Baltic Way 1998 · Problem 15

Geometry

Given an acute triangle $A B C$ . Point $D$ is the foot of the perpendicular from $A$ to $B C$ . Point $E$ lies on the segment $A D$ and satisfies the equation

\frac{|A E|}{|E D|}=\frac{|C D|}{|D B|}

Point $F$ is the foot of the perpendicular from $D$ to $B E$ . Prove that $\angle A F C=90^{\circ}$ .

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Topics

Triangles and centers · Combinatorial geometry and dissections

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Solution

Official solution diagram for Baltic Way 1998 Problem 15 (Figure 6).

Figure 6

Complete the rectangle $A D C P$ (see Figure 6). In view of

\frac{|A E|}{|E D|}=\frac{|C D|}{|D B|}=\frac{|A P|}{|D B|}

the points $B, E, P$ are collinear. Therefore $\angle D F P=90^{\circ}$ , and so $F$ lies on the circumcircle of the rectangle $A D C P$ with diameter $A C$ ; hence $\angle A F C=90^{\circ}$ .

Contest context

Results from Baltic Way 1998

11 teams

Mean score: 3.3 / 5
Scores of 4 or 5: 7 / 11
Estonia: 5 / 5

Score distribution

All team scores

Team	Score
Latvia	5 / 5
Estonia	5 / 5
Poland	5 / 5
Finland	0 / 5
St. Petersburg	4 / 5
Sweden	1 / 5
Denmark	0 / 5
Iceland	5 / 5
Norway	1 / 5
Germany	5 / 5
Lithuania	5 / 5

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