In the diagram, the triangle has side lengths 6,8 and 10. Three semi-circles are drawn using the sides of the triangle as diameters. A large circle is drawn so that it just touches each of the three semi-circles. What is the radius of the large circle?

![](https://cdn.mathpix.com/cropped/2024_04_17_ff0a25f37b4b6ded9838g-3.jpg?height=520&width=525&top_left_y=865&top_left_x=1212)

## PART B

For each question in Part B, your solution must be well-organized and contain words of explanation or justification. Marks are awarded for completeness, clarity, and style of presentation. A correct solution, poorly presented, will not earn full marks.
Since $6^{2}+8^{2}=36+64=100=10^{2}$, then the given triangle is right-angled.

We label the triangle as $\triangle A B C$ with $A B=8, B C=6$ and $A C=10$.

We put the diagram in the $x y$-plane with $B$ at the origin $(0,0), A$ on the positive $y$-axis at $(0,8)$, and $C$ on the positive $x$-axis at $(6,0)$.

Let $X, Y$ and $Z$ be the midpoints of sides $A B, B C$ and $A C$, respectively.

Since $A B, B C$ and $A C$ are the diameters of the semi-circles, then $X, Y$ and $Z$ are the centres of the semi-circles.

Note also that since $A B=8$, then $A X=X B=4$, which means that the radius of the corresponding semi-circle is 4 .

This also means that the coordinates of $X$ are $(0,4)$.

Since $B C=6$, then $B Y=Y C=3$, which means that the radius of the corresponding semicircle is 3 .

This also means that the coordinates of $Y$ are $(3,0)$.

Since $A C=10$, then $A Z=Z C=5$, which means that the radius of the corresponding semicircle is 5 .

The coordinates of $Z$, the midpoint of $A(0,8)$ and $C(6,0)$, are $\left(\frac{1}{2}(0+6), \frac{1}{2}(8+0)\right)$ or $(3,4)$.

Let $M(s, t)$ be the centre of the large circle and $r$ its radius.

Let the points $U, V, W$ be the points where the semi-circles with centres $X, Y$ and $Z$, respectively, touch the larger circle. Join $M$ to $U, V$ and $W$.

![](https://cdn.mathpix.com/cropped/2024_04_17_4c20b40ed845ad599500g-04.jpg?height=626&width=729&top_left_y=1975&top_left_x=1376)

It turns out that $M U, M V$ and $M W$ pass through $X, Y$ and $Z$, respectively.

Imagine drawing a line tangent to the large circle at $U$.

Since the large circle and semi-circle touch at $U$, then this line is also tangent to the semi-circle.

Since radii are perpendicular to tangents, then $X U$ and $M U$ (both radii) are perpendicular to this tangent line.

Since both $X U$ and $M U$ are perpendicular to the tangent line, then they must lie on top of each other, and so $M U$ passes through $X$.

Using a similar argument, we can see that $M V$ and $M W$ pass through $Y$ and $Z$, respectively.

Since $M U=M V=M W=r$ (the radius of the large circle), and $X U=4, Y V=3$, and $Z W=5$ (the radii of the semi-circles), then

$$
\begin{aligned}
M X & =M U-X U=r-4 \\
M Y & =M V-Y V=r-3 \\
M Z & =M W-Z W=r-5
\end{aligned}
$$

The coordinates of $M, X, Y, Z$ are $M(s, t), X(0,4), Y(3,0), Z(3,4)$.

Using the formula for the length of a line segment, we obtain the equations

$$
\begin{aligned}
& (s-0)^{2}+(t-4)^{2}=(r-4)^{2} \\
& (s-3)^{2}+(t-0)^{2}=(r-3)^{2} \\
& (s-3)^{2}+(t-4)^{2}=(r-5)^{2}
\end{aligned}
$$

Subtracting the third equation from the first, we obtain $s^{2}-(s-3)^{2}=(r-4)^{2}-(r-5)^{2}$ or $s^{2}-s^{2}+6 s-9=r^{2}-8 r+16-r^{2}+10 r-25$ and so $6 s-9=2 r-9$ or $r=3 s$.

Subtracting the third equation from the second, we obtain $t^{2}-(t-4)^{2}=(r-3)^{2}-(r-5)^{2}$ or $t^{2}-t^{2}+8 t-16=r^{2}-6 r+9-r^{2}+10 r-25$ and so $8 t-16=4 r-16$ or $r=2 t$.

Substituting $s=\frac{1}{3} r$ and $t=\frac{1}{2} r$ into the first equation, we obtain the following equivalent equations:

$$
\begin{aligned}
\left(\frac{1}{3} r\right)^{2}+\left(\frac{1}{2} r-4\right)^{2} & =(r-4)^{2} \\
\frac{1}{9} r^{2}+\frac{1}{4} r^{2}-4 r+16 & =r^{2}-8 r+16 \\
\frac{1}{9} r^{2}+\frac{1}{4} r^{2}+4 r & =r^{2} \\
4 r^{2}+9 r^{2}+144 r & \left.=36 r^{2} \quad \text { (multiplying through by } 36\right) \\
144 r & =23 r^{2} \\
0 & =r(23 r-144)
\end{aligned}
$$

Therefore, $r=0$ (which is impossible) or $r=\frac{144}{23}$, which is the desired answer.

ANSWER: $\frac{144}{23}$

## Part B