3
$\begingroup$

Consider a ∆ABC. Let the triangle formed by the points of contact of the sides of a given triangle with the excircles corresponding to these sides be called the extouch triangle; and let the triangle formed by joining, the points of contact of the three sides with the inscribed circle be called the intouch triangle. Prove that these two triangles have equal areas, and that the midpoint of their centroids is the centroid of ∆ABC.

Attached is a proof I found: Check Page 3 of this PDF

I've been trying really hard for an analytical proof, one using only coordinate geometry - but haven't been able to get anywhere.

Could someone please help me with a proof using coordinate geometry only?

Assuming coordinates for ∆ABC and then finding the vertices of the intouch and extouch triangles will definitely solve the problem, but it's getting really lengthy and cumbersome. (The areas can be proved equal using the determinant, and calculation of centroid is easy)

Is there anything I'm missing? Could someone help me find the vertices of the intouch and extouch triangles? (the result should be interesting, for the midpoint of the two centroids coincides with the centroid of ∆ABC)

A detailed solution, or guidance in the right direction would help! (The former, preferably) Thanks!

P.S. You can find a neat diagram on Page 4 of the PDF, I'll add it here to the question if need be.

$\endgroup$
  • $\begingroup$ Doing this using co-ordinates only is an awful mess. Even the proof that a centroid exists, using only co-ordinates, is so bad that it was used to test an early computer program for manipulating basic algebra . Some of the intermediate expressions covered half a page. $\endgroup$ – DanielWainfleet Mar 16 '18 at 7:36
  • 1
    $\begingroup$ If the coordinates of $A$, $B$, $C$ are given, then the derivation is quite messy. However, a reasonably-uncomplicated argument follows from taking "intouch triangle" vertices as given, situating them conveniently on an origin-centered incircle. $\endgroup$ – Blue Mar 16 '18 at 12:05
1
$\begingroup$

Elaborating on my comment: A coordinate derivation isn't terribly difficult if the intouch triangle's vertices are placed conveniently.


enter image description here

Let the origin-centered circle of radius $r$ be the incircle of $\triangle ABC$, and locate intouch vertices $D$, $E$, $F$ thusly: $$D := r(\cos 2\beta,-\sin 2\beta) \qquad E := r( \cos 2\alpha, \sin 2\alpha) \qquad F := r(1,0)$$

Since $\overline{OF}\perp\overline{AB}$, we see immediately that $$A = r(1,\tan\alpha) \qquad B = r(1,-\tan\beta)$$ For $C$, we easily write the normal forms of the equations of the tangent lines at $D$ and $E$ ... $$\begin{align} \overleftrightarrow{BC}:&\quad x \cos 2\beta - y \sin 2\beta = r \\ \overleftrightarrow{CA}:&\quad x \cos 2\alpha + y \sin 2 \alpha = r \end{align}$$ ... and calculate the intersection: $$C = \frac{r(\cos(\alpha-\beta), \sin(\alpha-\beta))}{\cos(\alpha+\beta)}$$

Paying attention to similar triangles $\triangle OAF \sim \triangle AC^\prime F^\prime$ and $\triangle OBF \sim \triangle B C^\prime F^\prime$, we have $$\frac{|\overline{AF^\prime}|}{|\overline{BF^\prime}|} = \frac{|\overline{C^\prime F^\prime}| \cot\alpha}{|\overline{C^\prime F^\prime}|\cot\beta} = \frac{|\overline{OF}| \tan\beta}{|\overline{OF}| \tan\alpha} = \frac{|\overline{BF}|}{|\overline{AF}|}\;\to\; |\overline{AF}|=|\overline{BF^\prime}|\;\text{and}\; |\overline{BF}|=|\overline{AF^\prime}|$$ This is exactly the "isotomic" property discussed in Dalcín's article. Since, as a consequence, $\overline{AB}$ and $\overline{FF^\prime}$ have a common midpoint (likewise for other sides), we have $$A + B = F + F^\prime \qquad B + C = D + D^\prime \qquad C + A = E + E^\prime \tag{1}$$ Therefore,

$$\frac12\left(\;\frac13(D+E+F)+\frac13(D^\prime+E^\prime+F^\prime)\;\right) = \frac13(A+B+C) \tag{$\star$}$$

which is the centroid property. Confirming $|\triangle DEF| = |\triangle D^\prime E^\prime F^\prime|$ is a little messier, but can be done using $(1)$ to get explicit coordinates for $D^\prime$, $E^\prime$, $F^\prime$; for now, I'll leave that as an exercise to the reader.

$\endgroup$
1
$\begingroup$

Once you know that $BX=s-b$, $CY=s-c$, $AZ=s-a$ and so on, it is easy to find the coordinates of the points of contact of the three sides with the inscribed circle:

$$ X=B+{s-b\over a}(C-B)={s-c\over a}B+{s-b\over a}C, $$ and the analogous expressions for $Y$ and $Z$: $$ Y={s-a\over b}C+{s-c\over b}A, \quad Z={s-b\over c}A+{s-a\over c}B. $$ In the same way you can find the coordinates of the other contact points $A_b$, $A_c$, etc.

enter image description here

$\endgroup$
  • $\begingroup$ Exactly, could you help me find the contact points? Possibly include them in your solution? That's my specific difficulty. $\endgroup$ – arya_stark Mar 16 '18 at 11:49
  • $\begingroup$ I don't understand: aren't the formulas above enough? $\endgroup$ – Aretino Mar 16 '18 at 11:54
  • $\begingroup$ Sure they are, but isn't the calculation cumbersome? $\endgroup$ – arya_stark Mar 16 '18 at 11:58
  • $\begingroup$ You can choose the coordinates of $A$, $B$ and $C$ in a clever way, for instance: $B=(0,0)$, $C=(1,0)$, $A=(x,y)$. Computing $X$, $Y$, $Z$ is not so complicated, getting an expression for the area might be. $\endgroup$ – Aretino Mar 16 '18 at 13:52

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.