## Tuesday, March 8, 2011

### 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:

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

$$h = 2p$$

Subscribe to:
Post Comments (Atom)

## No comments:

## Post a Comment