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Baltic Way 1994 · Problem 11

Geometry

Let NSN S and EWE W be two perpendicular diameters of a circle C\mathcal{C}. A line ll touches C\mathcal{C} at point SS. Let AA and BB be two points on C\mathcal{C}, symmetric with respect to the diameter EWE W. Denote the intersection points of ll with the lines NAN A and NBN B by A′A^{\prime} and B′B^{\prime}, respectively. Show that ∣SA′∣⋅∣SB′∣=∣SN∣2\left|S A^{\prime}\right| \cdot\left|S B^{\prime}\right|=|S N|^{2}.

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Topics

Circles and tangency · Angles and distances

Solutions

Solution

Solution:

We have ∠NAS=∠NBS=90∘\angle NAS = \angle NBS = 90^{\circ} (see Figure 1). Thus, the triangles NA′SNA'S and NSANSA are similar. Also, the triangles B′NSB'NS and SNBSNB are similar and the triangles NSANSA and SNBSNB are congruent. Hence, the triangles NA′SNA'S and B′NSB'NS are similar which implies SA′SN=SNSB′\frac{SA'}{SN} = \frac{SN}{SB'} and SA′⋅SB′=SN2SA' \cdot SB' = SN^2.

Diagram for the mathnet 00y5 1 of bw-1994-11. Figure 1

Contest context

Results from Baltic Way 1994

9 teams

Mean score
4.2 / 5
Scores of 4 or 5
7 / 9
Estonia
5 / 5

Score distribution

00
11
20
31
41
56
All team scores
TeamScore
St. Petersburg3 / 5
Latvia5 / 5
Poland5 / 5
Sweden5 / 5
Denmark1 / 5
Estonia5 / 5
Finland5 / 5
Lithuania4 / 5
Iceland5 / 5