Let $ABC$ be a triangle and k be a positive real number less than $1$. Take $ A_1, B_1, C_1 $ points on sides $\overline{BC}, \overline{AC},$ and $\overline{AB} $ so that:
$\frac{\overline{A_1B}}{\overline{BC}} = \frac{\overline{B_1C}}{\overline{AC}} = \frac{\overline{C_1A}}{\overline{AB}} = k$
a) Calculate as a function of k the ratio between the areas of triangles $ A_1B_1C_1 $ and $ABC.$
b) More generally, for every $n \geq 1$ the triangle $A_ {n + 1} B_ {n + 1} C_ {n + 1}$ is constructed, so that $A_ {n + 1}, B_ {n + 1 }$ and $C_ {n + 1}$ are dots on the sides $\overline{B_nC_n}, \overline{A_nC_n}$ and $\overline{A_nB_n}$ satisfying
$\frac{\overline{A_ {n + 1} B_ {n}}}{\overline{B_nC_n}} = \frac{\overline{B_ {n + 1} C_n}}{\overline{A_nC_n}} = \frac{\overline{C_ {n + 1} A_n}}{\overline{A_nB_n}} = k$
Determine the values of k so that the sum of the areas of all triangles $A_nB_nC_n$, for $n = 1, 2, 3, …,$ equals $ \frac {1}{3} $ of the area of triangle $ ABC$
What I thought: I don't have much idea except why the perimeters of similar triangles are, exemple: $\frac {2p_1}{2p_2} = k$
Parallel or anti-parallel lines to have a ratio between areas; I don't know how to apply it, but I think it always helps in something
Parallel or anti-parallel lines to have a ratio between areas so that certain sides are congruent with each other and thus have a ratio play; I don't know how to apply it, but I think it always helps in something
Best Answer
You are almost there. You just need to apply Heron's formula $$A=\sqrt{p(p-a)(p-b)(p-c)}$$ where $$p=\frac{a+b+c}2$$ If $$\frac{p_1}{p_2}=\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$$ then $$\frac{A_1}{A_2}=k^2$$
For the second part, then $$k^2+(k^2)^2+(k^2)^3+...=\frac{k^2}{1-k^2}=\frac 13$$ This yields $k=\frac 12$