Question

Let O be the center of a circle with radius 1. Draw another circle with center O and Radius 2. Select an arbitrary point Z on the circle with center O and radius 1. Draw a 120-degree to OZ and call the point where it meets the smaller circle N (so OZ = ON = 1 and ∠ZON = 120°). Draw a line perpendicular to OZ Where they meet the top of the circle is called B, and where it meets the bottom is called E. Draw another line perpendicular to ON, and where it meets the top is called C and where it meets the bottom is called D. Let BE and CD intersect at Y. Express your answer as (BY, CY).

a) (2√3, √3)
b) (√3, 2√3)
c) (3√3, √3)
d) (√3, 3√3)

Answer 1

The question involves constructing a geometric figure with circles and perpendicular lines, in which lengths BY and CY are needed. Using properties of equilateral triangles and rectangles, we can determine that BY is half the diameter of the larger circle, which is 2, and CY is half the longer diagonal of the smaller equilateral triangle, which is √3.

Step-by-step explanation:

We are tasked with finding the lengths of BY and CY in a geometric figure involving circles and perpendicular lines. First, let us consider the triangle OZN, which is an equilateral triangle since OZ = ON and ∠ZON = 120°. The line perpendicular to OZ at point O will go through the center of both circles, and the points B and E will be the intersection points with the circles' circumference. Point B is on the larger circle, so OB = 2, and point E is on the smaller circle, so OE = 1. Similarly, consider the perpendicular line on ON, which defines points C and D. The distance OC would be the radius of the smaller circle, which is 1. Since OZN is equilateral, the distance from Z to the midpoint of ON, and hence from C to the midpoint of ON, is √3/2 since the side length of the triangle OZN is 1.

Because points B, E, C, and D are on the circles, and all the perpendiculars from the diameter will intersect at the center O, we conclude that BE and CD will create a rectangle whose opposite sides are equal. Thus Y must be the center of this rectangle, equidistant to all sides. Therefore, BY is half the length of BE, which is the diameter of the larger circle, so BY = 2. Likewise, CY is half the distance of CD, which is √3, meaning CY = √3/2. In conclusion, BY and CY would be half the diameters of the respective circles, yielding the length values (2√3, √3).