This is an exercise from Morris Kline's "Calculus: An Intuitive and Physical Approach".
An object slides down an inclined plan $OP'$ (Fig. 3-9) starting from rest at $0$. Show that the point $Q$ reached in the time $t_1$ required to fall straight down to $P$ lies on a circle with $OP$ as diameter.
The text gives a proof by contradiction for this problem:
From the formula $s = 16t^2$ we find that the time to fall the distance $OP$ is $t_1 = \sqrt{OP}/4$. For the motion along $OP'$ we use (35), that is, $s = 16t^2 \sin{A}$. The distance $OQ$ that the object slides in time $t_1$ is $OQ = 16(OP/16) \sin{A}$. Then $ \sin{A} = OQ / OP$. Suppose $Q$ is not on the circle but R on $OP'$ is. Then $\angle OPR$ is $ \angle A$ by the use of right triangles. Then $\sin A = OR/OP$. But $\sin{A} = OQ/OP$. Hence $Q = R$ and $Q$ lies on the circle.
This proof is confusing to me because I do not know the logical form of the statement we are trying to prove. What if $Q$ does not intersect with the circle at all? What is the justification for the bold portions of the proof?
Is it possible to directly show that $Q$ lies on the circle?
My approach is to let $C$ be the center of the circle. If we can show that the length of $CO$ is equal to the length of $CQ$, then $Q$ will be on the circle.
Although, I keep getting stuck trying to show that $CQ = \frac{OP}{2}$.
Best Answer
This part supposes $Q\not=R$ where $R$ is the intersection point of the circle with $OP'$ where $R\not=O$. (note that $Q$ is on $OP'$.)
Since $\triangle{OPP'}$ and $\triangle{PRP'}$ are right triangles, we get $$\angle{OPR}=\angle{OPP'}-\angle{RPP'}=90^\circ-\angle{RPP'}=(180^\circ-\angle{PRP'})-\angle{RPP'}=\angle A$$
The last step of the proof is as follows :
It follows from $\sin A=\frac{OQ}{OP}$ and $\sin A=\frac{OR}{OP}$ that $\frac{OQ}{OP}=\frac{OR}{OP}\implies OQ=OR\implies Q=R$ which contradicts the supposition that $Q\not=R$. So, we see that $Q=R$, and that $Q$ lies on the circle.
By the way, I think we can prove that without using a proof by contradiction as follows :
(After getting $\sin A=\frac{OQ}{OP}$) Let us define $R$ as the intersection point of the circle with $OP'$ where $R\not=O$. Then, we get $\sin A=\frac{OR}{OP}$. It follows that $\frac{OQ}{OP}=\frac{OR}{OP}\implies OQ=OR\implies Q=R$. So, $Q$ is on the circle.
Applying the law of cosines to $\triangle{OQC}$, we get $$\begin{align}CQ&=\sqrt{OQ^2+OC^2-2OQ\cdot OC\cos\angle{QOC}} \\\\&=\sqrt{(OP\sin A)^2+\bigg(\frac{OP}{2}\bigg)^2-2\cdot OP\sin A\cdot\frac{OP}{2}\cos(90^\circ -A)} \\\\&=\sqrt{OP^2\sin^2A+\bigg(\frac{OP}{2}\bigg)^2- OP^2\sin^2A} \\\\&=\frac{OP}{2}\end{align}$$