11.16. The lengths of three pairwise perpendicular segments $O A, O B$, and $O C$ are $a, b$, and $c$, respectively, with $a \leqslant b \leqslant 208 \leqslant c$. What is the greatest and the least value that the sum of the distances from points $A, B$, and $C$ to a line $l$ passing through point $O$ can take?

## § 4. Various Problems
11.16. Let the angles between the line $l$ and the lines $O A, O B$ and $O C$ be $\alpha, \beta$ and $\gamma$. Then $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$ (Problem 1.21), and therefore, $\quad \sin ^{2} \alpha+\sin ^{2} \beta+$ $+\sin ^{2} \gamma=2$. The sum of the distances from points $A, B$ and $C$ to the line $l$ is $a \sin \alpha+$ $+b \sin \beta+c \sin \gamma$. Let $x=$ $=\sin \alpha, y=\sin \beta, z=\sin \gamma$. The problem requires finding the maximum and minimum values of the quantity $a x+b y+c z$ under the following conditions: $x^{2}+y^{2}+z^{2}=2, \quad 0 \leqslant x, \quad y$, $z \leqslant 1$. These conditions define a curved triangular region on the surface of the sphere $x^{2}+$ $+y^{2}+z^{2}=2$ 

![](https://cdn.mathpix.com/cropped/2024_05_21_3f2bc5f5a1af26514cfag-216.jpg?height=414&width=489&top_left_y=689&top_left_x=507)

Fig. 83 triangle (Fig. 83).

Let the plane $a x+b y+c z=p_{0}$ be tangent to the surface of the sphere $x^{2}+y^{2}+z^{2}=2$ at the point $M_{\hat{\delta}}$ with coordinates ( $x_{0}, y_{0}, z_{0}$ ), where $x_{0}, y_{0}, z_{0} \geqslant 0$. Then $x_{0}=\lambda . a, y_{0}=\lambda b, z_{0}=\lambda c$ and $\lambda^{2}\left(a^{2}+\right.$ $\left.+b^{2}+c^{2}\right)=2$, and $p_{0}=\gamma\left(a^{2}+b^{2}+c^{2}\right)=\sqrt{2\left(a^{2}+b^{2}+c^{2}\right)}$. If $z_{0} \leqslant 1$ (i.e., $c^{2} \leqslant a^{2}+b^{\underline{b}}$), then $M_{0}$ belongs to the specified curved triangular region, and thus, in this case, $p_{0}$ is the maximum value of the function $a x+b y+c z$. Now let $z_{0}>1$, i.e., $c^{2}>a^{2}+b^{2}$. The plane $a x+b y+$ $+c z=p$, where $p<p_{6}$, intersects the considered sphere in a circle. We are interested in the values of $p$ for which this circle does not intersect the specified curved triangular region. The maximum of such $p$ corresponds to the value $z_{0}^{\prime}=1$. Finding $x_{0}^{\prime}$ and $y_{0}^{\prime}$ reduces to the problem: for what $x$ and $y$ does the expression $a x+b y$ achieve its maximum value if $x^{2}+$ $+y^{2}=1$. It is easy to verify that $x_{0}^{\prime}=a / \sqrt{a^{2}+b^{2}}$ and $y_{0}^{\prime}=$ $=b / \sqrt{a^{2}+b^{2}}$, i.e., in this case, the sought maximum value of $p$ is $\sqrt{a^{2}+b^{2}}+c$.

We will now show that the minimum value of the expression $a x+$ $+b y+c z$ on the specified triangle is achieved at the vertex $x_{1}=y_{1}=1, z_{1}=0$. Indeed, since $0 \leqslant x, y, z \leqslant 1$, then $x+y+z \geqslant x^{2}+y^{2}+z^{2}=2$, and thus, $y+z-1 \geqslant 1-x$. Both sides of this inequality are non-negative, so $b(y+$ $+z-1) \geqslant a(1-x)$. Therefore, $a x+b y+c z \geqslant a x+$ $+b y+b z \geqslant a+b$.