Example 79. A sphere touches the base of a regular triangular pyramid $S A B C$ at point $B$ and its lateral edge $S A$. Let's find the radius of the sphere, if $|A B|=a$ and $|S A|=b$.

Construction of the image. Let the quadrilateral $S A B C$ with its diagonals (Fig. 88) be the image of the given pyramid. This image is complete and metrically defined (verify this independently). We cannot construct the image of the sphere since its radius is unknown (a kind of "vicious circle" arises: to construct the image of the sphere, we need to know its radius, and to find the radius, it is desirable to have the image of the sphere). Let's try to solve the problem using an image where the sphere is not constructed.

Given: $\quad S A B C$ - regular

![](https://cdn.mathpix.com/cropped/2024_05_21_318d1c34bf6979fc9790g-139.jpg?height=685&width=674&top_left_y=1665&top_left_x=357)

Fig. 88 pyramid, $S$ - vertex, $|A B|=a,|S A|=b, \Omega$ - sphere, $B$ - point of tangency of the sphere $\Omega$ with the plane $A B C, \Omega$ touches $[S A]$.

Find: radius of the sphere $\Omega$.
S o l u t i o n. First of all, note that if we have the image of the center of the sphere, point $P$, then $|P B|$ will be the desired length of the radius of the sphere $\Omega$. Thus, having an image of the sphere itself is not necessary for calculating its radius. We will construct point $P$ based on the following considerations.

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Since the sphere $\Omega$ touches the plane $A B C$ at point $B$, point $P$ lies on the perpendicular to the plane $A B C$ erected at point $B$. In our drawing, we already have the image of the height of the pyramid. Through point $B$, draw a line $m \|(S O)$. Then, since $(S O) \perp (A B C)$, we have $m \perp (A B C)$. Therefore, $P \in m$ and $[P B]$ is the radius of the sphere $\Omega$.

If the sphere touches the edge $S A$ at some point $D$, it is clear that $|A B|=|A D|$ (as the lengths of the tangent segments to the sphere $\Omega$ drawn from one point). But $|A B|=a$. Therefore, to construct point $D$, we need to lay off a segment $A D$ from point $A$ on the segment $S A$ such that $|A D|=a$. Using the fact that the lengths of the edges $A B$ and $S A$ are given in general form, take $D \in [S A]$ arbitrarily and assume that $|A D|=|A B|$. (If $a$ and $b$ are given specific numerical values, of course, we cannot take point $D$ arbitrarily. For example, if $a=15, b=20$, then $|A D|:|S A|=15:20$, or $|A D|:|S A|=3:4$, from which the construction of point $D$ is clear.)

Next, construct $[P D]$. Since the edge $S A$ touches the sphere $\Omega$, we have $[P D] \perp [S A]$ and $[P D]$ is the radius of the sphere $\Omega$, i.e., $|P D|=|P B|$. To calculate $|P B|$, perform the following constructions:

1) $[S P]$,
2) $[O B]$ - the segment of the median of $\triangle A B C$,
3) $[P K] \|[O B]$.

Let $|P B|=x$ and form an equation.

Since $|S A|=b$ and $|A D|=a$, we have $|S D|=b-a$.

From the right triangle $\triangle S D P$, we have:

$$
|S P|^{2}=(b-a)^{2}+x^{2} .
$$

From the right triangle $\triangle S B O$, we have:

$$
|B O|=\frac{a \sqrt{3}}{3} \text{ and } |S O|=\sqrt{b^{2}-\frac{a^{2}}{3}} .
$$

Since $m \|(S O)$ and $[P K] \|[O B]$, we have $|O B|=|P K|=\frac{a \sqrt{3}}{3}$ and $|P B|=|K O|=x$. Then $|S K|=\sqrt{b^{2}-\frac{a^{2}}{3}}-x$.

Now, from the right triangle $\triangle S P K$, we get:

$$
\begin{gathered}
|S P|^{2}=|S K|^{2}+|P K|^{2}, \text{ or } \\
(b-a)^{2}+x^{2}=\left(\sqrt{b^{2}-\frac{a^{2}}{3}}-x\right)^{2}+\left(\frac{a \sqrt{3}}{3}\right)^{2} .
\end{gathered}
$$

Solving this equation, we find:

$$
x=\frac{\sqrt{3} a(2 b-a)}{2 \sqrt{3 b^{2}-a^{2}}}
$$

Investigation. According to the problem, the sphere $\Omega$ touches the edge $S A$, i.e., $D \in [S A]$. Thus, by the meaning of the problem, $b>a$. It is not difficult to verify that if this inequality is satisfied, the inequality $x>0$ will also be satisfied.

Therefore, $|P B|=\frac{\sqrt{3} a(2 b-a)}{2 \sqrt{3 b^{2}-a^{2}}}$, where $b>a$.

Remark. We once again draw the reader's attention to the fact that the investigation is conducted based on the meaning of the problem, not on the formula obtained for the desired quantity. In the considered example, the investigation of the found value of $x$ according to the obtained formula would have boiled down to solving the system of inequalities:

$$
\left\{\begin{array}{l}
2 b-a>0, \\
3 b^{2}-a^{2}>0
\end{array}\right.
$$

(we omitted the inequalities $a>0$ and $b>0$ as self-evident). From this system, we would have obtained that $b>\frac{a \sqrt{3}}{3}$.

However, this relationship is sufficient only for the existence of the pyramid itself: in the right triangle $S A O$, the hypotenuse will be greater than the leg. For the existence of the sphere touching the edge $S A$, the relationship $b>\frac{a \sqrt{3}}{3}$ is insufficient. Indeed, let $b=0.8 a$ (in this case, $b>\frac{a \sqrt{3}}{3}$). As noted above, $|A D|=a$. Thus, we will have $|A D|>|S A|$, i.e., the sphere will touch not the segment $S A$, but its extension.