Given that the three smaller circles have radius $1$, what is the radius of the larger semicircle?
I managed to solve this question by assigning coordinates and equations of the circles, but this method took too long and wasn’t elegant in any sense. I’d appreciate if anyone can help me arrive at the answer through a purely geometrical approach. Here’s how I did it:
Take the center of the middle circle to be the origin. Then the equations of the rightmost circle and the bigger semicircle are, in order:
$$(x-2)^2 + y^2 =1 \\ x^2 +(y+1)^2 =r^2$$ Let the upper point where they touch be $\left(h,\sqrt{1-(h-2)^2}\right)$. So, we must have $$h^2 +\left( \sqrt{1-(h-2)^2} +1\right)^2 = r^2 \hspace{1 cm} \mathbf{(1)}$$
Further, equating their derivatives at that point gives $$\frac{2-h}{\sqrt{1-(h-2)^2}} = \frac{-h}{\sqrt{1-(h-2)^2}+1} \\ \implies h =\frac{2}{\sqrt 5} +2$$
Plugging this value in $\mathbf{(1)}$ we get $$r=1+\sqrt 5$$
As you can see, this involved a few calculations. Can anyone propose an alternate, relatively simpler solution?
Construct a line segment $AO$ from the center of the middle circle to the center of the semicircle. This has length $1$. Now construct a line segment $AB$ from the center of the middle circle to the center of the right circle. This has length $2$. Finally, construct a line $OT$ from the center of the semicircle to the point where the right circle and the semicircle meet at a tangent. Note that this line passes through the center $B$ of the right circle (try to prove if you were not already aware).
So, there is a right triangle formed by the first line, the second line, and the line segment $OT-BT$ between the semicircle's center and the center of the right circle. The length of this last segment is equal to $r-1$ (since it is the distance from the semicircle's center to the tangent point MINUS the distance from the tangent point to the right circle's center) and also equal by the Pythagorean Theorem to $\sqrt{1^2+2^2}=\sqrt{5}$.
$$\sqrt{5}=r-1$$ $$r=\sqrt{5}+1$$
Let me know if you have any questions.