A point P is given on the circumference of a circle of radius r. Chord QR is parallel to the tangent

Q: Why do we maximize $A^2$ instead of $A$?

Great question! Differentiating the square root function $\sqrt{r^2 - k^2}$ can lead to messy algebraic expressions. Since $A$ is always positive, maximizing $A^2$ is mathematically equivalent to maximizing $A$, but it simplifies the derivative significantly, making the calculation much cleaner.

Q: Why do we reject the root $k = r$?

If $k = r$, the chord $QR$ coincides with the tangent at $P$. In this case, the points $Q$ and $R$ both merge with $P$, and the triangle $PQR$ collapses into a single point. A point has an area of $0$, which is the minimum, not the maximum we are looking for.

Q: Is there a geometric way to solve this without calculus?

Yes, absolutely! You can use trigonometry by expressing the coordinates of $Q$ and $R$ in terms of an angle $\theta$ from the center. The area becomes a function of $\theta$, and you can maximize it using trigonometric identities. Calculus is just one of the many powerful tools in your arsenal.

Question

A point $P$ is given on the circumference of a circle of radius $r$. Chord $QR$ is parallel to the tangent at $P$. Determine the maximum possible area of the triangle $PQR$.

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of Symmetry

Imagine you are standing on the edge of a perfect circle of radius $r$. You place a point $P$ at the very top. Now, you want to draw a chord $QR$ parallel to the tangent at $P$.

Because the chord is parallel to the horizontal tangent, the triangle $PQR$ must be isosceles. The vertical axis passing through $P$ acts as a mirror, splitting our triangle into two identical right-angled triangles. This symmetry is our greatest ally.

### The Coordinate Setup

To unlock the secrets of this triangle, we place our circle on the Cartesian plane, centered at the origin $(0,0)$. The equation of our circle is $x^2 + y^2 = r^2$.

With $P$ at the top, its coordinates are $(0, r)$. The tangent at $P$ is the horizontal line $y = r$. Since the chord $QR$ is parallel to this tangent, it must be a horizontal line defined by $y = k$, where $k$ is a variable height.

### The Area Function

To find the area, we need the base and the height. The base is the length of the chord $QR$. Substituting $y = k$ into $x^2 + y^2 = r^2$, we find $x^2 = r^2 - k^2$, so $x = \pm\sqrt{r^2 - k^2}$.

The base length is $2\sqrt{r^2 - k^2}$. The height of the triangle is the vertical distance from the chord at $y = k$ to the point $P$ at $y = r$, which is $r - k$.

Thus, the area $A$ is given by:

$$A(k) = \frac{1}{2} \cdot (2\sqrt{r^2 - k^2}) \cdot (r - k) = (r - k)\sqrt{r^2 - k^2}$$

### The Optimization Trick

Now, we face a square root in our derivative. Instead of dealing with that, let us maximize the square of the area, $f(k) = [A(k)]^2 = (r - k)^2(r^2 - k^2)$.

This is a brilliant move because it eliminates the radical. We can factor $r^2 - k^2$ as $(r - k)(r + k)$, transforming our function into:

$$f(k) = (r - k)^3(r + k)$$

### The Calculus Masterclass

Now, we differentiate $f(k)$ with respect to $k$ using the product rule:

$$f'(k) = 3(r - k)^2(-1)(r + k) + (r - k)^3(1)$$

Factoring out $(r - k)^2$, we get:

$$f'(k) = (r - k)^2 [-3(r + k) + (r - k)] = (r - k)^2 (-2r - 4k)$$

Setting $f'(k) = 0$, we find $k = r$ (which gives zero area) or $k = -\frac{r}{2}$.

### The Equilateral Revelation

Substituting $k = -\frac{r}{2}$ back into our area formula, we get:

$$A_{max} = \left(r - \left(-\frac{r}{2}\right)\right)\sqrt{r^2 - \left(-\frac{r}{2}\right)^2} = \left(\frac{3r}{2}\right)\sqrt{\frac{3r^2}{4}} = \frac{3\sqrt{3}r^2}{4}$$

This is the maximum area! Interestingly, this occurs when the triangle is equilateral. Symmetry has guided us to the optimal solution once again.

Animated Solution for Mathematics - Differentiation: A point $P$ is given on the circumference of a circle of radius $r$ . Chord $QR$ is parallel to the tangent at $P$ . Determine the maximum possible area of the triangle $P QR$ .

Visualized Solution

Visualizing the Setup

Coordinate Geometry Setup

Defining Chord $QR$

Coordinates of $Q$ and $R$

Dimensions of $△ P QR$

Area Function $A (k)$

Squaring the Area Function

Differentiating $f (k)$

Finding Critical Points

Maximum Area Calculation

Final Result & Geometric Insight

The Sigma Insight: Maxima and Minima

Analyzing the Geometry of Symmetry

The Coordinate Setup

The Area Function

The Optimization Trick

The Calculus Masterclass

The Equilateral Revelation

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