${(a - r)^2} = {r^2} + \frac{{{a^2}}}{4} \Leftrightarrow $ $\boxed{r=\frac{3a}{8}}$ If $a=1,$ then $r=\dfrac{3}{8}$

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Why $A,I $ and the touchpoint of the incircle and the circle passing through $B,D$ are collinear?? $I$ is the centre of the incircle.

ayan_mathematics_king wrote: Why $A,I $ and the touchpoint of the incircle and the circle passing through $B,D$ are collinear?? $I$ is the centre of the incircle. HBCSE people write by symmetry

Yes they do they are because touch point of those circles is there excimilicentre of two circles are it is collinear with line joining their centres also you can say as on their touch point line joining their centres to touch point would be perpendicular so these three points are collinear

Tearfulstage wrote: Yes they do SYMMETRY= touch point of those circles is there excimilicentre of two circles are it is collinear with line joining their centres also you can say as on their touch point line joining their centres to touch point would be perpendicular so these three points are collinear So we got a definition of symmetry

an alternative way to prove that AE = BE = 1/2 (note that I am referring to the figure of the official solution): let r be the radius of Γ. in triangles OEA, OEB we have angle (OEA) = angle (OEB) = 90 degrees (since A tangent to a circle is perpendicular to the radius at the point of tangency) ; OA = OB = 1- r ; OE = OE. So by RHS criteria of congruence, it follows that triangle OEA is congruent to triangle OEB. Which implies AE = BE . Thus, AE = BE =1/2 (as their sum is 1). Hence proved

Let $O$ be the centre of $\Gamma$, and $P$,$Q$ be the touchpoints of $\Gamma$ and the two quarter circles such that $P$ is closer to $A$ than $B$. Let $X$ be the touchpoint of $\Gamma$ and $AB$. Let $r$ be the radius of $\Gamma$, we have that $OQ=1-r$, and $OX=r$. By symmetry, we have that $AX=BX$. Using pythagoras theorem in $\triangle ABX$, we have that $\frac{1}{4}+r^2=r^2-2r+1$. This gives $r=\frac{3}{8}$, and we are done!

???symmetry