1
$\begingroup$

I cannot solve this problem using synthetic geometry, mostly because I have not much knowledge of other types of geometry.

Let $I$ be the incenter of $\triangle{ABC}.$ Prove that for any point $X,$ $$a \cdot AX^2 + b \cdot BX^2 + c \cdot CX^2 = (a + b + c) \cdot IX^2 + a \cdot AI^2 + b \cdot BI^2 + c \cdot CI^2.$$

I specified synthetic geometry because I tried a coordinate bash of this with vertices of $\triangle{ABC}$ at the following points:

\begin{align} A &= (a, 0) \\ B &= (b, 0) \\ C &= (0, c). \end{align}

I think these are very easy coordinates to work with, but the coordinates for the incenter are horrific, and I don't want to proceed with this because I don't have a spare three days to work on this.

I have made some starts:

I know Stewart's Theorem: $man + dad = bmb + cnc,$ but I'm not sure how to apply it.

I know Law of Cosines, but I get nasty cosine values and I don't know how to get rid of them.

I would like some small hints, little baby steps in the right direction. I know this is a place for question and answer (or, in this case, problem and solution), but I still want to grapple with the bulk of the problem myself.


Yes, I know, this article covers the theorem, but then my question would be "what are barycentric coordinates and how do you use them?" since I know barycentric coordinates by name only.

$\endgroup$
2
$\begingroup$

Let me try. We have $a\vec{IA} + b\vec{IB} + c\vec{IC} = \vec{0}$.

So $$(a+b+c)\vec{IX} = a\vec{AX} + b\vec{BX} + c\vec{CX}.$$

Now we have $$aAX^2 + bBX^2 + cCX^2 = \sum a(\vec{AI} + \vec{IX})^2$$ $$= \sum aAI^2 + (a+b+c)IX^2 + 2(a\vec{AI} + b\vec{BI} + c\vec{CI})\vec{IX}$$ $$= \sum aAI^2 + (a+b+c)IX^2.$$

$\endgroup$
  • $\begingroup$ Why do we have $a \cdot \overrightarrow{IA} + b \cdot \overrightarrow{IB} + c \cdot \overrightarrow{IC} = 0?$ Also, why do we have $(a + b + c) \cdot \overrightarrow{IX} = a \cdot \overrightarrow{AX} + b \cdot \overrightarrow{BX} + c \cdot \overrightarrow{CX}?$ Sorry, I know about vectors in name only. $\endgroup$ – Reality Check Apr 28 '17 at 16:07
  • $\begingroup$ Check the page 2 of article you linked. You can check it yourself. For second question, note that $\vec{IA} = \vec{IX} + \vec{XA}$. $\endgroup$ – GAVD Apr 29 '17 at 14:20
  • $\begingroup$ Why is the $\sum{}$ blank? Usually, I see sums like this: $$\sum_{p = q}^{r}{s}.$$ $\endgroup$ – Reality Check May 2 '17 at 20:34
1
$\begingroup$

You may use the parallel axis theorem. If we assume that the mass of $A$ is $a$, the mass of $B$ is $b$ and the mass of $C$ is $c$, then $$ a\cdot AX^2+ b\cdot BX^2 + c\cdot CX^2 $$ is the (moment of) inertia of the system made by the massive points $A,B,C$ with respect to an axis through $X$. The center of mass of such system is the incenter of $ABC$.

$\endgroup$

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.