# Prove that $P$ lies on line $CC'$.

Semicircles diameters $$PA$$ and $$PB$$ are drawn such that they intersect at point $$Q$$ $$(P \not\equiv Q)$$ and $$PA \perp PB$$. $$M$$ and $$N$$ are points lying on line segment $$PA$$ and $$PB$$ respectively. $$MQ$$ and $$NQ$$ intersect semicircles diameters $$PB$$ and $$PA$$ at $$M'$$ and $$N'$$ in that order $$(N' \not\equiv Q \not \equiv M')$$. Given midpoints $$C$$ and $$C'$$ of respectively $$MN$$ and $$M'N'$$ $$(CP > CQ)$$. Prove that $$P$$ lies on line $$CC'$$.

This problem is said to be an application of Newton's line, which I found to be reluctant. Perhaps there might more points to be set up that I didn't know.

As the answer by Oldboy suggests, there's a condition that needs to satisfied for the problem to be correct but I haven't found it out yet. I hope someone can help me.

As the comment by Blue points out, a possible condition is that $$\left(\frac{PA}{PB}\right)^2 = \left|\frac{AM \cdot PN}{BN \cdot PM}\right|$$

• What is that C and c' satisfy? Midpoints?
– Moti
Jul 20 '19 at 4:06
• How does the phrasing imply that "M' is the intersection of MQ and the PB semicircle"? Aug 2 '19 at 20:11
• @LêThànhĐạt: Since the question, as originally stated, is flawed, please justify the change you've made to it. Does the source specifically state (or include a figure showing) that $\overline{MM'}$ and $\overline{NN'}$ pass through $Q$? Or does your figure, and your "fix", assume this property of $M'$ and $N'$? (Could it be that resolving the flaw in the problem is to re-interpret the locations of $M'$ and $N'$?)
– Blue
Aug 3 '19 at 0:48
• Yes. The problem and the figure showed do specifically state and draw out $MM'$ and $NN'$ passing through $Q$. I have looked at it thoroughly. Aug 3 '19 at 2:29
• Let $a:=\frac12|PA|$, $b:=\frac12|PB|$, $m:=|PM|$, $n:=|PN|$. If I've done things correctly, the condition to make $C$, $C'$, $P$ collinear is $$\frac{2a-m}{a^2m}=\frac{2b-n}{b^2n} \quad\to\quad \frac{b\;|AM|}{a\;|PM|}=\frac{a\;|BN|}{b\;|PN|}$$ It's not clear what the intended interpretation of this condition might be.
– Blue
Aug 3 '19 at 3:44

Something is deeply flawed here. I tried to draw the picture again in Geogebra and by following the same instructions I got this:

Points $$C,C',P$$ are clearly not collinear. Of course, you can get this condition satisfied for some special positions of points $$M$$ and $$N$$ but the OP did not mention any additional restriction:

Are we sure that the text of the problem is correct? Something is missing here, I think.

• The condition $CP > CQ$ has been added. The problem should be correct now. Jul 21 '19 at 9:21
• @LêThànhĐạt That condition is clearly satisified in both my pictures. But in the first case the statement is not true. Jul 21 '19 at 13:03
• What do you suggest? Jul 21 '19 at 14:38
• @LêThànhĐạt What is the source of this problem? Jul 21 '19 at 16:46
• A book in Vietnamese. - It's called "Tài liệu chuyên Toán - (Bài tập) Hình học 10". Jul 22 '19 at 0:37

This is the only way I found after struggling for three days. I will post a simpler and purely geometric answer when I found it.

First, we need to prove some theorems before solving this problem.

Theorem A: Consider a point $$P$$ and a line $$l$$. $$Q$$ is a point on line $$l$$, such that line segment $$PQ$$ is perpendicular to line $$l$$. Construct circle $$c$$ with $$P$$ as one end of the diameter and an arbitrary point $$R$$ on line $$l$$. Then,

1, Circle $$c$$ will always pass through point $$Q$$.

2, Locus of an arbitrary point $$D$$ on the circle $$c$$ (such that arc $$RD$$ is making an arbitrary angle $$2θ$$ at the center $$S$$ of the circle) is an straight line passing through point $$Q$$.

Image

Proof:

1, Any point that makes a right angle with diameter of the circle will be incident on the circle. $$PQ$$ was perpendicular to line $$l$$, and $$R$$ lies on line $$l$$. Hence, $$Q$$ will be on circle regardless of the position of $$R$$.

2, As Points $$P$$ and $$Q$$ are on same circle, the arc $$RD$$ will make an angle $$θ$$ (180-$$θ$$, when $$Q$$ is in between $$R$$ and $$D$$).

Line segment $$RD$$ always makes a constant angle $$θ$$ (180-$$θ$$, when $$Q$$ is in between $$R$$ and $$D$$) with line $$l$$. This is the locus of a straight line passing through $$Q$$ and making an an angle of $$θ$$ with line $$l$$.

Check this geogebra link for an interactive model

Lemma A1: Ratios of the lengths $$PD$$ and $$PR$$ will remain constant as long as θ remains constant.

Proof: As the triangle $$PDR$$ is a right angled triangle, $$PD/PR = cosθ$$

This ratio will remain constant as long as $$θ$$ remains constant.

Finally to the problem. Now, we can see that the two (semi)circles in the problem belong to the family of circles that are formed by rotating the diameter around point $$P$$ with other end resting on the line $$AB$$.

The diameters of two circles $$AP$$ and $$BP$$ with centers $$D$$ and $$E$$ are perpendicular.

The angles $$ADN'$$ and $$M"DA$$ are represented by $$α$$ and $$β$$ respectively.

Lemma B1: The arcs $$BN''$$ and $$M'B$$ make angles $$\alpha$$ and $$\beta$$ at Center $$E$$

Proof:

The arc $$AN'$$ makes an angle of '$$\alpha/2$$' at point $$Q$$. [half the angle at center.]

$$\angle {N''QB} = \angle AQN' = \alpha/2$$

$$\angle N''EB = \alpha$$

Similarly, $$\angle M'EB = \beta$$

• This is the converse of the Theorem A2.

Theorem B: Let, points $$C,C'$$ are midpoints of $$MN$$ and $$M'N'$$. The line $$CC'$$ will pass through $$P$$, if $$\alpha = \beta$$.

Let, $$PA = 2a; PB = 2b; PM = m; PN = n.$$

From Lemma A1,

$$PN'' = (2b)cos\alpha/2; PM' = (2b)cos\beta/2; PN' = (2a)cos\alpha/2; PM'' = (2a)cos\beta/2.$$

Writing the equation of line passing through points $$A$$ and $$B$$ in polar form.

$$r(\theta) = sec(\theta - \angle APQ) .PQ$$

$$r$$ = distance from $$P$$ (origin)

$$\theta$$ = angle made with $$PA$$ (x-axis)

$$r(\theta) = sec(\theta - \angle PBA) .PQ$$ [Similar triangles]

$$Let, \angle PBA = \angle B ;\angle BAP = \angle A.$$

$$r(\theta) = sec(\theta - \angle B) .PQ$$

$$N$$ is obtained when $$N''$$ slided on the line $$N''Q$$ by a making an angle of $$-\alpha/2$$ at $$P$$.

The corresponding diameter can be found by sliding point $$B$$ on line $$AB$$ while rotating $$-\alpha/2$$

The length of the corresponding diameter is,

$$r(\pi/2 - \alpha/2)$$ = $$sec(\pi/2 - \alpha/2 - \angle B). PQ$$

From Lemma A1,

$$PN = (PN''. r(\pi/2 - \alpha/2))/r(\pi/2)$$

$$n = ((2b)cos\alpha/2.sec(\pi/2 - \alpha/2 - \angle B). PQ)/2b$$

$$n = cos\alpha/2.sec(\alpha/2 - \angle A). PQ$$

Similarly,

$$m = cos\beta/2.sec(\beta/2 - \angle B). PQ$$

Coordinates of $$M =(m,0); N = (0,n)$$

$$C = (m/2,n/2)$$

Coordinates of $$M'= [(2b)cos\beta/2].(sin\beta/2 , cos\beta/2);$$

$$N'=[(2a)cos\alpha/2].(cos\alpha/2, sin\alpha/2)$$

$$C' = \frac{M'+N'}{2}$$

For $$P,C,C'$$ be collinear, slope of $$CP$$ must be equal to slope of $$C'P$$

$$\frac{cos\alpha/2.sec(\alpha/2 - \angle A)}{cos\beta/2.sec(\beta/2 - \angle B)} = \frac{bcos^2(\beta/2) + acos\alpha/2 sin\alpha/2}{acos\beta/2 sin\beta/2 + acos^2(\alpha/2)}$$

We know that $$\angle A + \angle B = \pi/2$$ and $$tan\angle B = \frac{a}{b}$$

Therefore, $$\frac{(cos\alpha/2)(bcos\beta/2 + asin\beta/2)}{(cos\beta/2)(acos\alpha/2 + bsin\alpha/2)} = \frac{bcos^2(\beta/2) + acos\alpha/2 sin\alpha/2}{acos\beta/2 sin\beta/2 + acos^2(\alpha/2)}$$

It can be seen that the equality is true for $$\alpha=\beta$$

• For more rigorous solution, we must form the polynomial of powers of $$a,b$$ and equate all coefficients to zero.

Corollary B1: $$P,C,C'$$ will be collinear if $$\frac{2a-m}{m.a^2} = \frac{2b-n}{n.b^2}$$

Proof: $$PQ = 2b.cos\angle B = 2a.cos\angle A$$

$$m = cos\beta/2.sec(\beta/2 - \angle B). PQ$$

$$tan\beta/2 = \frac{2b}{m} - cot\angle B$$

$$tan\beta/2 = \frac{b.(2a-m)}{m.a}$$

Similarly, $$tan\alpha/2 = \frac{a.(2b-m)}{n.b}$$

As $$\alpha = \beta$$ from Theorem B, Right Hand Side of both equations are equal.

Equating Left Hand Sides of both equations,

$$\frac{2a-m}{m.a^2} = \frac{2b-n}{n.b^2}$$