Use the derivative to find the size of an angle of a right triangle so that the radius of the inscribed circle is maximum, for a constant hypotenuse.
In what follows, \( C \) represents the constant of integration where applicable.
Problem
ABC is a right triangle and \( r \) is the radius of the inscribed circle.
a) Express \( r \) in terms of angle \( x \) and the length of the hypotenuse \( h \).
b) Assume that \( h \) is constant and \( x \) varies; find \( x \) for which \( r \) is maximum.
Show Solution to Problem
a) Let M, N, and P be the points of tangency of the circle with the sides of the triangle. \( OM \), \( ON \), and \( OP \) are perpendicular to \( CB \), \( CA \), and \( AB \) respectively.
Triangles COM and CON are right triangles and have two congruent sides (\( CO \) and radii \( OM = ON = r \)); the two triangles are therefore congruent. We denote the size of angle \( MCN \) by \( x \) and write:
\[ \tan\left(\frac{x}{2}\right) = \frac{r}{CM} \]Triangles BOM and BOP are right triangles and have two congruent sides (\( BO \) and radii \( OM = OP = r \)); the two triangles are therefore congruent. We denote the size of angle \( MBP \) by \( y \) and write:
\[ \tan\left(\frac{y}{2}\right) = \frac{r}{BM} \]Note that \( y + x = 90^\circ \), which gives \( \dfrac{y}{2} = 45^\circ - \dfrac{x}{2} \).
Substitute \( \dfrac{y}{2} \) into the equation for \( BM \):
\[ \tan\left(45^\circ - \frac{x}{2}\right) = \frac{r}{BM} \]Solving for \( CM \) and \( BM \):
\[ CM = \frac{r}{\tan\left(\dfrac{x}{2}\right)}, \quad BM = \frac{r}{\tan\left(45^\circ - \dfrac{x}{2}\right)} \]Using the fact that \( h = CM + BM \):
\[ h = \frac{r}{\tan\left(\dfrac{x}{2}\right)} + \frac{r}{\tan\left(45^\circ - \dfrac{x}{2}\right)} = r \left[ \frac{1}{\tan\left(\frac{x}{2}\right)} + \frac{1}{\tan\left(45^\circ - \frac{x}{2}\right)} \right] \]Using the trigonometric identity for tangent difference:
\[ \tan\left(45^\circ - \frac{x}{2}\right) = \frac{\tan(45^\circ) - \tan\left(\frac{x}{2}\right)}{1 + \tan(45^\circ)\tan\left(\frac{x}{2}\right)} = \frac{1 - \tan\left(\frac{x}{2}\right)}{1 + \tan\left(\frac{x}{2}\right)} \]Substitute this back into the formula for \( h \):
\[ h = r \left[ \frac{1}{\tan\left(\frac{x}{2}\right)} + \frac{1 + \tan\left(\frac{x}{2}\right)}{1 - \tan\left(\frac{x}{2}\right)} \right] \]Wait, let's check the reciprocal term:
\[ \frac{1}{\tan\left(45^\circ - \frac{x}{2}\right)} = \frac{1 + \tan\left(\frac{x}{2}\right)}{1 - \tan\left(\frac{x}{2}\right)} \]Using \( \tan\left(\dfrac{x}{2}\right) = \dfrac{\sin(x/2)}{\cos(x/2)} \), we simplify to:
\[ h = r \left[ \frac{\cos\left(\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} + \frac{\cos\left(\frac{x}{2}\right) + \sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)} \right] = r \left[ \frac{1}{\sin\left(\frac{x}{2}\right) \left(\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)\right)} \right] \]Solving for \( r \):
\[ r = h \sin\left(\frac{x}{2}\right) \left(\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)\right) = h \left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right)\right) \]Using half-angle and product identities (\( \sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) = \dfrac{1}{2}\sin x \) and \( \sin^2\left(\frac{x}{2}\right) = \dfrac{1 - \cos x}{2} \)):
\[ r = \frac{h}{2} (\sin x + \cos x - 1) \]b) Assume \( h \) is constant. Find the first derivative of \( r \) with respect to \( x \):
\[ \frac{dr}{dx} = \frac{h}{2} (\cos x - \sin x) \]Set \( \dfrac{dr}{dx} = 0 \) to find critical points:
\[ \frac{h}{2} (\cos x - \sin x) = 0 \implies \cos x = \sin x \]Since \( x \) is an acute angle in the right triangle, solving \( \cos x = \sin x \) gives:
\[ x = \frac{\pi}{4} = 45^\circ \]
Since \( \dfrac{dr}{dx} > 0 \) for \( x < \dfrac{\pi}{4} \) and \( \dfrac{dr}{dx} < 0 \) for \( x > \dfrac{\pi}{4} \), \( r \) has an absolute maximum at \( x = \dfrac{\pi}{4} = 45^\circ \). In this case, triangle ABC is isosceles.