Let ABCD be a unit square. Draw a quadrant of a circle with A as centre and B;D as end points of the arc. Similarly, draw a quadrant of a circle with B as centre and A;C as end points of the arc. Inscribe a circle ? touching the arc AC internally, the arc BD internally and also touching the side AB. Find the radius of the circle ?.
Problem
Source: RMO 2012
Tags: geometry
18.12.2017 13:49
${(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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08.10.2019 15:50
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.
08.10.2019 19:52
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
08.10.2019 20:45
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
08.10.2019 20:59
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
29.09.2020 20:25
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
16.09.2024 16:44
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!
20.09.2024 20:21
???symmetry