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Winter Mathematical Competition

Geometry Difficulty 6.2 National Olympiad Prove it Bulgaria

Let $ABC$ be a right triangle with $\angle ACB = 90^\circ$ , $AC = 1$ and $BC = 2$ . Given a point $A_1 \in BC$ such that $A_1C \neq \frac{1}{3}$ we construct a sequence of points $A_n \in BC$ , $n \ge 2$ in the following way. Let $B_1$ be the intersection point of $AC$ and the line through $A_1$ and parallel to $AB$ , and $C_1$ be the foot of the perpendicular from $B_1$ to $AB$ . Then $A_2$ is the intersection point of $BC$ and the line through $C_1$ and parallel to $AC$ . Using $A_2$ we construct $A_3$ in the same way, etc. Find:
a) $\frac{3A_2C - 1}{3A_1C - 1}$ ;
b) $\lim_{n \to \infty} S_{A_n B_n C_n}$ .

Solution

a) Set $A_nC = x_n$ . Then it follows from $\triangle A_nB_nC \sim \triangle BAC$ that $B_nC = \frac{x_n}{2}$ . Since $\triangle AB_nC_n \sim \triangle ABC$ , we find that $AC_n = \frac{2-x_n}{2\sqrt{5}}$ . Hence $A_{n+1}C = x_{n+1} = \frac{2-x_n}{5}$ and therefore $\frac{3x_{n+1}-1}{3x_n-1} = -\frac{1}{5}$ .

b) Since $A_nB_n = \frac{\sqrt{5}}{2}x_n$ , $B_nC_n = \frac{2-x_n}{\sqrt{5}}$ and $A_nB_n \perp B_nC_n$ we get $S_{A_nB_nC_n} = \frac{x_n(2-x_n)}{4}$ . But a) implies that $\{x_n - \frac{1}{3}\}_{n=1}^\infty$ is a geometric progression with ratio $-\frac{1}{5}$ . Hence $\lim_{n \to \infty} x_n = \frac{1}{3}$ and therefore $\lim_{n \to \infty} S_{A_nB_nC_n} = \frac{5}{36}$ .

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