# Let $G, S, I$ be respectively centroid, circumcentre, incentre of triangle $ABC$. If $R, r$ are circumradius and inradius respectively then...

Let $$G, S, I$$ be respectively centroid, circumcentre, incentre of triangle $$\triangle ABC$$. If $$R, r$$ are circumradius and inradius respectively then which of the following is INCORRECT ?

• (A) $$SI^2 = R^2 (1 - \cos A \cos B \cos C) ; A,B,C$$ being angles of triangle.
• (B) $$SI^2 = R^2 - 2Rr$$
• (C) $$SG^2 = R^2 - \frac{a^2 + b^2 + c^2}9 ; a, b,c$$ being sides of triangle
• (D) $$SG ≤ SI$$

Let area of triangle be $$A$$ and semi-perimeter be $$s$$.

$$R=\dfrac{abc}{4A} \;\; \text{and} \; \; r=\dfrac As$$

So, RHS for option B) $$=\dfrac{a^2b^2c^2}{16A^2}-\dfrac{abc}s=abc\cdot\dfrac{abcs-16A^2}{16sA^2}=abc\cdot\dfrac{abcs-16s(s-a)(s-b)(s-c)}{16s^2(s-a)(s-b)(s-c)}$$

$$=abc\cdot\dfrac{abc-16(s-a)(s-b)(s-c)}{16s(s-a)(s-b)(s-c)}$$

Also, $$A=rs\implies R=\dfrac{abc}{4rs}=\dfrac{abc}{2r(a+b+c)}$$

For option C), we may write: $$\dfrac{a^2+b^2+c^2}3\ge(a^2b^2c^2)^{1/3}\implies\dfrac{a^2+b^2+c^2}9\ge\dfrac{(abc)^{2/3}}3=\dfrac{(2rR(a+b+c))^{2/3}}3$$

Not able to conclude anything.

## 1 Answer

• For $$(A)$$, try it on an equilateral triangle to show that it's false.

• $$(B)$$ is the well-known Euler's identity in geometry.

• For $$(C)$$, use the Leibnitz theorem for $$S$$, $$3R^{2}=3SG^{2}+(GA^{2}+GB^{2}+GC^{2})$$ and an identity about the centroid (the proof follows from applying Apollonius’ Theorem to the all three sides of the triangle): $$3(GA^{2}+GB^{2}+GC^{2})=a^{2}+b^{2}+c^{2}$$

• $$(D)$$ is a corollary of $$(B)$$ and $$(C)$$

As an exercise one could try to find the correct version of $$(A)$$:

$$OH^{2} = R^{2}(1 - 8 \cos A \cos B \cos C)$$ where $$H$$ denotes the orthocenter.