**Problem 2 of the Canadian Mathematical Olympiad 1981**

Given a circle of radius r and a tangent line l to the circle through a given point P on the circle. From a variable point R on the circle, a perpendicular RQ is drawn to l with Q on l. Determine the maximum of the area of triangle PQR.

**Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Devan Vo)**

This problem is being solved using college calculus.

Let O be the center of the circle. Link and extend PO to meet the circle at S. Now let ε = ∠QPR and θ = ∠QRP. We then also have ε = ∠PSR and θ = ∠RPS. Let d = PR and denote (PQR) the area of triangle PQR. We have (PQR) = ½ PQ × RQ.

But PQ = d·sinθ and RQ = d·sinε and (PQR) = ½ d²·sinθ sinε (*)

In triangle PRS sinε = d/(2r) and sinθ = RS/(2r). Substitute them into (*), we have (PQR) = d³·RS/(8r²) (**)

With radius r being a constant, to find the maximum value of (PQR) we now need to find the maximum value of function f_d = d³ RS, and we’re stuck. But there’s a trick. Recall that a function reaches its extreme (maximum or minimum) point when its derivative is zero, but the function f_d above has two variables d and RS. Now let’s try to reduce it to a single variable d by relating RS to variable d and to eliminate RS. We have RS = \sqrt{4r² - d²} so now f_d = d³ \sqrt{4r² - d²} and f_d’ is the derivative of f_d with respect to the changing variable d, and it is the derivative of the product of two differentiable functions. We have the formula

D_x(u·v) = u·D_xv + v·D_xu

Therefore, f_d’ = [d³ \sqrt{4r² - d²}]’ = d³ [ \sqrt{4r² - d²}]’ + [ \sqrt{4r² - d²}] (d³)’

= [ ½d³/ \sqrt{4r² - d²}] D_d(4r² - d²)+ 3d² \sqrt{4r² - d²} = [½d³/ \sqrt{4r² - d²}](0 – 2d)+3d² \sqrt{4r² - d²}

= - d^4/ \sqrt{4r² - d²} + 3d² \sqrt{4r² - d²}

The derivative f_d’ = 0 when d^4/ \sqrt{4r² - d²} = 3d² \sqrt{4r² - d²} or when d² = 3 (4r^{2} - d^{2}) or when d² = 3r² or d = r \sqrt{3}.

We know the minimum of (PQR) occurs when it’s a degenerate triangle either by having R at P (R ≡ P and d = 0) or R at S (R ≡ S and d = 2r) and (PQR) = 0. Neither is the case when d = r \sqrt{3} when (PQR) is maximum.

When d = r \sqrt{3} , sinε = d/(2r) = \sqrt{3}/2, i.e., ε = 60° as seen on the graph, and

### Second solution

Let D be the diameter of the circle. Then

PR = D \cdot cos\theta

QP = PR\cdot sin\theta = D cos\theta sin\theta

QR = PR\cdot cos\theta = D cos^{2}\theta.

Twice the area of the triangle PQR equals

QP·QR = D cos^{3}\theta sin\theta = D\cdot f(\theta).

The question is thus reduced to maximizing function f(\theta ), for 0 \lt \theta \lt \pi /2 .

But f'(\theta) = cos^{4}\theta - 3cos^{2}\theta sin^{2}\theta.

The only 0 in the above range is attained when cos^{2}\theta - 3sin^{2}\theta = 1 - 4sin^{2} \theta = 0, i.e. for sin \theta = 1/2, meaning \theta = 30^\circ. As the other condidates are \theta = 0 or \theta = \pi/2, for both of which f(\theta) = 0, \theta = 30^\circ is the only possible maximum.