Reflection in a Triangle Tangent to Parabola
Experimental Mathematics

What Might This Be About?

10 July 2014, Created with GeoGebra


Let the three sidelines $BC,CA,AB$ of $\Delta ABC$ be tangent to a parabola at $A_1,B_1,C_1,$ respectively. Let $A'$ be the reflection of $A$ in the midpoint $M$ of $BC.$

parabola tangent to triangle - problem

Prove the three points $B_1,C_1,A'$ are collinear.


In a Cartesian coordinate system, let the parabola have equation:


The points of tangency are then expressed as $A_1=(a,pa^2),$ $B_1=(b,pb^2),$ $C_1(c,pc^2),$ for some real $a,b,c.$ From here we obtain the equations of the sideline of $\Delta ABC$ as the tangents at points $A_1,B_1,C_1$:

$\begin{align} l_{BC}:\,&y=2pax-pa^2\\ l_{CA}:\,&y=2pbx-pb^2\\ l_{AB}:\,&y=2pcx-pc^2 \end{align}$

Solving these in pairs we obtain the Cartesian coordinates of $A,B,C:$ $\displaystyle A=(\frac{b+c}{2},pbc),$ $\displaystyle B=(\frac{a+c}{2},pac)$, $\displaystyle C=(\frac{a+b}{2},pab).$

Now, me can find first $\displaystyle M=M_a(\frac{a}{2}+\frac{b+c}{4}, \frac{pac+pab}{2})$ and then $A'=(a, pab+pac-pbc).$


$\begin{align} B_{1}A'=(a-b,pb^2-pab-pac+pbc)=p(a-b)(1,-b-c),\\ C_{1}A'=(a-c,pc^2-pab-pac+pbc)=p(a-c)(1,-b-c), \end{align}$

which are obviously collinear.

Extra Problem

Let $B_2$ be the reflection of $B_1$ in $C;$ $C_2$ be the reflection of $C_1$ in $B.$ Then the four points $A',$ $A_1,$ $B_2,$ and $C_2$ are collinear.


The theorem, its proof and the extra problem have been posted by Dao Thanh Oai (Vietnam) at the CutTheNotMath facebook page. Observe that the theorem is a clear generalization of a lemma by Archimedes, which is the above statement when $M=A_1.$

Conic Sections > Parabola

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