Question

Let A be the point (0,a), and B be the point (0,a+b) on the xy-plane. Let P denote a variable point (x,0), where x > 0. Find the value of x in terms of a and b that gives the largest angle ∠APB.

Answer 1

Answer:

  x = √(a(a+b))

Step-by-step explanation:

We can also assume a > 0 and b > 0 without loss of generality. (If a and a+b have opposite signs, the maximum angle is 180° at x=0.)

We choose to define tan(α) = -(b+a)/x and tan(β) = -a/x. Then the tangent of ∠APB is ...

  tan(∠APB) = (tan(α) -tan(β))/(1 +tan(α)tan(β))

  = ((-(a+b)/x) -(-a/x))/(1 +(-(a+b)/x)(-a/x))

  = (-bx)/(x^2 +ab +a^2)

This will be maximized when its derivative is zero.

  d(tan(∠APB))/dx = ((x^2 +ab +a^2)(-b) -(-bx)(2x))/(x^2 +ab +a^2)^2

The derivative will be zero when the numerator is zero, so we want ...

  bx^2 -ab^2 -a^2b = 0

  b(x^2 -(a(a+b))) = 0

This has solutions ...

  b = 0

  x = √(a(a+b))

The former case is the degenerate case where ∠APB is 0, and the value of x can be anything.

The latter case is the one of interest:

  x = √(a(a+b)) . . . . . . the geometric mean of A and B rotated to the x-axis.

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Comment on the result

This result is validated by experiments using a geometry program. The location of P can be constructed in a few simple steps: Construct a semicircle through the origin and B. Find the intersection point of that semicircle with a line through A parallel to the x-axis. The distance from the origin to that intersection point is x.