Library / Bulgaria competition problems /35 of 44

Bulgaria competition problems, 2022

Geometry Difficulty 6.7 National Olympiad Prove it Bulgaria

An isosceles triangle $ABC$ with sides $AB = 4$ , $BC = AC = 6$ is given.
The points $X_1, X_2, X_3, \dots$ in that order are taken on side $AB$ , such that the lengths of segments $AX_1, X_1X_2, X_2X_3, \dots$ form an infinite geometric series with first term $3$ and common ratio $1/4$ .
The points $Y_1, Y_2, Y_3, \dots$ in that order are taken on side $BC$ , such that the lengths of segments $CY_1, Y_1Y_2, Y_2Y_3, \dots$ form an infinite geometric series with first term $3$ and common ratio $1/2$ .
The points $Z_1, Z_2, Z_3, \dots$ in that order are taken on side $BC$ , such that the lengths of segments $AZ_1, Z_1Z_2, Z_2Z_3, \dots$ form an infinite geometric series with the first term $3$ and common ratio $1/2$ .
Find all triples $(a, b, c)$ of natural numbers for which the line segments $AY_a$ , $BZ_b$ and $CX_c$ are concurrent.

Solution

Firstly we will prove the following:
Lemma. Let $n > 1$ be a natural number. On the line segment $PQ$ with length $\frac{a_1}{1 - \frac{1}{n}}$ are taken the points $T_1, T_2, \dots$ , such that $PT_1, T_1T_2, T_2T_3, \dots$ form an infinite geometric series with first term $a_1$ and common ratio $\frac{1}{n}$ . Then $\frac{PT_s}{T_sQ} = n^s - 1$ .

$\textit{Proof.} \text{ From } PT_s = \frac{a_1(1 - (\frac{1}{n})^s)}{1 - \frac{1}{n}} \text{ and } PQ = \frac{a_1}{1 - \frac{1}{n}} \text{ we get } \frac{\frac{a_1(1 - (\frac{1}{n})^s)}{1 - \frac{1}{n}}}{T_sQ} = \frac{a_1}{1 - \frac{1}{n} - \frac{a_1(1 - (\frac{1}{n})^s)}{1 - \frac{1}{n}}} = n^s - 1$

From the lemma above and Ceva's theorem we get
$1 = \frac{AX_c}{X_c B} \cdot \frac{BY_a}{Y_a C} \cdot \frac{CZ_b}{Z_b A} = \frac{4^c - 1}{(2^a - 1)(2^b - 1)}$
and therefore $4^c - 1 = (2^a - 1)(2^b - 1) \Leftrightarrow 4^c = 2^{a+b} - 2^a - 2^b + 2$ . But $a, b, c \ge 1$ and if $a \le b$ (without loss of generality) it follows $a = 1$ . Then $4^c = 2^b$ , $b = 2c$ and so $(a, b, c) = (1, 2k, k)$ or $(a, b, c) = (2k, 1, k)$ for $k \in \mathbb{N}$ .

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