2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $$S = [ABC] = [ABJ] + [CBJ] - [BCJ] = \frac{AB \cdot R}{2} + \frac{BC \cdot R}{2} - \frac{AC\cdot R}{2} = \frac{AB+BC-AC}{2} \cdot R = \left(\frac{AB+BC+AC}{2} - AC\right) \cdot R = \left(s-(\alpha+\beta)\right) \cdot R = R(s-\alpha-\beta)$$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $S = r\tau$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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: $$ 2\Delta = \frac{(AI_B+CI_B)\,AI_B\,CI_B\,r}{AI_B\,CI_B-r^2}.$$

$\endgroup$
0
$\begingroup$

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 $B$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.