Mathematical and Physical Journal
for High Schools
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Problem A. 560. (April 2012)

A. 560. Given a right circular cone with vertex O, and a fixed point P in the base plane of the cone. Draw a line x through P which intersects the base circle of the cone at two points, X1 and X2. Show that \tan\frac{\angle POX_1}2
\cdot \tan\frac{\angle POX_2}2 does not depend on the choice of the line x.

(5 pont)

Deadline expired on May 10, 2012.

Solution. Let G be the sphere with center O that contains the base circle of the cone. Let P' be the intersection of G with the ray OP, and let I be the point of G, diametrically opposite with P'.

Apply inversion to the sphere with center I and radius IP'. The image of G is the tangent plane S at point P'. The image of the base circle is some circle k in the plane S.

Let X1' and X2' be the images of X1 and X2, respectively. The points P',X1',X2' are collinear, moreover X1',X2' lie on k. Then

\tg \frac{POX_1\sphericalangle}2 \cdot \tg \frac{POX_2\sphericalangle}2
= \tg P'IX_2' \sphericalangle \cdot \tg P'IX_2' \sphericalangle
= \frac{P'X_1'}{IP'} \cdot \frac{P'X_2'}{IP'}
= \frac{P'X_1' \cdot P'X_2'}{IP'^2}.

The numerator, being the power of P' with respect to k, does not depend on the points X1',X2'.


8 students sent a solution.
5 points:Ágoston Tamás, Janzer Olivér, Mester Márton, Omer Cerrahoglu, Strenner Péter, Szabó 789 Barnabás, Szabó 928 Attila.
4 points:Gyarmati Máté.

Problems in Mathematics of KöMaL, April 2012