Triangle and circle

Question

There is a right angle traingle circumscribed about a circle. Another circle is sandwiched between the sides of triangle and the bigger circle. What is the ratio of radius of the two circles ? Note: the circles touch each other externally and touch the sides too. There is no intersection of curves and lines. My approach : let The sides of triangle be 3 , 4 and 5. So in radius = area / semiperimeter=1=radius of bigger circle. Can't process further. Help !

Answer 1

We just have two touching circles in a wedge. If the opening angle of the wedge is $2\alpha$ then one immediately verifies that $${R-r\over R+r}=\sin\alpha\ .$$ This implies $${r\over R}={1-\sin\alpha\over 1+\sin\alpha}\ .$$

Answer 2

As a generic right triangle, this problem has a degree of freedom; it's possible for the incircle to have any radius $1<R<3+2\sqrt{2}$. To show this, we will construct the triangle for any given radius.

For a given $R$, construct a right triangle with "opposite" leg $R-1$ and hypotenuse $R+1$; this triangle when drawn will have the ends of its hypotenuse at the centers of the circles and the "adjacent" leg parallel to the vertical line in your diagram.

The length of the adjacent leg here is thus $\sqrt{(R+1)^2-(R-1)^2}=2\sqrt{R}$.

Now, we can use this triangle to measure the larger triangle; first we will find the length of the vertical side, using similarity.

The vertical side is divided into three parts: the bottom part, between the right angle and the contact point of the large circle, clearly has length $R$; the middle, between the two circle contact points, has radius $2\sqrt{R}$. For the top, note that the little triangle we made must be similar to the triangle that includes the top point, the contact point of the unit circle, and the center of the unit circle. To resize the triangle to fit, we must multiply all the sides by $\frac{1}{R-1}$ -- so the top part is $\frac{2\sqrt {R}}{R-1}$.

Add that all up and we get $R+2\sqrt{R}+\frac{2\sqrt {R}}{R-1}=R+\frac{2R\sqrt{R}}{R-1}$ for the length of the vertical leg.

Now, to find one other leg; I choose the hypotenuse. To find this, we will need some bigger guns: it's trigonometry time.

Let's name the edges: for the large triangle, I'll call the vertical line $A$, the horizontal line $O$, and the hypotenuse $H$; for the small triangle I'll call them $a$, $o$ and $h$ respectively.

$h$ extends through the top point, and because it's through the center of a circle tangent to the lines through the top point, it bisects the angle up there. So we'd like to double the existing angle. We want the hypotenuse and have the adjacent, so we'll use the cosine double angle formula: $\cos (2 \theta) = \cos(\theta)^2-\sin (\theta)^2$, or using actual edges, $\frac {A}{H}=\frac{a^2-o^2}{h^2}$. This we can solve for $H$:

$$H=\frac{Ah^2}{a^2-o^2}=\frac{(R+\frac{2R\sqrt{R}}{R-1})(R+1)^2}{(2\sqrt{R})^2-(R-1)^2}$$