Tough Geometry I

!---Just for personal reference

Diagram shows a straight line IJK\text{IJK}, a circle, 22 congruent quadrants BEF\text{BEF} and CGH\text{CGH}, and straight lines extended from the ends of each arc. The top centre circle is tangent to the upper line at J\text{J} and the two quadrants at the marked points.

AB\text{AB} and CD\text{CD} are collinear


p=Distance of FGp = \text{Distance of FG}

q=Distance of BC=Distance if EH q = \text{Distance of BC} = \text{Distance if EH}

r=Radius of each quadrant r = \text{Radius of each quadrant}

s=Perpendicular distance from IJK to AB or CDs = \text{Perpendicular distance from IJK to AB or CD}

R= Radius of the circleR = \text{ Radius of the circle}

The relation between p,q,r,s,andR p, q, r, s, \text{and} R:

(s+rR)2+(p2+r)2=(R+r)2 (s + r - R)^2 + (\dfrac{p}{2} + r)^2 = (R + r)^2

Using Pythagorean Theorem

Note :The figure is symmetrical about the vertical diameter line of the circle (ignoring extra lines and points)

Note by Lin Shun Hao
1 year, 1 month ago

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