Prove the following:

The intersection points of any three tangents to a parabola given by the formula $y(y-y_0)=2p(x-x_0)$ are vertices of a triangle whose orthocenter belongs to the directrix of the parabola and the circumcircle of the triangle passes through the focus of the parabola.

My attempt:

Orthocenter problem part- edited (old notes deleted to be less chaotic and I believe there is still some space for improvement of the first part):

In the meantime, I realized it would be better to just use the condition of tangency and plug some of its parameters into the formula for a line in the $xy$ plane.

Let $y=k_ix + l_i,\ i=1,2,3$ be a tangent line to a parabola. Then $$p=2k_il_i\implies l_i=\frac{\frac{p}2}{k_i}$$

Now, our equation becomes: $$\boxed{y=k_ix+\frac{\frac{p}2}{k_i}}$$ We can substitute $\frac{p}2$ by $\alpha$, so $$\boxed{y=k_ix+\frac{\alpha}{k_i}}$$ This way, our computations are getting easier. The equations of the three arbitrary tangents to a parabola are: $$\begin{aligned}y&=k_1x+\frac{\alpha}{k_1}\\y&=k_2x+\frac{\alpha}{k_2}\\y&=k_3x+\frac{\alpha}{k_3}\end{aligned}$$ The intersection point of the $i-$th and $j-$ th tangent line: $$k_ix+\frac{\alpha}{k_i}=k_jx+\frac{\alpha}{k_j}\implies x=\alpha\frac{\frac1{k_j}-\frac1{k_i}}{k_i-k_j}=\frac{\alpha}{k_ik_j}$$ $$y=k_i\cdot\frac{\alpha}{k_ik_j}+\frac{\alpha}{k_i}=\alpha\left(\frac1{k_i}+\frac1{k_j}\right)$$ $$\boxed{S_{ij}=\left(\frac{\alpha}{k_ik_j},\alpha\frac{k_i+k_j}{k_ik_j}\right)}$$

Now, we have to find the line perpendicular to the $k-$th tangent line passing through the point $S_{ij}$. $k_\perp=-\frac1{k_k}$ E.g., one altitude of the formed triangle belongs to the line: $$\begin{aligned}y-y_{S_{1,2}}&=k_\perp(x-x_{S_{1,2}})\iff &y&=-\frac1{k_3}x+\frac{\alpha}{k_1k_2k_3}+\alpha\left(\frac1{k_1}+\frac1{k_2}\right)\\&&y&=-\frac1{k_3}x+\alpha\left(\frac1{k_1}+\frac1{k_2}+\frac1{k_1k_2k_3}\right)\end{aligned}$$

So, the $x$ coordinate of the intersection of the three tangent lines: $$\begin{aligned}y&=-\frac1{k_3}x+\alpha\left(\frac1{k_1}+\frac1{k_2}+\frac1{k_1k_2k_3}\right)\\y&=-\frac1{k_2}x+\alpha\left(\frac1{k_1}+\frac1{k_3}+\frac1{k_1k_2k_3}\right)\\y&=-\frac1{k_1}x+\alpha\left(\frac1{k_2}+\frac1{k_3}+\frac1{k_1k_2k_3}\right)\end{aligned}$$ $$\begin{aligned}-\frac1{k_3}x+\alpha\left(\frac1{k_1}+\frac1{k_2}+\frac1{k_1k_2k_3}\right)&=-\frac1{k_2}x+\alpha\left(\frac1{k_1}+\frac1{k_3}+\frac1{k_1k_2k_3}\right)\\\left(\frac1{k_2}-\frac1{k_3}\right)x&=\left(\frac1{k_3}-\frac1{k_2}\right)\alpha\\x&=-\alpha=-\frac{p}2\end{aligned}$$

End of the first part. The rest remains the same not to be off-topic.

Since the points $P_1,P_2, P_3$ are close to the directrix, $\triangle ABC$ in my picture is obtuse and its orthocenter is outside the triangle, but it doesn't have to be so at all.

Let $A,B,C$ be the intersection points of the tangents. $P_2\in\overline{AC}$ due to $P_1\preceq P_2\preceq P_3$.

Let $A'\in\overline{BC}$ s.t. $AA'\perp BC$, $B'\in\overline{AC}$ s.t. $BB'\perp AC$ and $C'\in\overline{AB}$ s.t. $CC'\perp AB$.

The center $S$ of the circumscribed circle $q$ of $\triangle ABC$ is the intersection point of the bisectors $s_1,s_2,s_3$ of the sides $\overline{AB},\overline{BC}$ and $\overline{AC}$ respectively. Furthermore, $\underline{\text{each side bisector is parallel to one of the sides of the triangle}}$, i.e., $$s_1\parallel\overline{AA'}\ \&\ s_2\parallel\overline{BB'}\ \&\ s_3\parallel\overline{CC'} $$

If the orthocenter $T$ is an orthogonal projection of the point $P_2$ onto the directrix $x=-\frac{p}2$ of the parabola, and if the circumscribed circle $q$ really passes through the focus $$\boxed{F\left(\frac{p}2,0\right)\ \text{or}\ F\left(\frac{p}2+x_0,y_0\right)}$$ then $|TP_2|=|P_2F|$.

Picture: enter image description here

zoomed: zoomed just in case According to the notation in the picture: $$\begin{aligned}\measuredangle AA'B&=\measuredangle A'CT=\measuredangle BFL\\\measuredangle B'TA&=\measuredangle ACA'=\measuredangle AFB\end{aligned}$$ I can see: $$\triangle AB'T\sim\triangle BB'C\sim\triangle A'AC\sim AFB'$$ In particular: $\boxed{\triangle AB'T\cong\triangle AFB'\implies\ |TB'|=|B'F|\implies\triangle TFP_2\ \text{is isosceles}\ \implies |TP_2|=|P_2F|\ }$

Also: we can prove the orthogonal projections of the focus onto the three tangents belong to the tangent passing through the vertex of the parabola, meaning those projections are collinear, which, by the Simson theorem, implies the focus belongs to the circumcircle of the triangle.

May I ask for advice on solving this task and improve the parts I might have done correctly to be concise as possible? Thank you in advance!

P. S. I found a related paper, but almost none of the information has been covered in our official literature.

Just in case, I found an answer by @JeanMarie recalling the fact the orthocenter of the observed triangle lies on the directrix of the parabola, but I couldn't think of proof.

Update on a special case:

I've also read the directrix is the set of all the points in the plane we can draw two mutually perpendicular tangents from (which can be proven via Vieta's formulae and the condition of tangency). In a right-triangle, the orthocenter is obviously in the vertex opposite to the hypothenuse.


1 Answer 1


The second request easily follows from this theorem (see here for a proof):

The exterior angle between any two tangents is equal to the angle which either segment of tangent subtends at the focus.

It follows that $\angle CFP_2=\angle BCA$ and $\angle AFP_2=\angle CAB$, whence: $$ \angle CFA=\angle BCA+\angle CAB=\pi-\angle ABC. $$ Focus $F$ lies then on circle $ABC$.

enter image description here

  • 3
    $\begingroup$ (+1) I discovered that this interesting result is called Lambert theorem $\endgroup$
    – Jean Marie
    Commented May 3, 2022 at 7:25
  • 1
    $\begingroup$ See as well here and here and $\endgroup$
    – Jean Marie
    Commented May 3, 2022 at 7:32

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