Miscellaneous · Q9: Shortest hypotenuse through a point at distances $a$⁠, $b$ from the legs: $\left(a^{2/3}+b^{2/3}\right)^{3/2}$

Basics you need

Solution

1. Set up the figure and choose the variable

   Let the right angle be at $C$⁠, with the legs along $CA$ and $CB$⁠. Drop perpendiculars from $P$⁠: $PM$ to $CB$ and $PN$ to $CA$⁠. Then

   $$PM = a, \qquad PN = b.$$

   All the triangles in the question share the right angle at $C$ and the point $P$ (it stays at distances $a$ and $b$ from the two legs). What changes is the direction of the hypotenuse through $P$⁠. Measure it by

   $$\theta = \angle CAB,$$

   the angle between the hypotenuse and the leg $CA$⁠.

   Once $\theta$ is chosen, the line through $P$ is fixed, so $A$⁠, $B$ and the length $AB$ are fixed too. That makes $AB$ a function of the single variable $\theta$⁠, which is what the derivative tests need.
   Distance from a point to a line is measured along the perpendicular. Basics: building the function

2. The allowed range of $\theta$

   $$0 < \theta < \frac{\pi}{2}.$$

   Every $\theta$ in this range gives a genuine triangle: the line through $P$ going down to the right at angle $\theta$ meets the ray $CA$ at a point $A$ beyond $N$⁠, and going up to the left meets the ray $CB$ at a point $B$ beyond $M$⁠.

   $\theta$ is an acute angle of a right triangle, so it lies strictly between 0 and $\frac{\pi}{2}$⁠. At $\theta = 0$ the line through $P$ would be parallel to $CA$ and never meet it; at $\theta = \frac{\pi}{2}$ it would be parallel to $CB$⁠. Neither end gives a triangle, so the range is open.
   The two acute angles of a right triangle add up to $\frac{\pi}{2}$⁠, so each lies in $\left(0, \frac{\pi}{2}\right)$⁠. Basics: interval notation

3. The hypotenuse as a function of $\theta$

   The piece $PA$⁠. Triangle $PNA$ is right-angled at $N$⁠, with angle $\theta$ at $A$⁠. The side opposite $\theta$ is $PN = b$ and the hypotenuse is $PA$⁠, so

   $$\sin\theta = \frac{b}{PA} \quad\Longrightarrow\quad PA = \frac{b}{\sin\theta}.$$

   The piece $PB$⁠. $PM$ is parallel to $CA$ (both are perpendicular to $CB$⁠), so the hypotenuse makes the same angle $\theta$ with $PM$⁠: $\angle BPM = \theta$⁠. Triangle $BMP$ is right-angled at $M$⁠; the side next to $\theta$ is $PM = a$ and the hypotenuse is $PB$⁠, so

   $$\cos\theta = \frac{a}{PB} \quad\Longrightarrow\quad PB = \frac{a}{\cos\theta}.$$

   $P$ lies between $A$ and $B$⁠, so

   $$L(\theta) = AB = PB + PA = \frac{a}{\cos\theta} + \frac{b}{\sin\theta},$$

   that is, $L(\theta) = a\sec\theta + b\,\mathrm{cosec}\,\theta$ for $0 < \theta < \frac{\pi}{2}$⁠.

   Splitting $AB$ at $P$ turns one awkward length into two easy ones, each the hypotenuse of a small right triangle in which we know one side ($a$ or $b$⁠) and the angle $\theta$⁠. The equal angles come from parallel lines cut by the hypotenuse (corresponding angles).
   In a right triangle, hypotenuse = opposite $\div \sin\theta$ = adjacent $\div \cos\theta$⁠. Basics: sin and cos

4. Differentiate

   Using the reciprocal rule (Basic 3) on each term:

   $$\begin{aligned} L'(\theta) &= \frac{a\sin\theta}{\cos^2\theta} - \frac{b\cos\theta}{\sin^2\theta} \\ &= \frac{a\sin^3\theta - b\cos^3\theta}{\sin^2\theta\,\cos^2\theta}. \end{aligned}$$

   $L$ is differentiable at every $\theta$ in $\left(0, \frac{\pi}{2}\right)$⁠, because $\sin\theta \ne 0$ and $\cos\theta \ne 0$ there.

   Putting the two fractions over the common denominator $\sin^2\theta\cos^2\theta$ leaves a single numerator, $a\sin^3\theta - b\cos^3\theta$⁠, and the denominator is positive on the whole range. So the sign of $L'$ is decided by the numerator alone, which is what Step 6 needs.
   $\frac{d}{d\theta}(\cos\theta)^{-1} = -(\cos\theta)^{-2}\cdot(-\sin\theta)$⁠: chain rule with inner function $\cos\theta$ and outer function $u^{-1}$ (Theorem 4, Section 5.3.1). Basics: the chain rule

5. The only critical point

   $$\begin{aligned} L'(\theta) = 0 &\iff a\sin^3\theta = b\cos^3\theta \\ &\iff \tan^3\theta = \frac{b}{a} \\ &\iff \tan\theta = \left(\frac{b}{a}\right)^{1/3}. \end{aligned}$$

   On $\left(0, \frac{\pi}{2}\right)$⁠, $\tan\theta$ increases strictly from 0 towards $+\infty$ and takes every positive value exactly once. So there is exactly one critical point $\theta_0$⁠, with

   $$\tan\theta_0 = \frac{b^{1/3}}{a^{1/3}}.$$
   Dividing by $\cos^3\theta$ is allowed because $\cos\theta > 0$ on the range. Since $L$ is differentiable everywhere on the range (Step 4), points where $L' = 0$ are the only critical points (Note after Theorem 2, Section 6.4).
   $\frac{\sin^3\theta}{\cos^3\theta} = \tan^3\theta$⁠, and the real cube root of a positive number is a single positive number. Basics: solving equations

6. It is the least value on the whole range

   Take $\cos^3\theta$ out of the numerator:

   $$L'(\theta) = \frac{\cos^3\theta\,\left(a\tan^3\theta - b\right)}{\sin^2\theta\,\cos^2\theta}.$$

   On $\left(0, \frac{\pi}{2}\right)$ the factors $\cos^3\theta$ and $\sin^2\theta\cos^2\theta$ are positive, so $L'$ has the sign of $a\tan^3\theta - b$⁠:

   • for $0 < \theta < \theta_0$⁠: $\tan\theta < \tan\theta_0$⁠, so $a\tan^3\theta < b$ and $L'(\theta) < 0$⁠;
   • for $\theta_0 < \theta < \frac{\pi}{2}$⁠: $\tan\theta > \tan\theta_0$⁠, so $a\tan^3\theta > b$ and $L'(\theta) > 0$⁠.

   By the First Derivative Test (Theorem 3 (ii)), $\theta_0$ is a point of local minima. More than that: for any $\theta < \theta_0$⁠, $L$ is continuous on $[\theta, \theta_0]$ with $L' < 0$ inside, so $L(\theta) > L(\theta_0)$⁠; for any $\theta > \theta_0$⁠, $L' > 0$ on $(\theta_0, \theta)$ gives $L(\theta) > L(\theta_0)$⁠. Hence

   $$L(\theta) > L(\theta_0)\ \text{ for every other } \theta \text{ in } \left(0, \tfrac{\pi}{2}\right),$$

   and $L(\theta_0)$ is the minimum length of the hypotenuse.

   A local test only says $L(\theta_0)$ is smallest near $\theta_0$⁠. The question asks for the minimum over all such triangles, so we need the sign of $L'$ on the whole range: decreasing all the way up to $\theta_0$⁠, increasing all the way after. There are no endpoints to compare with, because the range is open, and near each end $L$ blows up: as $\theta \to 0^+$⁠, $\frac{b}{\sin\theta} \to \infty$⁠; as $\theta \to \frac{\pi}{2}^-$⁠, $\frac{a}{\cos\theta} \to \infty$⁠.
   Theorem 1 (b), Section 6.3, applied on $[\theta, \theta_0]$⁠. The strict $L(\theta) > L(\theta_0)$ comes from the book's Mean Value argument for part (a) with the signs reversed: $L(\theta_0) - L(\theta)$ $= L'(c)(\theta_0 - \theta) < 0$ for some $c$ between $\theta$ and $\theta_0$⁠. Lesson: increasing and decreasing functions

7. Evaluate $L(\theta_0)$

   Draw a right triangle with angle $\theta_0$⁠, opposite side $b^{1/3}$ and adjacent side $a^{1/3}$ (that is what $\tan\theta_0 = \frac{b^{1/3}}{a^{1/3}}$ says). Its hypotenuse is $\sqrt{a^{2/3}+b^{2/3}}$⁠. Write $S = a^{2/3}+b^{2/3}$⁠. Then

   $$\cos\theta_0 = \frac{a^{1/3}}{\sqrt{S}}, \qquad \sin\theta_0 = \frac{b^{1/3}}{\sqrt{S}}.$$

   Substitute into $L$⁠:

   $$\begin{aligned} L(\theta_0) &= \frac{a\sqrt{S}}{a^{1/3}} + \frac{b\sqrt{S}}{b^{1/3}} \\ &= \sqrt{S}\left(a^{2/3} + b^{2/3}\right) \\ &= S^{1/2}\cdot S = S^{3/2}. \end{aligned}$$

   So the minimum length of the hypotenuse is

   $$\left(a^{2/3}+b^{2/3}\right)^{3/2}.$$
   $\frac{a}{a^{1/3}} = a^{1 - 1/3} = a^{2/3}$⁠, and likewise for $b$⁠. The sum $a^{2/3}+b^{2/3}$ is exactly $S$ again, so the answer collapses to $S \cdot S^{1/2} = S^{3/2}$⁠. The triangle trick avoids messy formulas for $\sec\theta_0$ and $\mathrm{cosec}\,\theta_0$⁠; both $\sin\theta_0$ and $\cos\theta_0$ are positive because $\theta_0$ is acute.
   Laws of indices: $\frac{a}{a^{1/3}} = a^{2/3}$ and $S^{1/2}\cdot S = S^{3/2}$⁠. Pythagoras: $\left(a^{1/3}\right)^2 + \left(b^{1/3}\right)^2$ $= a^{2/3}+b^{2/3}$⁠.

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