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$ABC$ is a triangle and $A_1, B_1, C_1$ are points on $BC, CA, AB$ such that $$\frac{BA_1}{A_1C}=\frac{CB_1}{B_1A}=\frac{AC_1}{C_1B}=\lambda$$

If $A_2, B_2, C_2$ are points on $B_1C_1, C_1A_1$, and $A_1B_1$ such that $$\frac{B_1A_2}{A_2C_1}=\frac{C_1B_2}{B_2A_1}=\frac{A_1C_2}{C_2B_1}=\frac{1}{\lambda}$$

prove that $\triangle ABC$ is similar to $\triangle A_2B_2C_2$ and find the ratio of simlitude.

I'm missing something obvious, but I don't know what.

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  • $\begingroup$ have you made an Image? $\endgroup$ Commented Feb 5, 2018 at 17:50
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    $\begingroup$ hint: prove that $A_2B_2 // AB$, then the claim is immediate. $\endgroup$
    – dezdichado
    Commented Feb 5, 2018 at 17:58
  • $\begingroup$ @Dr.SonnhardGraubner no I haven't. Apologies for the tedious work :) $\endgroup$
    – QFTheorist
    Commented Feb 5, 2018 at 18:03
  • $\begingroup$ By using barycentric coordinates, the problem is straightforward. $\endgroup$ Commented Feb 5, 2018 at 19:22
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    $\begingroup$ @U.Basumatary: from the best for the best - web.evanchen.cc/handouts/BMC_Bary/BMC_Bary.pdf $\endgroup$ Commented Feb 5, 2018 at 19:36

3 Answers 3

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$$\vec{A_2B_2}=\vec{A_2C_1}+\vec{C_1B_2}=\frac{1}{1+\frac{1}{\lambda}}\vec{B_1C_1}+\frac{\frac{1}{\lambda}}{1+\frac{1}{\lambda}}\vec{C_1A_2}=$$ $$=\frac{\lambda}{1+\lambda}\vec{B_1C_1}+\frac{1}{1+\lambda}\vec{C_1A_2}=\frac{\lambda}{1+\lambda}\left(\vec{B_1A}+\vec{AC_1}\right)+\frac{1}{1+\lambda}\left(\vec{C_1B}+\vec{BA_1}\right)=$$ $$=\frac{\lambda}{1+\lambda}\left(\frac{1}{1+\lambda}\vec{CA}+\frac{\lambda}{1+\lambda}\vec{AB}\right)+\frac{1}{1+\lambda}\left(\frac{1}{1+\lambda}\vec{AB}+\frac{\lambda}{1+\lambda}\vec{BC}\right)=$$ $$=\frac{\lambda}{(1+\lambda)^2}\left(\vec{BC}+\vec{CA}\right)+\frac{1+\lambda^2}{(1+\lambda)^2}\vec{AB}=\frac{1-\lambda+\lambda^2}{(1+\lambda)^2}\vec{AB},$$ which says $A_2B_2||AB.$

By the same way we can prove that $A_2C_2||AC$ and $B_2C_2||BC,$

which says that $$\Delta ABC\sim\Delta A_2B_2C_2$$ and $$\frac{A_2B_2}{AB}=\frac{1-\lambda+\lambda^2}{(1+\lambda)^2}.$$

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A more thorough hint:

Use the theorem of sines on triangles $BB_2C_1$ and $BB_2A_1$, to conclude that: $$\dfrac{\sin\angle B_2BA}{\sin\angle B_2BC} = \dfrac{A_1C}{BC_1}.$$

Then the sine version of Ceva's theorem will let you conclude that $AA_2, BB_2, CC_2$ are concurrent. This will greatly help you prove the $A_2B_2//AB$, for example.

Let me know if you still have trouble after this.

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We have $B_1 = \frac{1}{1+\lambda} A +\frac{\lambda}{1+\lambda} C$ and cyclic identities.
We have $B_2 = \frac{\lambda}{1+\lambda} A_1 +\frac{1}{1+\lambda} C_1$ and cyclic identities.
By combining them $$ B_2 = \frac{\lambda}{1+\lambda}\left(\frac{1}{1+\lambda}C+\frac{\lambda}{1+\lambda}B\right)+\frac{1}{1+\lambda}\left(\frac{1}{1+\lambda}B+\frac{\lambda}{1+\lambda}A\right)$$ $$ B_2 = \frac{\lambda}{(1+\lambda)^2}(A+B+C)+\color{red}{\frac{\lambda^2-\lambda+1}{(\lambda+1)^2}} B $$ where the red term is the wanted ratio of similitude.

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