four tangent circles

Given four circles: $L_1,L_2,L_3,L_4$ such that $L_1$ is tangent to $L_2$ at $X$, $L_2$ is tangent to $L_3$ at $Y$, $L_3$ is tangent to $L_4$ at $Z$ and $L_4$ is tangent to $L_1$ at $W$. All tangencies are outside (the circles do not overlap each other other than at the points of tangency).

Show that $X,Y,Z,W$ all lie on a circle.

Excircle and circumcircle

In $\triangle ABC$, the $L_1$ be the excircle with respect to $AB$. It touches $CA$ and $CB$ and $M$ and $N$. The circle centered at $C$ and passing through $M$ and $N$ intersects $AB$'s extension at $P,Q$. Show that $L_2$ the circumcircle of $\triangle CPQ$ is tangent to $L_1$, and that the point of tangency between $L_1$ and $L_2$, the point of tangency between $L_1$ and $\triangle ABC$, and $C$ all lie on one line.

Solution Let $X$ be the tangency point of $L_1$ and $AB$. Let $CX$ meets $L_1$ in $Y_1$ and $CX$ meets $L_2$ in $Y_2$. We shall establish that $Y_1 = Y_2$.

Let $C'$ be the diameter of $L_2$ and let $CC'$ meets $AB$ (or its extension) at $D$. Because $CP = CQ$ then it's easy to see that $CD \perp PQ$. Let $s = CM = CN = CP = CQ$.

By power point on $L_1$: $$CX.CY_1 = CM^2 = s^2$$.

By similarity of $\triangle CXD \sim \triangle CC'Y_2$ and $\triangle CC'P \sim \triangle CDP$ : $$CX.CY_2 = CD.CC' = CP^2 = s^2$$

Thus $CY_1 = CY_2$ which means $Y_1 = Y_2$. Now we know that $L_1$ meets $L_2$ in at least one point $Y = Y_1 = Y_2$. Suppose they meet in another point $Z \neq Y$

Let $CZ$ meets $L_1$ in $CZ$ and $X_1 \neq X$. By power point we again have:$CX_1.CZ = s^2$.

Let $CZ$ meets $AB$ in $X_2$. Similar as above, because $Z \in L_2$, we can repeat the similarity argument to establish: $CX_2.CZ = s^2$. This means that $CX_1 = CX_2$, thus $X_1 = X_2 \in L_1, AB$ but that is a contradiction because $L_1$ is only tangent to $AB$, and can only meet $AB$ at $X$.

Now that we've established the tangency between $L_1$ and $L_2$, it's clear that $C,X,Y_1/Y_2$ all lie on one line from the proof.

Technology Review March/April 2011 Problem 3

The figure contains three semicircles and one circle. The semicircles all have horizontal diameters, and their centers are shown. Their radii are $R, r_1, r_2$ with $r_1 + r_2 = R$. The circle is constructed tangent to the three semicircles. Find $p$ the radius of the circle, and show that the distance from its center to the baseline is $2p$.

(Note: I couldn't draw them as semicircles and I was too lazy to crop the image. But you get the idea).

Solution

Suppose the circle centered in $A$ has radius $r_1$, and the circle centered in $B$ has radius $r_2$. We can readily see that $BC = r_1$ and $AC = r_2$. Also because the third circle is tangent to all circles:
$OA = p+r_1$
$OB = p+r_2$
$OC = r_1 + r_2 -p$

From Stewart's Theorem:

$$OC^2 AB = OA^2 BC + OB^2 AC - AB.BC.AC$$
$$(r_1 + r_2 - p)^2 (r_1+r_2) = r_1(p+r_1)^2 +r_2(p+r_2)^2 - r_1 r_2 (r_1 + r_2)$$

The coefficient of $p^2$ on both sides of that equation is $r_1 + r_2$, thus this is not a quadratic equation, but merely a linear equation in $p$. Expanding and canceling both sides gives us:

$$ p = \frac{r_1r_2(r_1+r_2)}{(r_1+r_2)^2-r_1r_2} = \frac{r_1r_2R}{R^2-r_1r_2}$$

To compute the distance $h$ from $O$ to $AB$, first we calculate the area of triangle $OAB$. Since $OA = p+r_1, OB = p+r_2, AB = r_1+r_2$, we can use Heron's formula:

$$S_{OAB} = \sqrt{(p+r_1+r_2)pr_1r_2} = \frac{1}{2} h AB = \frac{1}{2} h R$$
$$4(p+r_1+r_2)pr_1r_2 = h^2 R^2$$
$$h^2 = \frac{4(p+r_1+r_2)pr_1r_2}{R^2}$$

Now note that $p+r_1+r_2 = p+R = \frac{r_1r_2R}{R^2-r_1r_2} + R = \frac{R^3}{R^2-r_1r_2}$

Plug this into the equation above:

$$h^2 = \frac{4(p+R)pr_1r_2}{R^2} = \frac{4}{R^2} \frac{R^3}{R^2-r_1r_2} \frac{r_1r_2R}{R^2-r_1r_2} r_1r_2 = 4 \frac{(r_1r_2R)^2}{(R^2-r_1r_2)^2}$$
$$h = 2p$$