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Brazilian Mathematical Olympiad

Geometry Difficulty 6.2 National Olympiad Prove it Brazil

Given a triangle $ABC$ and a point $P_0$ on the side $AB$ . Construct points $P_i$ , $Q_i$ , $R_i$ as follows: $Q_i$ is the foot of the perpendicular from $P_i$ to $BC$ , $R_i$ is the foot of the perpendicular from $Q_i$ to $AC$ and $P_i$ is the foot of the perpendicular from $R_{i-1}$ to $AB$ . Show that the points $P_i$ converge to a point $P$ on $AB$ and show how to construct $P$ .

Solution

It is clear from the diagram that
$Q_n Q_{n+1} = P_n P_{n+1} \cos \angle B$
$R_n R_{n+1} = Q_n Q_{n+1} \cos \angle C$
$P_n P_{n+1} = R_{n-1} R_n \cos \angle A$
Hence $P_n P_{n+1} = k^n P_0 P_1$ , where $|k| = |\cos A \cos B \cos C| < 1$ . If we take the direction $A$ to $B$ as positive, then the signed distance $P_n P_{n+1}$ may be positive or negative, but the series $1 + |k| + |k|^2 + |k|^3 + \cdots$ converges to $\frac{1}{1-|k|}$ , so $P_0 P_1 + P_1 P_2 + P_2 P_3 + \cdots$ is absolutely convergent and hence convergent. So the points $P_i$ converge to a point $P$ on the line $AB$ .

Take any point $P'$ on $AB$ . Take $Q'$ as the foot of the perpendicular from $P'$ to $BC$ . Now take $R'$ as the intersection of the lines through $P'$ perpendicular to $AB$ and through $Q'$ perpendicular to $AC$ . Now $P'Q'R'$ is similar to the desired triangle $PQR$ . Since $B, P', P$ are collinear and $B, Q', Q$ are collinear, it follows that $B, R', R$ must be collinear. Thus extend $BR'$ to meet $AC$ at

$R$ . It is now straightforward to construct $Q$ , then $P$ .

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