Let $ABC$ be a triangle and $k < 1$ a positive real number. Let $A_1$, $B_1$, $C_1$ be points on the sides $BC$, $AC$, $AB$ such that $$\frac{A_1B}{BC} = \frac{B_1C}{AC} = \frac{C_1A}{AB} = k.$$ (a) Compute, in terms of $k$, the ratio between the areas of the triangles $A_1B_1C_1$ and $ABC$. (b) Generally, for each $n \ge 1$, the triangle $A_{n+1}B_{n+1}C_{n+1}$ is built such that $A_{n+1}$, $B_{n+1}$, $C_{n+1}$ are points on the sides $B_nC_n$, $A_nC_n$ e $A_nB_n$ satisfying $$\frac{A_{n+1}B_n}{B_nC_n} = \frac{B_{n+1}C_n}{A_nC_n} = \frac{C_{n+1}A_n}{A_nB_n} = k.$$Compute the values of $k$ such that the sum of the areas of every triangle $A_nB_nC_n$, for $n = 1, 2, 3, \dots$ is equal to $\dfrac{1}{3}$ of the area of $ABC$.
Problem
Source: Rio de Janeiro Mathematical Olympiad 2018, Level 4 #1
Tags: ratio, geometry, algebra
BobThePotato
17.11.2019 22:41
(a) WLOG $[ABC]=1$ so that the ratio is just $[A_1B_1C_1].$ We have $[A_1B_1C_1]=1-[AB_1C_1]-[A_1BC_1]-[A_1B_1C].$ Note that, for example, $$[AB_1C_1]=[AB_1B]\cdot\frac{AC_1}{AB}=[ACB]\cdot\frac{AC_1}{AB}\cdot\frac{AB_1}{AC}=k(1-k),$$so the desired area is $[A_1B_1C_1]=1-3k(1-k)=\boxed{3k^2-3k+1.}$ (b) We have $$\sum_{n=1}^{\infty}[A_nB_nC_n]=\sum_{n=1}^{\infty}(3k^2-3k+1)^n=\frac{3k^2-3k+1}{3k-3k^2}.$$Setting this equal to $1/3$ gives $3k^2-3k+1=k-k^2\implies(2k-1)^2=0\implies\boxed{k=1/2.}$
Math5000
17.11.2019 23:01
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 $k=\frac 12$
BobThePotato
17.11.2019 23:16
Wellingson wrote: If $$\frac{p_1}{p_2}=\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$$then $$\frac{A_1}{A_2}=k^2$$ Watch out—the ratios given in the problem are $A_1B/BC$ and so forth, not $A_1B_1/BC.$
Math5000
17.11.2019 23:50
BobThePotato wrote: Wellingson wrote: If $$\frac{p_1}{p_2}=\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$$then $$\frac{A_1}{A_2}=k^2$$ Watch out—the ratios given in the problem are $A_1B/BC$ and so forth, not $A_1B_1/BC.$ Thanks for the correction