The angle \(A\) of triangle \(ABC\) is a right angle and the sides \(BC\), \(CA\) and \(AB\) are of lengths \(a\), \(b\) and \(c\) respectively. Each side of the triangle is tangent to the circle \(S_1\) which is of radius \(r\). Show that \(2r=b+c-a\).

Let the points where \(S_1\) touches the triangle be \(U\), \(V\) and \(W\) as shown in the diagram. Let the circle have centre \(I\).

The circle \(S_1\) is the *incircle* of triangle \(ABC\); it’s conventional to label its centre as \(I\).

We have lots of tangents to a circle… we know that a tangent to a circle is perpendicular to the circle’s radius at the point of touching.

And we know that if we pick a point not on the circle, and draw the two tangents to the circle from that point, then the distances from the point to the touching points are equal.

Since \[a = CU + BU = b+c-2r,\] we have the required result.

Each vertex of the triangle lies on the circle \(S_2\). The ratio of the area of the region between \(S_1\) and the triangle to the area of \(S_2\) is denoted by \(R\). Show that \[\pi R=-(\pi-1)q^2+2\pi q-(\pi+1),\] where \(q=\dfrac{b+c}{a}\).

The line segment \(BC\) is the hypotenuse of a right-angled triangle, and so it is the diameter of the circumcircle \(S_2\) of triangle \(ABC\).

Strictly speaking, we are using here the converse of Thales’s theorem (which says that the angle in a semicircle is a right angle).

Thus the radius of the circle \(S_2\) is \(\frac{1}{2}a\). The area of the right-angled triangle \(ABC\) is \(\frac{1}{2}bc\). Circle \(S_1\) has radius \(r\), so has area \(\pi r^2\). Circle \(S_2\) has radius \(\frac{1}{2}a\), so has area \(\frac{1}{4}\pi a^2\).

Remembering the definition of \(R\), this gives \[R=\frac{\frac{1}{2}bc - \pi r^2}{\frac{1}{4}\pi a^2} = \frac{2bc - 4\pi r^2}{\pi a^2}.\]

The question defines \(q = \dfrac{b+c}{a}\), so \(aq=b+c\). This gives \[r=\frac{1}{2}(b+c-a)=\frac{1}{2}a(q-1)\] and so \[\pi R=\frac{2bc - \pi a^2(q-1)^2}{a^2}.\] Now we need to eliminate \(2bc\). Since \(ABC\) is a right-angled triangle, we know \(a^2=b^2+c^2\).

This is Pythagoras’s theorem.

and so \[\frac{2bc}{a^2} = q^2 - 1.\]

Thus \[\begin{align*} \pi R &= \frac{2bc}{a^2} - \pi(q-1)^2\\ &= q^2 - 1 - \pi(q-1)^2\\ &=-(\pi-1)q^2+2\pi q-(\pi+1) \end{align*}\]as required.

Alternatively, we could have substituted \(q=\dfrac{b+c}{a}\) directly into \(-(\pi-1)q^2+2\pi q-(\pi+1)\) and checked that it was equal to \(\pi R\).

Deduce that \[R\leq\frac{1}{\pi(\pi-1)}.\]

The are a number of ways to tackle this.

#### Approach 1 — Using the discriminant

We know that \(\pi R =-(\pi-1)q^2+2\pi q-(\pi+1)\). We can write this as a quadratic in q: \[(\pi-1)q^2-2\pi q+(\pi+1)+\pi R=0.\] For this to have real values for q, the discriminant must be non-negative. This gives us \[(2\pi)^2 \geq 4(\pi -1)(\pi R +\pi+1)\] which rearranges to give \[R \leq \dfrac{1}{\pi(\pi-1)}\] as required.

#### Approach 2 — Completing the square

Complete the square on the expression for \(\pi R\): we have \[\begin{align*} \pi R &= -(\pi-1)\left[q^2-\frac{2\pi}{\pi-1}q+\frac{\pi+1}{\pi-1}\right]\\ &= -(\pi-1)\left[\left(q-\frac{\pi}{\pi-1}\right)^2+\frac{\pi+1}{\pi-1}-\frac{\pi^2}{(\pi-1)^2}\right]\\ &= -(\pi-1)\left[\left(q-\frac{\pi}{\pi-1}\right)^2-\frac{1}{(\pi-1)^2}\right] \end{align*}\]and so \[R = \dfrac{1}{\pi(\pi-1)}-(1-\dfrac{1}{\pi})\left(q-\dfrac{\pi}{\pi-1}\right)^2.\]

We can see that \(R\) has a maximum at \(q=\dfrac{\pi}{\pi-1}\), when \(R = \dfrac{1}{\pi(\pi-1)}\) and thus \(R\leq\dfrac{1}{\pi(\pi-1)}\).

#### Approach 3 — Differentiation

*Note — you may not have covered differentiation yet…*

We can deduce that this stationary point is a maximum because we have a quadratic in \(q\) where the \(q^2\) term has a negative coefficient.