Let $\triangle ABC$ be equilateral with the side length $1$, $P$ be the midpoint of $\overline{AB}$ and $Q\in\overline{AC}$ s. t. $\overrightarrow{AQ}=\frac13\overrightarrow{AC}$. Let $T$ be a point satisfying $\overrightarrow{CT}=\lambda\overrightarrow{CP}$. Find all the parameters $\lambda\in[0,1]$ s. t. $\measuredangle BTQ=90^\circ$.

My attempt:

If $\measuredangle{BTQ}=90^\circ$, we can construct a circle $k$ with a diameter $\overline{BQ}$.

$\overrightarrow{CT}=\lambda\overrightarrow{CP}\implies \overline{CT}\in CP$, so there are two possibilities $T_1$ and $T_2$ where one of the two points will be inside $\triangle ABC$ and the other one will be outside.

Let $I$ and $R$ be the other intersection points of $AB$ and circle $k$ and $BC$ and $k$ respectively. Then $I$ is the foot of the altitude of $\triangle ABQ$ and $R$ is the foot of the altitude of $\triangle BCQ$.

According to the notation in the picture below: $$\begin{aligned}\measuredangle BIT_1&=\measuredangle BQT_1=\measuredangle BT_2T_1\\\measuredangle IT_1T_2&=\measuredangle IBT_2=\measuredangle T_1T_2Q\\\measuredangle T_1QI&=\measuredangle QT_1P=\measuredangle T_1BA=QBR\end{aligned}$$ Then $$\triangle AIQ\sim\triangle APC\implies\frac{|AI|}{|AP|}=\frac{|AQ|}{|AC|}\implies|AI|=\frac{|AQ|\cdot|AP|}{|AC|}=\frac16\implies|IB|=\frac56$$

Let $M$ be the intersection point of $CP$ and $BQ$. $$\triangle PBM\sim\triangle IBQ$$ From $\triangle AIQ$ we have $|IQ|=\frac{\sqrt{3}}{6}$. $$|BQ|=\sqrt{|IQ|^2+|BI|^2}=\frac{\sqrt{7}}3\implies r_k=|BS|=\frac{|BQ|}2=\frac{\sqrt{7}}6$$ Let $O$ be the intersection point of $T_1Q$ and $AB$. Also: $$\begin{aligned}\triangle IT_1O&\sim\triangle QOB\\\triangle QT_1M&\sim\triangle T_2MB\\\triangle IT_1P&\sim\triangle QT_1B\sim\triangle T_2PB\\\triangle MT_1B&\sim\triangle QMT_2\end{aligned}$$ enter image description here

However, I couldn't find $|CT_1|$ and $|CT_2|$.

May I ask for advice on how to solve this task?

Thank you in advance!


Let $\vec{CA}=\vec{a}$ and $\vec{CB}=\vec{b}.$

Thus, $$\vec{TQ}=-\lambda\left(\frac{1}{2}\vec{a}+\frac{1}{2}\vec{b}\right)+\frac{2}{3}\vec{a},$$ $$\vec{TB}=-\lambda\left(\frac{1}{2}\vec{a}+\frac{1}{2}\vec{b}\right)+\vec{b}$$ and $$\vec{TQ}\cdot\vec{TB}=0.$$ Can you end it now?

I got $\lambda=\frac{1}{3}$ or $\lambda=\frac{4}{3}.$

  • 1
    $\begingroup$ Thank you very much! I can end it. $\endgroup$
    – Invisible
    Jun 17 '20 at 18:37
  • $\begingroup$ I apologize for following up the question, but is there any other efficient method apart from placing the vertex $C$ in the origin and considering $\vec{a}=\begin{bmatrix}\cos 0\\\sin 0\end{bmatrix},\ \vec{b}=\begin{bmatrix}\cos\frac{\pi}3\\\sin\frac{\pi}3\end{bmatrix}$ so as to compute the dot product? $\endgroup$
    – Invisible
    Jun 29 '20 at 21:26

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