The circle x² + y² = 1 cuts the x-axis at P and Q. Another circle with centre at Q and variable radi

Q: Wait, why is $QS = QR$? I thought $S$ was just some point on the line segment $PQ$.

That is a great question! Remember the definition of the circle $C_2$. It is centered at $Q$ and passes through $R$. By definition, any point on this circle is at a distance $QR$ from the center $Q$. Since $S$ is also on this circle (and on the line segment $PQ$), the distance $QS$ must be equal to the radius $QR$. It is a beautiful geometric constraint!

Q: How did we get $\angle RQS = \frac{\pi - \theta}{2}$? This part is confusing.

Let's look at the slope. The slope of $QR$ is $\frac{\sin \theta - 0}{\cos \theta - 1}$. Using the identity $\sin \theta = 2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})$ and $\cos \theta - 1 = -2 \sin^2(\frac{\theta}{2})$, the slope simplifies to $-\cot(\frac{\theta}{2})$. This slope represents the tangent of the angle the line $QR$ makes with the $x$-axis. Since $Q$ is at $(1,0)$, the angle $\angle RQS$ is derived from this slope relative to the $x$-axis, leading to $\frac{\pi - \theta}{2}$.

Q: Why can we ignore $\sin t = 0$ when setting the derivative to zero?

Excellent observation. We are looking for the maximum area of a triangle, which requires a non-zero area. Since $t = \frac{\theta}{2}$ and $0 < \theta < \pi$, we have $0 < t < \frac{\pi}{2}$. In this interval, $\sin t$ is strictly positive and cannot be zero. Therefore, we can safely divide by $\sin t$ without losing any valid solutions.

Question

The circle $x^2 + y^2 = 1$ cuts the $x$-axis at $P$ and $Q$. Another circle with centre at $Q$ and variable radius intersects the first circle at $R$ above the $x$-axis and the line segment $PQ$ at $S$. Find the maximum area of the triangle $QSR$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane, looking at the unit circle $x^2 + y^2 = 1$. This circle is our canvas, a perfect, symmetric shape that cuts the $x$-axis at two points: $P(-1, 0)$ and $Q(1, 0)$.

We are tasked with finding the maximum area of a triangle $QSR$, where $R$ is a point on the circle above the $x$-axis, and $S$ is a point on the line segment $PQ$ created by a second circle centered at $Q$.

### The Parametric Elegance

To tackle this, we define $R$ using the parametric form $R(\cos 	heta, \sin 	heta)$, where $0 < 	heta < \pi$. This choice is a strategic move that simplifies our trigonometric expressions.

Consider the second circle, $C_2$, centered at $Q(1, 0)$. This circle passes through $R$, so its radius is the distance $QR$. Since $S$ also lies on this circle and on the line segment $PQ$, the distance $QS$ must also be equal to the radius $QR$.

We have just discovered that $	riangle QSR$ is an **isosceles triangle**.

### The Distance and the Angle

Let us calculate the length of $QR$ using the distance formula between $Q(1, 0)$ and $R(\cos 	heta, \sin 	heta)$:

$$QR = \sqrt{(\cos 	heta - 1)^2 + \sin^2 	heta}$$

Expanding this, we find $QR^2 = \cos^2 	heta - 2 \cos 	heta + 1 + \sin^2 	heta$. Since $\cos^2 	heta + \sin^2 	heta = 1$, this simplifies to $2 - 2 \cos 	heta$.

Using the half-angle identity $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$, we find:

$$QR = 2 \sin\left(\frac{	heta}{2}ight)$$

Because $QS = QR$, we now have two sides of our triangle. By analyzing the geometry, the included angle $\angle RQS$ is found to be $\frac{\pi - 	heta}{2}$.

### The Calculus of Optimization

We assemble our area formula $A = \frac{1}{2} \cdot QS \cdot QR \cdot \sin(\angle RQS)$. Substituting our values, we get:

$$A = \frac{1}{2} \cdot \left(2 \sin\left(\frac{	heta}{2}ight)ight)^2 \cdot \sin\left(\frac{\pi - 	heta}{2}ight)$$

This simplifies to $A = 2 \sin^2(\frac{	heta}{2}) \cos(\frac{	heta}{2})$. Let $t = \frac{	heta}{2}$, so our area function becomes $A(t) = 2 \sin^2 t \cos t$.

To maximize this, we differentiate with respect to $t$:

$$\frac{dA}{dt} = 2(2 \sin t \cos^2 t - \sin^3 t)$$

Setting this to zero, we get $2 \cos^2 t = \sin^2 t$, or $	an^2 t = 2$. This implies $	an t = \sqrt{2}$, which leads to $\sin t = \sqrt{\frac{2}{3}}$ and $\cos t = \sqrt{\frac{1}{3}}$.

### The Final Triumph

Substituting these values back into our area function, we calculate the maximum area:

$$A_{max} = 2 \cdot \left(\sqrt{\frac{2}{3}}ight)^2 \cdot \sqrt{\frac{1}{3}} = 2 \cdot \frac{2}{3} \cdot \frac{1}{\sqrt{3}} = \frac{4}{3\sqrt{3}}$$

Rationalizing the denominator, we arrive at the final, elegant answer:

**$$\frac{4\sqrt{3}}{9} 	ext{ square units}$$**

Visualized Solution

Visualizing the Setup

Parametric Coordinates of $R$

The Second Circle $C_{2}$

Calculating the Radius $QR$

Simplifying the Radius

Finding Length $QS$

Finding Angle $∠ R QS$

Area Formula Setup

Simplifying the Area Function

Differentiating for Maxima

Solving for Critical Point

Calculating Maximum Area

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Parametric Elegance

The Distance and the Angle

The Calculus of Optimization

The Final Triumph

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