another good

[ins-]

It is given a circle k(O, R) and a segment AB outside this circle. Construct with compass and ruller all possible points M that are from k and the line OM is an angle bisector of angle AMB. (I don't know if it may be solved with compas and ruller but if it is not - find the locus of M).

[незваный гость]

Can you clarify it a little?

ins- писал(а):

Construct ‹…› all possible points M that are from k …

Do you mean $M \in k(O,R)$?

Also, do you mean circle (not disk)?

[ins-]

Yes to all questions.

[незваный гость]

I do not think it has geometric (compass and ruler) solution.

Without loss of generality we can consider the segment $[AB]$ to have coordinates $(0,0)$ and $(1, 0)$ correspondingly on the Cartesian plane. Let $O$ be $(x, y)$ and $M$ be $O + R (\cos \varphi, \sin \varphi)$. The condition that $OM$ is a bisector translates to (weaker) condition that distances from $O$ to $AM$ and $BM$ are equal. The equations for lines are easy to calculate, and, therefore, distances. It is also convenient to introduce $t = \tg(\varphi/2)$. Then we have a polynomial eqation for $t$. It has couple of good roots ($\frac{-R-\sqrt{R^2-y^2}}{y}$, $\frac{-R+\sqrt{R^2-y^2}}{y}$). The rest of the equation — generic polynomial of 4th degree, so the roots should not be geometric. The “good” roots correspond to $M=(x \pm \sqrt{R^2-y^2}, 0)$. These points (if exist, which mean if $|y| \leq R$) are intersection of the circle with the axis X. While distances from $O$ to $AM$ and $BM$ are equal, they are on the same side of $OM$, so the do not belong to solution.


Generally speaking, $M$ belongs to intersection of the circle with a curve of third degree.

Добавлено спустя 23 минуты 30 секунд:

Reflecting further, it is possible to describe the locus as subset of the intersection of the circle and a hyperbola (to which $O$ belongs). One of the hyperbola’s axis is parallel to the bisector ${\scriptstyle\angle} AOB$, another one is orthogonal to the bisector. The locus consists of no more than four points and contains at least two points.

The last fact can be proven by a very simple way: lets consider an intersection of circle with lines $OT$, where $T \in [AB]$. It consists of 2 sectors, e.g. . $P_1Q_1$ and $P_2Q_2$. Lets consider one of them, and let $P_1 \in OA$. Then for $P_1$ $\widehat{AP_1O} < \widehat{BP_1O}$. Similar $\widehat{AQ_1O} > \widehat{BQ_1O}$. As angles change continuously, $\exists M: \widehat{AMO} = \widehat{BMO}$. The similar reasoning is applicable to $P_2Q_2$.

[ins-]

It seems to be very good solution. You are right most probably. I know that it has maximum 4 solutions by using trygonometry. And from mathlinks I know it is not constructable with compas and ruler. I cannot confirm your solution because I don't know all terms used "Cartesian plane" for example. I guess something about intersection point of a hyperbola and a circle. If it is possible - try to solve the problem, using more elementary technics. I'm wondering also how to find the contitions - when there are 0, 1, 2, 3, 4 solutions. Are there any different opinions?

[незваный гость]

ins- писал(а):

I'm wondering also how to find the contitions - when there are 0, 1, 2, 3, 4 solutions. Are there any different opinions?

As I have shown at the end of the previous message, there is at least 2 points in the locus. Quite an elementary reasoning, BTW.

Insofar I was not able to construct an example with 4 points. On the second thought, I think I can prove that the locus never consists of more than 2 points. Again, the proof is fairly technical and based on an application of the Descartes' rule of signs to the equation of intersection of the hyperbola and the circle. But, because of the complexity of calculations there is a certain probability of errors.

I am ready to put this calcs online if there is anybody willing to verify/validate it.

[ins-]

If it is not a problem for you, please show your calculation. I've a friend finnished mathematical education he may try to check them and I believe - people from this forum also may check them - there are lots of excellent matematicians here.
Did you took a look at two cases - M is on different sides of O?

[незваный гость]

Here we are.

Let $R = 1$, axis $OX$ be bisector of the ${\scriptstyle \angle} AOB$, $1<|OA| = a \leq b = |OB|$, $\widehat{AOB} = 2 \varphi$. For a point $T$ we will designate its coordinates as $(T_x,T_y)$. Then $A_x = a \cos \varphi$, $A_y = a \sin \varphi$, $B_x = b \cos \varphi$, $B_y = -b \sin \varphi$. We will make an exception for $M$, having its coordinates to be $(x, y)$ for brevity.

For line with equation $y = \lambda x + \mu$ the square of the distance from $O(0,0)$ to it is $\frac{\mu^2}{1+\lambda^2}$. For $OM$ to be bisector of the ${\scriptstyle\angle} AMB$ we need distances from $O$ to $AM$ and to $BM$ to be equal and to have $A$ and $B$ to be on different sides of $OM$. The last condition boils down to $M$ belongs to arcs $P_1Q_1$ and $P_2Q_2$.

Having coordinates of points $A$, $B$, and $M$, the calculation of distance is easy. The condition of equedistance becomes $(-y A_x + $ $x A_y + $ $y B_x - $ $A_y B_x - $ $x B_y + $ $A_x B_y) (-x^2 y A_x - $ $y^3 A_x + $ $x^3 A_y + $ $x y^2 A_y - $ $x^2 y B_x - $ $y^3 B_x + $ $2 x y A_x B_x - $ $x^2 A_y B_x + $ $y^2 A_y B_x + $ $x^3 B_y + $ $x y^2 B_y - $ $x^2 A_x B_y + $ $y^2 A_x B_y - $ $2 x y A_y B_y) = 0$. The first factor is equation of line $AB$ and is of no interest for us. The second factor is more interesting. Using polar form of coordinates for $A$ and $B$ we get much more manageable expression: $2 a b x y - $ $ (a + b) y (x^2 + y^2) \cos\varphi - $ $(b - a) (x^2 + y^2) x \sin\varphi = 0$. Now we note that $M$ belongs to unit circle and therefore $x^2+y^2 = 1$. So finally $2 a b x y - $ $ (a + b) y \cos\varphi - $ $(b - a) x \sin\varphi = 0$. Or, $(x-\frac{a+b}{2 a b}\cos\varphi)(y-\frac{b-a}{2 a b}\sin\varphi)=\frac{b^2-a^2}{4 a^2 b^2} \sin\varphi\cos\varphi$ — an equation of a hyperbola.

Of course we need to consider case $a = b$ when the hyperbola reduces to pair of lines $y=0$ and $x = \frac{\cos\varphi}{ a}$. The horizontal line intersects arcs in points $(\pm 1,0)$, the vertical does not intersect arcs. So in this case we have exactly two points in the locus.

In following we will assume $1 < a < b$ — all inequalities are strict.

Some properties of this hyperbola:
1) Its axises are parallel to the coordinate axises.

2) The intersection (point $H$) of axises is always inside rectangle $P_1Q_1P_2Q_2$. As we assumed $a < b$, both $H_x > 0$ and $H_y > 0$.

3) Point $O$ always belongs to the hyperbola.

4) Immediate consequence is that the hyperbola intersects the arc $P_2Q_2$ in exactly one point.

5) Each intersection of the hyperbola and circle has unique $x$ coordinates (the function is monotonic in each segment $(-\infty, H_x)$ and $(H_x, +\infty)$).

Therefore, to find number of points in the locus we need to find how many intersections we have with the arc $P_1Q_1$, id est how many solutions of the system:
$\left\{\begin{array}{l}x^2+y^2=1 \\ (x-\frac{a+b}{2 a b}\cos\varphi)(y-\frac{b-a}{2 a b}\sin\varphi)=\frac{b^2-a^2}{4 a^2 b^2} \sin\varphi\cos\varphi \\ x > \cos\varphi \end{array} \right.$

First, we will solve second equation for $y$: $y = \frac{(b-a)x \sin\varphi}{2 a b x - (a+b)\cos\varphi}$. Next, we will substitute $y$ in the first equation with this expression and simplify result: $4 a^2 b^2 x^4 - $ $4 a b (a + b) \cos\varphi x^3+$ $(a^2 + b^2 - 4 a^2 b^2 + 2 a b \cos^2\varphi - 2 a b \sin^2 \varphi) x^2 +$ $4 a b (a + b) \cos\varphi x -$ $(a + b)^2 \cos^2\varphi = 0$. Thirdly, we will replace $x$ with $z + \cos \varphi$, thus making the last inequality equivalent to $z > 0$: $4a^2b^2 z^4 + $ $ 4a b (4a b - a - b) \cos\varphi z^3 + $ $((a^2+b^2-6 a^2b-6 a b^2+8 a^2b^2)+2ab(1-3a -3b+6ab)\cos2\varphi) z^2 + $ $2\cos\varphi((a^2+b^2-ab(a+b))+(2ab-3a^2b-3ab^2+4a^2b^2)\cos2\varphi )z - $ $a b(a-1)(b-1)\sin^22\varphi = 0$.

We are ready to apply the Descartes' rule of signs to this equation. The number of positive roots is equal to number of sign changes in the list of coefficients, or less than that by even number. Marking positive coefficients with + and negative with -, we have ++??-. The only way we can have more than 1 root is if ?? = -+, that is $((a^2+b^2-6 a^2b-6 a b^2+8 a^2b^2)+2ab(1-3a -3b+6ab)\cos2\varphi) < 0 $ and $2\cos\varphi((a^2+b^2-ab(a+b))+(2ab-3a^2b-3ab^2+4a^2b^2)\cos2\varphi ) > 0$. Because coefficients for $\cos2\varphi$ are positive, we have that $\cos2\varphi > \frac{a^2b + ab^2-a^2-b^2}{a b (2 - 3 a - 3 b + 4 a b)}$ and $\cos 2\varphi < \frac{6a^2b+6ab^2-a^2-b^2-8a^2b^2}{2ab(1-3a-3b+6ab)}$ simultaneously. Therefore $\frac{a^2b + ab^2-a^2-b^2}{a b (2 - 3 a - 3 b + 4 a b)} < \frac{6a^2b+6ab^2-a^2-b^2-8a^2b^2}{2ab(1-3a-3b+6ab)}$ which is never true for $1 < a < b$.

All together now! We have exactly one intersection on the arc $P_2Q_2$ and exactly one intersection on the arc $P_1Q_1$.

[незваный гость]

It looks promising to consider a point $S$ — intersection of the hyperbola with segment $P_1Q_1$. Coordinates of $S$ are $(\cos\varphi, \frac{b-a}{2 a b - b - a}}\sin\varphi)$, so it is always on the side of the rectangle.

Unfortunately, the analysis must be more subtle than for the left branch. Their the hyperbola is declinig and the circle is rising. Here both are declining.

But it is really tempting to throw away most of the calculations.