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Points $A_1$, $B_1$, $C_1$ divide sides $BC$, $CA$, $AB$ equilateral triangle $ABC$ in a ratio of $1: 2$. The line segments $AA_1$, $BB_1$, $CC_1$ determine the triangle $KLM$.

Is the triangle $KLM$ also an equilateral side?
In what relation are the area of ​​triangle $KLM$ the area of ​​triangle$ ABC$?

https://i.stack.imgur.com/QN00P.png

My attempt:

I can see that $KLM$ is an equilateral triangle. But, why the fraction $\frac{1}{7}$ is the area ratio of ​​the triangle $KLM$ to the ​​triangle $ABC$?

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    $\begingroup$ Look at the triangle $\triangle ABA_1$ and $\triangle AA_1C$. They share the same height from $A$ to $AB$ and their bases $BA_1$ and $A_1C$ are in the proportion $1:2$. Therefore, their areas are also in that proportion. Hence the area of $\triangle ABA_1$ is $1/3$ of the area of $\triangle ABC$. Use Menelaus' theorem to find the proportion $C_1L$ to $LC$ and apply the same idea to get the area of $\triangle AC_1L$. Then you have all areas in the problem. $\endgroup$ – user762847 Mar 25 '20 at 15:13
  • $\begingroup$ Any more hint about applyMenelaus theorem? $\endgroup$ – josf Mar 25 '20 at 15:32
  • $\begingroup$ height from A to AB?? $\endgroup$ – josf Mar 25 '20 at 15:39
  • $\begingroup$ From $A$ to $BC$. $\endgroup$ – user762847 Mar 25 '20 at 15:46
  • $\begingroup$ You can apply Menelaus, for example, to the triangle $\triangle C_1BC$ with the transversal line $AA_1$. Note that $\frac{AC_1}{AB}=\frac{AC_1}{AC_1+C_1B}=\frac{AC_1}{AC_1+2AC_1}=\frac{1}{3}$. You will need that quotient to apply it. When you do, you will get the ratio $\frac{CL}{LC_1}$. $\endgroup$ – user762847 Mar 25 '20 at 15:48
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Establish the area ratios

$$\frac{[ACL]}{[ACA_1]} = \frac{[ACC_1]}{[ACA_1C_1]} = \frac{\frac13[ABC]}{[ABC]-[BC_1A_1]}= \frac{\frac13[ABC]}{[ABC]-\frac13\frac23[ABC]} =\frac37 $$ which leads to $$[ACL] =\frac37[ACA_1]=\frac37\frac23[ABC]=\frac27[ABC] $$ Likewise, $[BCK]={[ABM]}=\frac27[ABC]$, Then $$\frac{[KLM]}{[ABC]} = \frac{[ABC]-[ACL]-[BCK]-[ABM]}{[ABC]} = 1-3\cdot\frac27=\frac17 $$ In general, $\frac{[KLM]}{[ABC]} =\frac{(r-1)^2}{r^2+r+1}$, where $1:r$ is the given ratio.

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Here is a proof without using Menelaus's thm. Let $A_2$, $B_2$, and $C_2$ be the midpoints of $CA_1$, $AB_1$, and $BC_1$, resp. Draw $MA_1$, $KA_2$, $KB_1$, $LB_2$, $LC_1$, $LC_2$, $MA_2$, $KB_2$, and $LC_2$. In this proof, $[\mathcal{P}]$ is the area of a polygon $\mathcal{P}$.

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Due to symmetry $\triangle KLM$ is equilateral. Therefore $\angle C_1LA=\angle KLM=60^\circ$. Since $AC_1=AB_2$, $\angle C_1B_2A=60^\circ$. Therefore $\square AC_1LB_2$ is cyclic so we see that $\angle B_2LA=\angle B_2C_1A=60^\circ$. That is $\angle KLB_2=180^\circ-\angle KLM-\angle B_2LA=60^\circ$.

By symmetry $\angle CKB_1=\angle C_1LA=60^\circ$ too. Hence $\angle CKB_1=\angle KLB_2$, so $KB_1\parallel LB_2$. Since $CB_1=B_1B_2$, we get $LK=KC$.

Turn $\triangle KB_1C$ around $C$ in the counterclockwise direction until $CB_1$ coincides with $CA_2$. Suppose $K_c$ is the image of $K$ under this transformation. Then $\triangle CKK_c$ is equilateral with side length $KC=KL$. This shows that $$[CB_1KA_2]=[CKK']=[KML].$$

Now $[KB_1B_2]=[KB_1C]$ because $\triangle KB_1B_2$ and $\triangle KB_1C$ share the same altitude from $K$ and have the same base legth $B_1B_2=B_1C$. Furthermore $\triangle KLB_2\cong \triangle CKA_2$ because $KL=CK$, $\angle KLB_2=60^\circ = \angle CKA_2$, and $LB_2=KB_2$ (by symmetry). Therefore $$[KB_1B_2L]=[KB_1B_2]+[KLB_2]=[KB_1C]+[CKA_2]=[CB_1KA_2].$$

By symmetry, we have $$[AC_1LB_2]=[BA_1MC_2]=[CB_1KA_2]$$ and $$[MA_1A_2K]=[KB_1B_2L]=[LC_1C_2M].$$ Therefore $[KLM]=\frac{1}{7}[ABC]$.


Here is another argument why $\triangle KML$ is equilateral (which is not just saying "due to symmetry"). Note that $\triangle ABA_1$ is obtained by a rotation of $\triangle CAC_1$ about the center of $\triangle ABC$ by $60^\circ$ in the counterclockwise direction. Therefore $CC_1$ and $AA_1$ make an angle of $60^\circ$. Hence $\angle KLM=60^\circ$. By the same argument $\angle LMK=60^\circ$ and $\angle MKL=60^\circ$ too.


If you want to use a Menelaus argument, you can do it like this. From Menelaus's thm (on $\triangle ABA_1$ with the traversal $C_1LC$) $$\frac{AC_1}{C_1B}\cdot\frac{BC}{CA_1}\cdot \frac{A_1L}{LA}=-1.$$ Here the length ratios are signed (the ratio of lengths measured in the same direction is positive, and the ratio of lengths measured in the opposite directions is negative). That is $$\frac{1}{2}\cdot\left(-\frac{3}{2}\right)\cdot\frac{AL_1}{LA}=-1\implies \frac{AL_1}{LA}=\frac{4}{3}.$$

Using Menelaus's thm on $\triangle AA_1C$ with traversal $BMB_1$ yields $$\frac{AM}{MA_1}\cdot \frac{A_1B}{BC}\cdot\frac{CB_1}{B_1A}=-1.$$ Hence $$\frac{AM}{MA_1}\cdot\left(-\frac{1}{3}\right)\cdot\frac{1}{2}=-1\implies \frac{A_1M}{MA}=\frac{1}{6}.$$ This shows that $A_1M:ML:LA=1:3:3$.

We have $$\frac{[KLM]}{[LA_1C]}=\frac{LM}{LA_1}\cdot \frac{LK}{LC}.$$ By symmetry $\frac{LK}{LC}=\frac{ML}{MA}=\frac12$, so $$\frac{[KLM]}{[LA_1C]}=\frac{3}{3+1}\cdot\frac{1}{2}=\frac{3}{8}.$$ Similarly $$\frac{[LA_1C]}{[C_1BC]}=\frac{CA_1}{CB}\cdot\frac{CL}{CC_1}=\frac{2}{3}\cdot\frac{3+3}{1+3+3}=\frac{4}{7}.$$ Finally $$\frac{[C_1BC]}{[ABC]}=\frac{BC_1}{BA}=\frac{2}{3}.$$ Therefore $$\frac{[KLM]}{[ABC]}=\frac{[KLM]}{[LA_1C]}\cdot \frac{[LA_1C]}{[C_1BC]}\cdot \frac{[C_1BC]}{[ABC]}=\frac{3}{8}\cdot\frac{4}{7}\cdot\frac{2}{3}=\frac17.$$

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According to Routh’s theorem the ratio of the areas in general case is

\begin{align} \frac{S_{KLM}}{S_{ABC}} &= \frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)} . \end{align}
The question is a special case, for which $x=y=z=\tfrac12$, so

\begin{align} \frac{S_{KLM}}{S_{ABC}} &= \frac{(\tfrac18-1)^2}{(\tfrac14+\tfrac12+1)^3} =\tfrac17 . \end{align}

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Let $O$ be the center of triangle $ABC$ and observe a rotation about $O$ for $120^{\circ}$. Then \begin{eqnarray}A &\mapsto& B\\B &\mapsto& C\\C &\mapsto& A \end{eqnarray}and thus \begin{eqnarray}A_1 &\mapsto& B_1\\B_1 &\mapsto& C_1\\C_1 &\mapsto& A_1 \end{eqnarray} and so intersection of segements $AA_1$ and $BB_1$ goes to intersection of segements $BB_1$ and $CC_1$ i.e. $M\mapsto K$ and similary $K\mapsto L$ and $L\mapsto M$ which meanst that $KLM$ is also equilateral.

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