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*1. Let ABCD be a unit square. Draw a quadrant of a circle with A as the center and B, D as the end points of the arc. Similarly draw a quadrant of a circle with B as the center and A, C as the end points of the arc. Inscribe a circle Γ touching the arcs AC and BD both externally and also touching the side CD.*

Solution:

Suppose O is the center of Γ. The point of tangency of Γ with CD be P, with arc AC be R and with BD be Q.

By symmetry DP = 1/2 unit.

Also note that AQO is a straight line. (If we draw a tangent line through Q, AQ will be perpendicular to the tangent line as A is the center of quadrant ABD. Also OQ will be perpendicular to the tangent line as O is center of Γ. OQ, AQ are perpendicular to same line - the tangent line - at the same point (Q). Hence they are the same straight line).

Length of AQO = AQ + QO = 1 + r (AQ = AB = 1 and let the radius of Γ be r)

Drop a perpendicular OM on AD.

Then OM = DP = 1/2

AM = 1-r.

Applying Pythagoras' Theorem on triangle AQM we have

=> r = 1/16.

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