Let $S$ be a circle with it s center at point $O$ and a radius of $1$. Let $\triangle$ABC be a triangle such that all its vertices are on $S$ and $AB:AC=3:2$. As shown in the figure, let $D$ be a point on the extension of side $BC$ and $k$ be the number where


Moreover, set

$\vec{OA}=\vec{a},$ $\vec{OB}=\vec{b},$ $\vec{OC}=\vec{c},$

Since the equality


holds, by expressing the inner product $\vec{a}\cdot\vec{b}$ in terms of the inner product $\vec{a}\cdot\vec{c}$, we have


Find F, G, H and I.

This is my scholarship exam practice assuming high school math knowledge.

The answer key provided is 9, 4, 5 and 4. I do not know how to begin here, could you please give me a hint to start on this question?


Hint: Expand $\lvert\vec{b}-\vec{a}\rvert^2=\frac94\lvert\vec{c}-\vec{a}\rvert^2$ and remember $\vec{a}^2=\vec{b}^2=\vec{c}^2=1$. The point $D$ plays no part here.

  • $\begingroup$ This question was cut to only a problematic part so there might be excessive info for this problem. Thank you so much for the help, now I got where I was missing. $\endgroup$ – Trey Anupong Jun 19 at 15:00
  • $\begingroup$ I think I would like some hint on my last question as well saying that: It follows that when the tangent to $S$ at the point $A$ passes through the point $D$, then $k=?$ $\endgroup$ – Trey Anupong Jun 19 at 15:09
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    $\begingroup$ Oops, yes, it should be $\vec{a}$ perpendicular to $\overrightarrow{AD}$ so with $\overrightarrow{AD}=\vec{b}-\vec{a}+\frac{2+k}2(\vec{c}-\vec{b})$ we get $k$ by dotting with $\vec{a}$. $\endgroup$ – user10354138 Jun 19 at 16:19
  • $\begingroup$ So we dot $\vec{a}$ with $\vec{AD}$ which is equal to $\vec{OD}$, am I correct? $\endgroup$ – Trey Anupong Jun 19 at 16:24
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    $\begingroup$ $\overrightarrow{AD}=\overrightarrow{OD}-\vec{a}$. We know $\overrightarrow{AD}\cdot\vec{a}=0$ which gives us an equation for $k$ in terms of $\vec{a}\cdot\vec{b}$ and $\vec{a}\cdot\vec{c}$. $\endgroup$ – user10354138 Jun 19 at 16:26

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