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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.

right triangle where one leg is the diameter of a circle as well

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

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  • $\begingroup$ Did you forget that $OP = 2CO$? $\endgroup$
    – YNK
    Jul 8, 2020 at 15:22
  • $\begingroup$ @YNK I'm still a bit confused on how to use that. Any more hints? $\endgroup$
    – Iyeeke
    Jul 8, 2020 at 16:05
  • $\begingroup$ I am sorry. My comment is wrong. Forget about it. Give me an hour or so. I will see what I can do. $\endgroup$
    – YNK
    Jul 8, 2020 at 16:26

5 Answers 5

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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 $$\begin{align}CQ&=\sqrt{OQ^2+OC^2-2OQ\cdot OC\cos\angle{QOC}} \\\\&=\sqrt{(OP\sin A)^2+\bigg(\frac{OP}{2}\bigg)^2-2\cdot OP\sin A\cdot\frac{OP}{2}\cos(90^\circ -A)} \\\\&=\sqrt{OP^2\sin^2A+\bigg(\frac{OP}{2}\bigg)^2- OP^2\sin^2A} \\\\&=\frac{OP}{2}\end{align}$$

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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.

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  • $\begingroup$ If we assume that $Q$ lies on the circle, how does it follow that $\angle PQO = 90$ degrees? $\endgroup$
    – Iyeeke
    Jul 9, 2020 at 2:13
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    $\begingroup$ Angle subtended at circumference is half of angle subtended at center of the circle. If $C$ is center of the circle then $\angle OCP=180°\implies \angle PQO=90°$. $\endgroup$
    – SarGe
    Jul 9, 2020 at 2:50
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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.

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enter image description here

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.

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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 $\overline {OP}$

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