Track / Stage 5 / 332 of 400 #932 of 2000

Problem 932

AIME late

Geometry Difficulty 5.8 Prove it

7. Prove that the radius of a circle is equal to the difference in lengths of two chords, one of which subtends an arc of $1 / 10$ of the circumference, and the other subtends an arc of $3 / 10$ of the circumference.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

66.7. Consider six consecutive points of division of a circle into 10 equal arcs: $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ , and $A_{6}$ . Then the line $A_{2} A_{5}$ is parallel to the diameter $A_{1} A_{6}$ and the line $A_{3} A_{4}$ , and the line $A_{3} A_{6}$ is parallel to the line $A_{4} A_{5}$ . Denote the intersection point of the lines $A_{2} A_{5}$ and $A_{3} A_{6}$ by $P$ and we get that $P A_{3} A_{4} A_{5}$ is a parallelogram, and therefore we need to prove that the length of the segment $A_{2} P$ is equal to the radius. But since $A_{2} O A_{6} P$ is also a parallelogram ( $O$ is the center of the circle), then $A_{2} P = O A_{6}$ , which is what we needed to prove.

Fig. 31

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