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The inscribed circle in $\triangle ABC$ touches $AB, BC, A$C respectively at $M, N, P$. How can I calculate the angles of $\triangle ABC$ if $\angle PNM:\angle PMN : \angle MPN = \angle BAC : \angle ACB : \angle ABC$.

Please only provide me with a hint. Thank you for understanding!

Inscribed circle in triangle

John Wayland Bales
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Math Student
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  • Would that possibly make AMNP, BNPM and CPMN parallelograms? – John Wayland Bales Apr 20 '19 at 19:51
  • Note that the triangles (with points $M,N,P$) are isosceles (why?) and you know one angle for each of them. – dnqxt Apr 20 '19 at 19:51
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    How is the proportionality defined? As the first triple being a multiple of the other, i.e. the existence of $c$, such that $c\angle PNM=\angle BAC$, $c\angle PMN=\angle ACB$ and $c\angle MPN=\angle ABC$? – user647486 Apr 20 '19 at 19:53
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    If the symbol has the interpretation above, then you have that $\angle PNM=\angle BAC$, $\angle PMN=\angle ACB$, and $\angle MPN=\angle ABC$. This is because each triple adds up to $\pi$ and since they are also proportional, then are term-wise equal.Now $\angle BAC+2\angle PNM=\pi$. From this and the previous observation it follows that $\angle BAC=\pi/3$. Likewise for all other angles. – user647486 Apr 20 '19 at 20:04
  • @user647486, I don't understand why these triangles are isosceles (why $\angle PNM = \angle BAC, \angle PMN = \angle ACB, \angle MPN = \angle ABC$) Sorry for my questions but I'm not very good at English and I don't fully understand what you've written. – Math Student Apr 20 '19 at 20:41
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    I don't really know what the notation $x_1:y_1:z_1=x_2:y_2:z_2$ means. If it means that there is a $c$ such that $cx_1=x_2$, $cy_1=y_2$ and $cz_1=z_2$, then note that in your case $x_1+y_1+z_1=\pi$ and $x_2+y_2+z_2=\pi$, because they are the interior angles of a triangle. Equating those equations and using the ones with the $c$ you get that $c\pi=\pi$. This is, $c=1$. – user647486 Apr 20 '19 at 20:45
  • @user647486, I think that this means there is $c$, such that $c\angle PNM = \angle BAC, c \angle PMN = \angle ACB, c\angle MPN = \angle ABC$. – Math Student Apr 20 '19 at 20:49
  • The answer is $60^\circ$ as you have written in your previous comment. Can you try to describe me clearer why the triangles are isosceles? – Math Student Apr 20 '19 at 20:51
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    From the proof that $c=1$ you get those equalities between each angle of the circumscribed triangle and the opposite angle of the inscribed triangle. But those angles also satisfy that $\angle BAC+2\angle PNM=\pi$. Note that $2\angle PNM=\angle POM$, where $O$ is the center of the circle. Also $\angle BAC+\angle POM=\pi$. You can see this by joining $AO$ and drawing $OM$, which will be perpendicular to $AM$. – user647486 Apr 20 '19 at 20:57
  • Let us continue this discussion in chat. – Math Student Apr 20 '19 at 21:03
  • @user647486, I got it. Thank you! – Math Student Apr 21 '19 at 08:21

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Hint: This condition means that triangles $\triangle ABC$ and $\triangle MNP$ are similar, and corresponding angles are equal.

Adam Latosiński
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