At first, there are a fixed conic curve $\Gamma$ and three fixed collinear points $A,B,C$. Draw a moving straight line $l$ through point C, intersecting $\Gamma$ at two points $P,Q$. Find the two intersections $S=AP\cap BQ$ and $T=AQ\cap BP$. As line $l$ move, $S$ and $T$ generate a locus $\Delta$. Due to symmetry, $S$ and $T$ must have a common locus.
I found that $\Delta$ will be a new conic curve. So the question is: why this is true?
Use knowledge harmonic point sequences, the intersection of $ST$ and $AB$ is a fixed point $D$ that form a harmonic point sequences $[A,B;C,D]$ together with points $A,B,C$.
Suppose that $C$ is outside of $\Gamma$. Make two tangent lines $CM,CN$ from $C$ to $\Gamma$. Then the line $MN$ is the polar of $C$ in $\Gamma$. Let $E$ be the intersection of $MN$ and $l$. Through the image we can guess these two are right:
- $E$ is also the intersection of $MN$ and $ST$.
- $DM,DN$ also tangent to $\Delta$.
The proof of the first guess is not too hard. Let $E'$ be the intersection of $ST$ and $l$, then all that need to do is to prove $E=E'$. According to the property of pole $C$ and polar $MN$, $[C,E;P,Q]$ is a harmonic point sequences; according to the property of complete quadrangle $[A,B;P,Q;S,T]$, $[C,E';P,Q]$ is also a harmonic point sequences; these two facts indecate $E=E'$.
The second guess suggest that $[C,\Gamma,P,Q]$ and $[D,\Delta,S,T]$ may be symmetric. This inspired me to turn to find that what's the relation between $\Gamma$ and $C,M,N$. Or, if three points $C,M,N$ have been given, what conditions should a point $P$ meets in order to the locus of $P$ is a conic curve tangenting $CM,CN$ at $M,N$. If this is finded, we can transfer the condition from $P$ or $Q$ to $S$ or $T$, then $S$ or $T$ must meet the same condition and the locus of $S,T$ must be a conic curve tangenting $DM,DN$ at $M,N$.
The ratio $\dfrac{\overline{ME}}{\overline{EN}}$ uniquely determines the position of $E$ on $MN$; the position of $E$ determine the line $l$; and the other ratio $\dfrac{\overline{CP}}{\overline{PE}}$ also uniquely determines the position of $P$ on $l$. So these two ratio must have a relation when $P$ moving on a fixed conic curve $\Gamma$. By attempting on some special curve, I find this theorem:
Theorem: Let $E$ be a moving point on $MN$, and $P$ a point on $CE$, then the locus of $P$ is a conic curve if and only if this equation holds:
$$ \left(\frac{\overline{CP}}{\overline{PE}}\right)^2\frac{\overline{ME}\overline{EN}}{\overline{MN}^2}=\text{Const}$$
If the Theorem is true, by Menelaus's Theorem on triangle $\triangle CED$ and transversal line $PSA$ we know
$\dfrac{\overline{CP}}{\overline{PE}}\times
\dfrac{\overline{ES}}{\overline{SD}}\times
\dfrac{\overline{DA}}{\overline{AC}}=-1$.
So there is $\dfrac{\overline{DS}}{\overline{SE}}=
-\dfrac{\overline{DA}}{\overline{AC}}
\times\dfrac{\overline{CP}}{\overline{PE}}$
in which $-\dfrac{\overline{DA}}{\overline{AC}}$ is contant.Then
$$
\left(\frac{\overline{DS}}{\overline{SE}}\right)^2
\frac{\overline{ME}\overline{EN}}{\overline{MN}^2}=
\left(\frac{\overline{DA}}{\overline{AC}}\right)^2
\left(\frac{\overline{CP}}{\overline{PE}}\right)^2
\frac{\overline{ME}\overline{EN}}{\overline{MN}^2}
=\text{Const}$$
and give what I want.
Now I have these questions:
- How to prove the above theorem?
- How to explain the above theorem?
- How to prove the same problem when $C$ is inside of $\Gamma$?, excluding use the space $\mathbb{CP}^2$. Is there another way to prove the problem
- Do I find this fact that collinear $A,B,C$ transform $\Gamma$ to a new conic curve $\Delta$ first? Is it been found by someone before?
Best Answer
I get a straight line segment rather than a conic (in a special case). We can place the line $ABC$ at infinity, and assume that $C$ is the horizontal direction, whereas A and B are the directions of $e^{\pi i/3}$ and $e^{2\pi i/3}$. The intersection points P and Q are symmetric with respect to the $y$-axis, and therefore the intersection point $AP\cap BQ$ will always lie on the $y$-axis. This is of course a degenerate case.
In general, one can argue as follows. Use a projective transformation to send the line $ABC$ to the line at infinity. Then use an affine transformation to send the ellipse to the unit circle. Furthermore, by a rotation we can assume that $C$ corresponds to the pencil of horizontal lines (including the $x$-axis in the plane). Then the points $A$ and $B$ correspond to pencils of parallel lines of slopes $m_1, m_2\not=0$.
A horizontal line cuts the unit circle in (at most) two points $(s,\sqrt{1-s^2})$ and $(-s,\sqrt{1-s^2})$. We need to show that the intersection point $(x,y)$ of a pair of lines of slopes $m_1$ and $m_2$ passing through such a pair of points satisfies a quadratic equation. Then by definition it will be a conic section (at least in the nondegenerate case).
The equations of the two lines are $y=\sqrt{1-s^2} + m_1 (x+s)$ and $y=\sqrt{1-s^2} + m_2(x-s)$. It follows that $x=\frac{m_1+m_2}{m_2-m_1} s$ or $s=mx$ where $m=\frac{m_2-m_1}{m_1+m_2}$. From the equation of the first line we obtain $\big(y-m_1(x+mx)\big)^2= 1- (mx)^2$. This is a quadratic equation in $x$ and $y$, as required.
This argument works in the case when the conic and the line $ABC$ have no common points (either finite or infinite).
Your argument is longer but has the advantage of being almost synthetic ("almost" because you also use Menelaus theorem).
If the conic $K$ and the line $ABC$ have a single intersection point, one can apply a similar argument. We move the line $ABC$ to the line at infinity, and then move $C$ to the pencil of horizontal lines and the point $ABC \cap K$ to the pencil of vertical lines. We can then apply a translation to make sure that the axis of symmetry of the parabola $K$ is the $y$-axis and the apex of the parabola is at the origin. Afterwards apply a scaling to make sure the parabola is the standard parabola $y=x^2$. A horizontal line cuts the parabola at $(\pm s, s^2)$. The equations of the two lines of slopes $m_1$ and $m_2$ are then $y=s^2+m_1(x+s)$ and $y=s^2+m_2(x-s)$. We obtain a similar relation $s=mx$ (for the same $m$ as above). In this case also we obtain a quadratic relation between $x$ and $y$, of the form $y=(mx)^2 + m(x+mx)$, as required.