Track / Stage 7 / 119 of 300 #1519 of 1964

Problem 1519

National olympiad second round; IMO P1/P4

Geometry Difficulty 7.2 Find the answer

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 $ABC$ .

The source for this one didn't record the answer, so there is nothing to check what you type against. Work it on paper and mark yourself against the solution below.

Official solution

### Part (a)
1. **Assume the area of $\triangle ABC$ is 1:**
$[ABC] = 1$
This simplifies the calculations as we are dealing with ratios of areas.

2. **Express the area of $\triangle A_1B_1C_1$ in terms of $k$ :**
$[A_1B_1C_1] = [ABC] - [AB_1C_1] - [A_1BC_1] - [A_1B_1C]$
Since $[ABC] = 1$ , we need to find the areas of $\triangle AB_1C_1$ , $\triangle A_1BC_1$ , and $\triangle A_1B_1C$ .

3. **Calculate the area of $\triangle AB_1C_1$ :**
$[AB_1C_1] = [AB_1B] \cdot \frac{AC_1}{AB} = [ACB] \cdot \frac{AC_1}{AB} \cdot \frac{AB_1}{AC}$
Given $\frac{A_1B}{BC} = \frac{B_1C}{AC} = \frac{C_1A}{AB} = k$ , we have:
$[AB_1C_1] = k \cdot (1 - k)$

4. **Calculate the area of $\triangle A_1BC_1$ :**
$[A_1BC_1] = [A_1BC] \cdot \frac{A_1B}{BC} = [ACB] \cdot \frac{A_1B}{BC} \cdot \frac{BC_1}{BC}$
Similarly, we get:
$[A_1BC_1] = k \cdot (1 - k)$

5. **Calculate the area of $\triangle A_1B_1C$ :**
$[A_1B_1C] = [A_1B_1B] \cdot \frac{A_1C}{AB} = [ACB] \cdot \frac{A_1C}{AB} \cdot \frac{B_1C}{AC}$
Again, we get:
$[A_1B_1C] = k \cdot (1 - k)$

6. Sum the areas of the smaller triangles:
$[AB_1C_1] + [A_1BC_1] + [A_1B_1C] = 3k(1 - k)$

7. **Find the area of $\triangle A_1B_1C_1$ :**
$[A_1B_1C_1] = 1 - 3k(1 - k)$
Simplify the expression:
$[A_1B_1C_1] = 1 - 3k + 3k^2$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.