2
$\begingroup$

A triangle $ABC$, whose symmedians intersect at $L$, has an altitude $BT$ drawn.

Given that $\widehat{TLA} = 2\cdot\widehat{TBA}$, I have to prove that $\widehat{TLC} = 2\cdot\widehat{TBC}$.

Any hints are welcome.

1 Answers 1

1

Let we rename $T$ as $H_B$. The midpoint $M_C$ of $AB$ is the center of the circumcircle of $H_B AB$, in particular $2\cdot\widehat{H_B BA}=\widehat{AM_C H_B}$ and the given relation implies that the symmedian point $L$ lies on the circumcircle of $AM_C H_B$. It is not difficult to prove$^{(*)}$ that in general the symmedian through $B$ always goes through the intersection of the circumcircles of $AH_B M_C$ and $CH_B M_A$. It follows that the given hyphotesis imply that $L$ lies on the circumcircle of $CH_B M_A$ too, so $\widehat{H_B L C}=2\cdot \widehat{H_B B C}$ is granted. enter image description here

(*) Assume that $J$ is the depicted (but unnamed) intersection of the circumcircles of $AH_B M_C$ and $CH_B M_A$. $J$ "sees" $M_C H_B$ under an angle equal to $\pi-\widehat{A}$ and $M_A H_B$ under an angle equal to $\pi-\widehat{C}$, hence $B M_C J M_A$ is a cyclic quadrilateral and $$ \frac{d(J,BA)}{d(J,BC)}=\frac{JM_C}{JM_A}=\frac{M_C H_B}{M_A H_B} =\frac{AM_C}{CM_A}=\frac{AB}{CB} $$ by the sine theorem applied to $JH_B M_C$ and $JH_B M_A$. The last identity proved that $J$ lies on the symmedian through $B$.

  • 0
    I am trying to prove your point about the symmedian going through the intersection of the two circles, but I am struggling - angle chase? – 2017-02-08
  • 0
    @Fermat: I applied a circular inversion with respect to a circle centered at $B$, but probably that can be achieved by pure angle chasing, too. – 2017-02-08
  • 0
    Hmm, not too familiar with inversion. Does the symmedian have a particular angle property which I can prove? My tactic is to let the intersection be X, and prove BX follows some law followed by the symmedian. – 2017-02-08
  • 0
    @Fermat: well, with a bit of trigonometry you may prove that, if $P$ is the intersection point of such circles, the distances of $P$ from the $BC$ and $BA$ sides are proportional to $a,c$. That proves $BP$ is a symmedian. – 2017-02-08
  • 0
    I have inverted the two circles about B, but what does that give me? – 2017-02-08
  • 0
    @Fermat: a nice thing, since $H_A,H_C,M_A,M_C$ all lie on the same circle (the nine point circle) hence by choosing a proper inversion radius they got exchanged in couples. – 2017-02-08
  • 0
    @Fermat: I found a proof avoiding inversion (and appended it to my previous answer). It is a bit lengthy but quite simple. – 2017-02-08
  • 0
    I think you may have put $M_B$ instead of $M_C$ and got that fraction the wrong way round...? – 2017-02-08
  • 1
    @Fermat: I made some typos, now fixed. – 2017-02-08
  • 0
    I get the whole proof except for how you apply the sine rule - could you please include a little bit of extra detail involving the angles you consider and why they are in fact in the same ratio? – 2017-02-09
  • 0
    @Fermat: $M_C H_B$ just depends on the circumradius of $AH_B M_C$ (that is known) and the angle at $A$ (that is known as well). So to compute the ratio $\frac{JM_C}{JM_A}$ boils down to computing the ratio $\frac{\sin\widehat{JAC}}{\sin\widehat{JCA}}$. Now you may apply the sine theorem to the triangles $JAH_B$ and $JCH_B$ that have the $JH_B$ side in common and a known ratio $\frac{H_B A}{H_B C}$. – 2017-02-09
  • 0
    Isn't this quite different to what you have in your answer? (Sorry for being pedantic/dim, I just am not quite seeing it) – 2017-02-09
  • 0
    @Fermat: just another way to skin the same cat :D There are so many cyclic quadrilaterals (and two circles tangent to the $M_A M_C$ line) that you may pick a couple of random triangles, and have a positive probability of finishing the proof by applying the sine theorem to them :D What is the source of the problem, by the way? I find it really interesting. – 2017-02-09
  • 1
    Ah thank you - I don't know what the source is - I was researching symmedians for a class presentation and bumped into this problem which I couldn't solve, but I have lost the webpage - sorry about that. If I find it again I will comment. By the way, I get your proof now, thank you. – 2017-02-09