3
$\begingroup$

Construct $ABC$ with straightedge and compass, given $G,I,Q_a$.

$G - $ the intersection point of medians;

$I -$ the center inscribed circle;

$Q_a -$ point of tangent inscribed circle to the side of $BC$.

I think $IQ_a \perp BC$, but I don't know the properties that combines these three points.

$\endgroup$
1
$\begingroup$

An interesting approach may be the following one: you may construct the Nagel point $N$ as $3G-2I$. You have the $BC$-line, i.e. the perpendicular to $IQ_a$ through $Q_a$.
You may assume that some $P_a\in BC$ is the contact point of the $A$-excircle, then:

  1. $A$ lies on $NP_a$;
  2. The midpoint of $M_a$ of $P_a Q_a$ is also the midpoint of $BC$;
  3. $A$ lies on $M_a G$ (so $A$ is fixed by $P_a$ through $A=NP_a\cap M_a G$);
  4. Assume that the perpendicular to $BC$ through $P_a$ meets the $AI$-line at $J_a$: if the distance of $J_a$ from $BC$ is the same as the distance of $J_a$ from $AB$ then $J_a$ is the $A$-excenter and the problem is solved.

enter image description here

Now $J_a$ lies on a hyperbola, while the external angle bisectors of the candidate $B,C$ vertices meet on a line, so the problem boils down to intersecting a line and a hyperbola, finding the actual $A$-excenter, then the contact point of the $A$-excircle, then the midpoint of $BC$, then $A$ as $NP_a\cap M_a G$, then $B$ and $C$ by drawing the tangents from $A$ to the incircle.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I do not understand how to find $J_A$. Why $J_{a}$ lies on a hyperbola? $\endgroup$ – Roman83 Jun 25 '16 at 17:16
  • $\begingroup$ @Roman83: because the map sending $J_a$ to $P_a$ is a projectivity. If you choose a coordinate system, that is straightforward to check. So the idea is to pick five differents $P_a$, reconstruct through them the foci of the hyperbola, then intersect the hyperbola with the appropriate line. $\endgroup$ – Jack D'Aurizio Jun 25 '16 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.