Motion down an inclined plane with leg equal to the diameter of a circle 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 $CQ = \frac{OP}{2}$.
 A: 
Suppose $Q$ is not on the circle but R on $OP'$ is.

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 $OP'$.)


Then $\angle OPR$ is $ \angle A$ by the use of right triangles.

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.


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 $CQ = \frac{OP}{2}$.

Applying the law of cosines to $\triangle{OQC}$, we get
$$\begin{align}CQ&=\sqrt{OQ^2+OC^2-2OQ\cdot OC\cos\angle{QOC}}
\\\\&=\sqrt{(OP\sin A)^2+\bigg(\frac{OP}{2}\bigg)^2-2\cdot OP\sin A\cdot\frac{OP}{2}\cos(90^\circ -A)}
\\\\&=\sqrt{OP^2\sin^2A+\bigg(\frac{OP}{2}\bigg)^2- OP^2\sin^2A}
\\\\&=\frac{OP}{2}\end{align}$$
A: Assume that $Q$ lies on the circle and then join $PQ$. We have $\angle PQO=90°$ and hence $\angle QPO=A$. Now in $\triangle OPQ$, $$\sin A=\frac{OQ}{OP}$$ This result is same as you derived without considering right angle at $Q$ (which is correct). So, our assumptions was correct and $Q$ lies on the circle.
A: With a few Physics.
The vertical acceleration $(0,-g)$ projected over $OP'$ is $-\sin A(\cos A,\sin A)g$
Now the movement along $OQ$ is given by
$$
(x,y) = (0,|OP|)-\frac 12m \sin A(\cos A,\sin A)gt^2
$$
but $t$ is the time needed to reach $P$ in the vertical fall so
$$
|OP| = \frac 12 m g t^2
$$
so eliminating $t$ we obtain
$$
(x,y) = (0,|OP|)-\sin A(\cos A,\sin A)|OP|
$$
and we have
$$
\cases{
x = -\sin A\cos A|OP|\\
y = (1-\sin^2 A) |OP|
}
$$
and $(x(A), y(A))$ describes a circle as expected.
A: 
The gravitational acceleration is $g$ from O straight down to P, while it is $g\cos\theta$ from O to Q along the ramp. Use the time-distance formula $d = \frac12 a t^2$ for an accelerating object to establish the equation of same time for the two cases,
$$t^2 = \frac{2OP}g = \frac{2OQ}{g\cos\theta}$$
Simplify to get $\cos\theta = \frac{OQ}{OP}$, which indicates that $\triangle$OQP is a right triangle. Hence, Q lies on the circumcircle of $\triangle$OQP with OP as the diameter.
A: For both paths.
Time = $\frac 12 \frac {\text {distance}}{\text {acceleration}}$
On the trip from $O$ to $Q,$ the acceleration is $g\sin A$
The distance covered is $\overline {OP} \sin A$
The alternative trip from $O$ to $P$ the accelartion is $g$ and the distance is $\overline {OP}$
