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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.

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  • 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.

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