jul 31 '15 at 15:04
  • $\begingroup$ @RobArthan Yes, there was indeed a typo in the question. Thanks for pointing this out. $\endgroup$ – Pantelis Sopasakis Jul 31 '15 at 15:07
  • $\begingroup$ @Sawarnik What are $a$ and $b$? It should be $a=s-\alpha=\beta+BE$ and $b=s-\beta=\alpha+BE$ but I can't see how this helps. $\endgroup$ – Pantelis Sopasakis Jul 31 '15 at 15:12
  • $\begingroup$ $$\cot C = \frac{\cot ^2 \frac{C}2 -1 }{2\cot \frac{C}2}=\frac{\frac{(s-c)^2-r^2}{r^2}}{2\frac{s-c}{r}}=\frac{\beta^2-r^2}{2r \beta}$$ Similarly, we can get $\cot B$, and use this formula, and done: $$\Delta = \frac{(\alpha +\beta)^2}{2(\cot B+ \cot C)}$$ $\endgroup$ – Sawarnik Jul 31 '15 at 17:34

Let $s,S,R$ be semiperimeter, area and $B$-exradius of triangle $ABC$ respectively. It is well-known that $rR=\alpha\beta$ and $rs=S=R(s-\alpha-\beta)$ (see below). From the latter we get $s=\frac{R(\alpha+\beta)}{R-r}$ which leads to $$S=rs=\frac{rR(\alpha+\beta)}{R-r}=\frac{r\alpha\beta(\alpha+\beta)}{rR-r^2}=\frac{r\alpha\beta(\alpha+\beta)}{\alpha\beta-r^2}.$$


For a proof that $rR=\alpha\beta$ denote the $B$-excenter by $J$. Let $X$ be a point of tangency of $B$-excircle with side $AC$. It is well-known that $AX=CD$. Note that triangles $JAX$ and $COD$ are similar as their angles are equal. Thus $$\frac{R}{\beta}=\frac{JX}{AX}=\frac{CD}{OD}=\frac{\alpha}{r}$$ yielding $rR=\alpha\beta$.

Now we'll prove that $S=R(s-\alpha-\beta)$. We'll use a notation $[\mathcal F]$ which denotes area of $\mathcal F$. We have $$S = [ABC] = [ABJ] + [CBJ] - [BCJ] = \frac{AB \cdot R}{2} + \frac{BC \cdot R}{2} - \frac{AC\cdot R}{2} = \frac{AB+BC-AC}{2} \cdot R = \left(\frac{AB+BC+AC}{2} - AC\right) \cdot R = \left(s-(\alpha+\beta)\right) \cdot R = R(s-\alpha-\beta)$$

  • $\begingroup$ Could you provide some references about $rR=\alpha\beta$ and $S=R(x-\alpha-\beta)$ (which means that $S=R\cdot BE$)? $\endgroup$ – Pantelis Sopasakis Jul 31 '15 at 17:52
  • $\begingroup$ these are easy facts, I'll sketch proofs of them in a sec $\endgroup$ – timon92 Jul 31 '15 at 19:37

Solution 1: Draw $AO$, $AO$ and $CO$. We know that these are the bisectors of $\angle BAC$, $\angle ABC$ and $\angle BCA$ respectively.

Let $\angle OAC = \frac{1}{2} \angle BAC =: \phi_A$ and $\angle OBE = \frac{1}{2} \angle ABC =: \phi_B$ and $\angle OCA = \frac{1}{2}\angle BCA =: \phi_C$.

We have that $\angle ODA$ is a right angle (and so is $\angle ODC$). Thus $\tan \phi_A = \frac{r}{\alpha}$ and $\tan \phi_C = \frac{r}{\beta}$. Therefore $\angle AOD = \frac{\pi}{2} - \phi_A$.

The two right-angled triangles $\vartriangle ADO$ and $\vartriangle AEO$ are equal (indeed, they share a common side $AO$ and $OE=OD=r$). Thus, $\angle EOA = \angle DOA = \frac{\pi}{2} - \phi_A$.

Similarly, $\angle COD = \frac{\pi}{2}-\phi_C$ and since $\vartriangle COD = \vartriangle COF$ it is $\angle COF = \angle COD = \frac{\pi}{2}-\phi_C$.

Using the fact that $$ \angle AOD + \angle DOC + \angle COF + \angle FOE + \angle EOA = 2\pi,\\ 2(\frac{\pi}{2}-\phi_A) + 2(\frac{\pi}{2}-\phi_C) + \angle FOE = 2\pi,\\ \angle FOE = 2(\phi_A + \phi_C). $$ Because of the equality of $\vartriangle BEO$ and $\vartriangle BFO$ we have $\angle EOB = \frac{\angle FOE}{2} = \phi_A + \phi_C$, so $$ \tan \angle EOB = \frac{BE}{r},\\ BE = r \tan(\phi_A + \phi_C) = r\frac{\tan\phi_A + \tan\phi_C}{1-\tan\phi_A\tan\phi_C}=\\ =r\frac{\frac{r}{\alpha}+\frac{r}{\beta}}{1-\frac{r}{\alpha}\frac{r}{\beta}}=r\frac{\frac{r(\alpha+\beta)}{\alpha\beta}}{\frac{\alpha\beta-r^2}{\alpha\beta}}=r^2\frac{\alpha+\beta}{\alpha\beta-r^2} $$

From the aforementioned triangle equalities we have $AE = \alpha$, $CF = \beta$ and $BF = BE$, therefore, the semiperimeter of $\vartriangle ABC$ is

$$ \tau = \alpha + \beta + BE = \alpha + \beta + r^2\frac{\alpha+\beta}{\alpha\beta-r^2}=\frac{\alpha\beta(\alpha+\beta)}{\alpha\beta-r^2}, $$

and the area of the triangle is

$$S = r\tau = r\frac{\alpha\beta(\alpha+\beta)}{\alpha\beta-r^2}$$

Solution 2: This is a solution without trigonometry and is less elegant. Since $\angle BOE = \phi_A + \phi_C$ there is a point $M$ on $BE$ so that $\angle MOE = \phi_A$ and $\angle MOB = \phi_C$ as in the following figure:

Figure 2: A more geometric solution

It is easy to see that the two triangles $\vartriangle MEO$ and $\vartriangle OEA$ are similar (they have all their angles equal to one another), so, $$ \frac{EM}{EO}=\frac{EO}{EA},\\ EM = \frac{r^2}{\alpha}. $$ We now need to determine $BM$. What is the same, $\vartriangle BMO$ and $\vartriangle BOC$ are similar (all angles equal to one another). Note that (Pythagorean theorem on $\vartriangle BEO$)

$$BO=\sqrt{BE^2 + r^2} = \sqrt{(BM+ME)^2 + r^2},$$

and (Pythagorean theorem on $\vartriangle OCD$)

$$OC = \sqrt{r^2 + \beta^2}.$$

We also have

$$ MO = \sqrt{r^2 + ME^2} = \sqrt{r^2 + \frac{r^4}{\alpha^2}} = r\sqrt{1+(r/a)^2} $$

Using the fact that $\vartriangle BMO$ and $\vartriangle BOC$ are similar (all angles equal to one another), we arrive at

$$ \frac{BM}{BO}=\frac{MO}{OC},\\ \frac{BM}{\sqrt{(BM+\frac{r^2}{\alpha^2})^2 + r^2}}=\frac{r\sqrt{1+\frac{r^2}{\alpha^2}}}{\sqrt{r^2 + \beta^2}},\\ \frac{BM^2}{(BM+\frac{r^2}{\alpha^2})^2 + r^2}=\frac{r^2 (1+\frac{r^2}{\alpha^2})}{r^2 + \beta^2}, $$

which is a quadratic equation one can solve to derive a formula for $BM$ (the non-positive solution should be discarded) in terms of $r$, $\alpha$ and $\beta$. Having determined $BM$ and $ME$ we can easily determine the semiperimeter of $\vartriangle ABC$, $\tau = \alpha + \beta + BM + ME$ and the area of $\vartriangle ABC$ is again $S = r\tau$.

  • $\begingroup$ Have you checked your answer against any examples? For an equilateral with sides of length $2\sqrt{3}$, you have $r = 1$ and $\alpha = \beta = \sqrt{3}$ and the area should come out as $3\sqrt{3}$. $\endgroup$ – Rob Arthan Jul 31 '15 at 16:00
  • $\begingroup$ @RobArthan I have made some modifications in my answer and wrote $S$ in a simpler form which agrees with the answers of other contributors. $\endgroup$ – Pantelis Sopasakis Jul 31 '15 at 17:41

If $I$ is the incenter of $ABC$ and $I_A,I_B,I_C$ are the projections of the incenter on the sides $BC,AC,AB$ (sorry, but to call the incenter $O$ gives me some itch), we have: $$ AI_B=AI_C=\frac{b+c}{2}\,\quad BI_A=BI_C=\frac{a+c}{2},\quad CI_A=CI_B=\frac{a+b}{2} $$ and $2\Delta = r(a+b+c)$, or $2\Delta = r(AI_B+BI_C+CI_A) $, so we just need to find $BI_C$ in terms of $r,AI_B,CI_B$. Pretty easy:

$$ BI_C = \frac{r}{\tan\frac{\widehat{B}}{2}} = r\tan\frac{\widehat{A}+\widehat{C}}{2}=r\cdot\frac{\tan\frac{\widehat{A}}{2}+\tan\frac{\widehat{C}}{2}}{1-\tan\frac{\widehat{A}}{2}\tan\frac{\widehat{C}}{2}}$$ hence: $$ BI_C = r^2\cdot\frac{AI_B+CI_B}{AI_B\cdot CI_B-r^2} $$ and: $$ 2\Delta = \frac{(AI_B+CI_B)\,AI_B\,CI_B\,r}{AI_B\,CI_B-r^2}.$$


I'll use the more-traditional notation with tangent points $D$, $E$, $F$ opposite respective vertices $A$, $B$, $C$. I'll also take $\alpha$, $\beta$, $\gamma$ (the first two given, the last unknown) to be the lengths of the tangent segments from vertices $A$, $B$, $C$. Finally, I'll reduce some notational clutter by writing $A_2$, $B_2$, $C_2$ for half-angles $A/2$, $B/2$, $C/2$. (So, $A_2 + B_2 + C_2 = 90^\circ$.)

Since $\alpha + \beta + \gamma$ is the semi-perimeter, we know the area of the triangle is given by $$|\triangle ABC| = r (\alpha + \beta + \gamma) \tag{$\star$}$$ Our task is to replace $\gamma$. Clearly, $$r = \alpha \tan A_2 = \beta \tan B_2 = \gamma \tan C_2$$

But, $$\tan C_2 = \tan(90^\circ - A_2 - B_2) = \cot(A_2 + B_2) = \frac{\cot A_2 \cot B_2 - 1}{\cot A_2 + \cot B_2} = \frac{\frac{\alpha}{r}\frac{\beta}{r}-1}{\frac{\alpha}{r}+\frac{\beta}{r}} = \frac{\alpha\beta-r^2}{r(\alpha+\beta)}$$ Therefore, $$\gamma = \frac{r}{\tan C_2} = \frac{r^2(\alpha+\beta)}{\alpha\beta-r^2}$$ so that, by $(\star)$, $$|\triangle ABC| = r\left(\alpha + \beta + \frac{r^2(\alpha+\beta)}{\alpha\beta-r^2}\right)= \frac{r\,\alpha\beta\,(\alpha+\beta)}{\alpha \beta - r^2}$$

Observation: Recall that the circle with diameter $\overline{AB}$ meets the perpendicular at $F$ in points $P$ and $Q$ such that $|\overline{AP}|^2 = |\overline{AQ}|^2 = |\overline{AF}||\overline{BF}| = \alpha \beta$. This says exactly that the "power" of point $F$ with respect to that circle is $\alpha\beta$. Moreover, since that perpendicular passes through $I$ (which, we'll say, lies between $F$ and $P$), we have $$|\overline{IP}||\overline{IQ}| = \left(|\overline{AP}| - |\overline{AI}|\right)\left(|\overline{AQ}| + |\overline{AI}|\right) = \left(\sqrt{\alpha\beta} - r\right)\left(\sqrt{\alpha\beta} + r\right) = \alpha \beta - r^2$$

Thus, the denominator of the area formula is the power of $I$ with respect to that circle. This seems significant ... or coincidental. (It may also be worth noting that that circle contains the feet of the altitudes from $A$ and $B$.)


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