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Give a vector based proof of Heron’s formula for computing the area of a triangle

by RoRi

Let S denote the area of a triangle with sides of lengths a,b,c. Heron’s formula states that

    \[ S = \sqrt{s (s-a)(s-b)(s-c)}, \qquad \text{where} \qquad s = \frac{a+b+c}{2}. \]

We prove this formula using vectors via the following steps. Assume the triangle has vertices O,A, and B with

    \[ \lVert A \rVert = a, \qquad \lVert B \rVert = b, \qquad \lVert B - A \rVert = c. \]

  1. Using the identities

        \[ \lVert A \times B \rVert^2 = \lVert A \rVert^2 \lVert B \rVert^2 - (A \cdot B)^2, \qquad -2A \cdot B = \lVert A-B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 \]

    prove the formula

        \[ 4S^2 = a^2 b^2 - \frac{1}{4} (c^2 - a^2 - b^2)^2 = \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2). \]

  2. Simplify the formula in part (a) to obtain the formula

        \[ S^2 = \frac{1}{16}(a+b+c)(a+b-c)(c-a+b)(c+a-b), \]

    and use this to deduce Heron’s formula.

  1. Proof. We know the area of the triangle (in terms of the vectors A and B) is

        \[ S = \frac{1}{2} \lVert A \times B \rVert. \]

    Therefore,

        \begin{align*} 4S^2 &= \lVert A \times B \rVert^2 \\ &= \lVert A \rVert^2 \lVert B \rVert^2 - (A \cdot B)^2 \\ &= a^2 b^2 - \frac{1}{4} ( \lVert A-B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 ) \\ &= a^2 b^2 - \frac{1}{4} (c^2 - a^2 - b^2)^2 \\ &= \frac{1}{4} (4a^2 b^2 - (c^2 - a^2 - b^2)^2) \\ &= \frac{1}{4} (2ab - (c^2 - a^2 - b^2))(2ab + (c^2 - a^2 - b^2)) \\ &= \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2). \qquad \blacksquare \end{align*}

  2. Proof. We simplify the formula in part (a),

        \begin{align*} && 4S^2 &= \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2) \\ \implies && S^2 &= \frac{1}{16} ((a+b)^2 - c^2)(c^2 - (a-b)^2) \\ \implies && S^2 &= \frac{1}{16} (a+b+c)(a+b-c)(c-a+b)(c+a-b) \\ \implies && S^2 &= \frac{a+b+c}{2} \cdot \frac{a+b-c}{2} \cdot \frac{c-a+b}{2} \cdot \frac{c+a-b}{2} \\[9pt] \implies && S^2 &= s(s-c)(s-a)(s-b) \\ \implies && S &= \sqrt{s (s-a)(s-b)(s-c)}. \qquad \blacksquare \end{align*}

Find a function whose graph has given properties

by RoRi

Let f(x) be a function on the interval [0,1] that is nonnegative and differentiable, and such that f(1) = 0. If for each a in the open interval (0,1) the line given by the equation x = a divides the ordinate set of f(x) into regions A and B where A denotes the leftmost region. If the areas of the two regions obey the equation

    \[ A - B = 2f(a) + 3a + b, \]

where b is a constant that does not depend on a, find f(x) and b.

Incomplete.

Compute the area of regions enclosed by a circular sector and tangent lines

by RoRi

In the figure below let S(X) be the area of the shaded region and T(x) the area of the triangle ABC. Compute

  1. T(x);
  2. S(x);
  3. \displaystyle{ \lim_{x \to 0^+} \frac{T(x)}{S(x)}}.

    4. We have the following diagram, with some additional markings from the diagram in Apostol,

    Rendered by QuickLaTeX.com

    1. First, to calculate the area of the triangle ABC we calculate the area of the polygon OACB and subtract the area of the triangle OAB. Since the polygon OACB is two congruent triangles (since the line from O to C bisects the polygon into two equal sized triangles) we have

          \begin{align*} \operatorname{Area}(OABC) &= 2 \cdot \operatorname{Area}(OAC) \\ &= 2 \cdot \left( \frac{1}{2} \right) (1) \left( \tan \frac{x}{2} \right) \\ &= \tan \frac{x}{2}. \end{align*}

      To calculate the area of the triangle OAB we use the auxiliary point D. The line from B to D is the height of the triangle OAB and it has length \sin x. Therefore,

          \[ \operatorname{Area}(OAB) = \left( \frac{1}{2} \right) (1) (\sin x) = \frac{1}{2} \sin x. \]

      Hence,

          \[ \operatorname{Area}(ABC) = \tan \frac{x}{2} - \frac{1}{2} \sin x. \]

    2. The area of the shaded region S(x) is the area of circular sector minus the area of the triangle OAB (which we calculated in part (a) as \frac{1}{2} \sin x). The area of the sector is given by

          \[ \pi r^2 \cdot \frac{x}{2 \pi} = \frac{1}{2} x. \]

      Hence,

          \[ S(x) = \frac{1}{2}x - \frac{1}{2} \sin x. \]

    3. For this we take the limit using our expressions for T(x) and S(x) obtained in parts (a) and (b) and use L’Hopital’s,

          \begin{align*} \lim_{x \to 0^+} \frac{T(x)}{S(x)} &= \lim_{x \to 0^+} \frac{ \tan \frac{x}{2} - \frac{1}{2} \sin x}{\frac{1}{2} x - \frac{1}{2} \sin x} \\[9pt] &= \lim_{x \to 0^+} \frac{ \frac{1}{2} \sec^2 \frac{x}{2} - \frac{1}{2} \cos x}{\frac{1}{2} - \frac{1}{2} \cos x} \\[9pt] &= \lim_{x \to 0^+} \frac{ \frac{1}{\cos^2 \frac{x}{2}} - \cos x}{1 - \cos x}. \end{align*}

      Then, we note that \cos^2 \frac{x}{2} = \frac{1}{2} (\cos x + 1) (from the trig identity \cos (2x) = \cos^2 x - 1). Therefore,

          \begin{align*} \lim_{x \to 0^+} \frac{T(x)}{S(x)} &= \lim_{x \to 0^+} \frac{ \frac{1}{\cos^2 \frac{x}{2}} - \cos x}{1 - \cos x} \\[9pt] &= \lim_{x \to 0^+} \frac{ \frac{2}{\cos x + 1} - \cos x}{1 - \cos x} \\[9pt] &= \lim_{x \to 0^+} \frac{2 - \cos^2 x - \cos x}{(1+\cos x)(1- \cos x)} \\[9pt] &= \lim_{x \to 0^+} \frac{1 - \cos^2 x + 1 - \cos x}{(1+\cos x)(1- \cos x)} \\[9pt] &= \lim_{x \to 0^+} \frac{(1- \cos x)(1+\cos x) + (1 - \cos x)}{(1+\cos x)(1 - \cos x)} \\[9pt] &= \lim_{x \to 0^+} \frac{1 + \cos x + 1}{1 + \cos x} \\[9pt] &= \lim_{x \to 0^+} \frac{2+ \cos x}{1+ \cos x} \\[9pt] &= \frac{3}{2}. \end{align*}