Let ABCD be a square of side of unit length.

Question

Let $ABCD$ be a square of side of unit length. Let a circle $C_1$ centred at $A$ with unit radius be drawn. Another circle $C_2$, which touches $C_1$ and the lines $AD$ and $AB$, is also drawn. Let a tangent line from the point $C$ to the circle $C_2$ meet the side $AB$ at $E$. If the length of $EB$ is $\alpha+\sqrt3\beta$, where $\alpha,\beta$ are integers, then $\alpha+\beta$ is equal to \_\_\_\_\_\_\_\_.

Answer: 1

Accepted Answer

Set $A=(0,0),B=(1,0),C=(1,1),D=(0,1)$. $C_1$: centre $A$, radius $1$.

$C_2$ has centre $(r,r)$, radius $r$, tangent internally to $C_1$: $r\sqrt2=1-r\Rightarrow r=\dfrac1{\sqrt2+1}=\sqrt2-1$.

A line through $C=(1,1)$ with slope $m$: $mx-y+(1-m)=0$. Distance from $(r,r)$ equals $r$:
$$(2-\sqrt2)|m-1|=(\sqrt2-1)\sqrt{m^2+1}.$$

Squaring and simplifying: $2(m-1)^2=m^2+1\Rightarrow m^2-4m+1=0\Rightarrow m=2\pm\sqrt3$.

For $m=2+\sqrt3$: the line meets $y=0$ at $x=1-\frac1m=\sqrt3-1$ (inside $AB$), giving $E=(\sqrt3-1,0)$, so
$$EB=1-(\sqrt3-1)=2-\sqrt3.$$

(The other slope gives $x<0$, outside $AB$.) So $\alpha=2,\beta=-1$, and $\alpha+\beta=1$.

Solutions · 1