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.


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

  • $\begingroup$ 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

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 $ABC$.


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