# Tough Geometry I

!---Just for personal reference

Diagram shows a straight line $\text{IJK}$, a circle, $2$ congruent quadrants $\text{BEF}$ and $\text{CGH}$, and straight lines extended from the ends of each arc. The top centre circle is tangent to the upper line at $\text{J}$ and the two quadrants at the marked points.

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

Given:

$p = \text{Distance of FG}$

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

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

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

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

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

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

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
3 months, 2 weeks ago

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