# Find all the parameters $\lambda\in[0,1]$ s. t. $\measuredangle BTQ=90^\circ$.

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}

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

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 $$\lambda=\frac{4}{3}.$$
• 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? Jun 29 '20 at 21:26