Track / Stage 6 / 61 of 400 #1061 of 1964

Problem 1061

National olympiad, first round

Geometry Difficulty 6.0 Prove it

From the endpoints of the diameter of a semicircle, $A$ and $B$ , we draw arbitrary chords $AC$ and $BD$ , which intersect at point $I$ . Show that

(a) the sum $AC \cdot AI + BD \cdot BI$ is constant, and

(b) $IK$ is the angle bisector of $\angle D K C$ , where $K$ is the projection of point $I$ onto the diameter $AB$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

(a) Since $I C B \angle=I K B \angle=90^{\circ}$ and $A D I \angle=I K A \angle=90^{\circ}$ , therefore $K I C B$ and $A D I K$ are cyclic quadrilaterals, and circles can be circumscribed around them. Therefore,

$B K \cdot B A=B I \cdot B D$

and

$A B \cdot A K=A I \cdot A C$

$A C \cdot A I+B D \cdot B I=A B \cdot(B K+A K)=\overline{A B}^{2}=\text{const.}$

(b) Since the inscribed angles subtended by equal arcs are equal, therefore

$I K C \angle=I B C \angle \quad \text{and} \quad I K D \angle=I A D \angle ;$

but

$I A D \angle=I B C \angle$

$I K C \angle=I K D \angle=\frac{1}{2} D K C \angle .$

(László Bánó, Budapest.)

The problem was also solved by: Bauer E., Bayer N., Ehrenfeld N., Ehrenstein P., Epstein K., Erdélyi I., Erdős V., Fodor H., Földes R., Freund E., Gádor Z., Heimlich P., Kirchknopf E., Kiss E., Kovács Gy., Lusztig M., Murarik A., Neubauer K., Paunz A., Pichler S., Sárközy P., Schuster Gy., Seligmann A., Spitzer L., Szilas O., Tandlich E., Tóth B.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.