Problem 92. A tangent $A B$ is drawn to two touching circles. Parallel to $A B$, a tangent to the smaller circle is drawn, intersecting the larger circle at points $D$ and $C$ (Fig. 93). Given that the radii of these circles are $R_{1}$ and $R_{2}$, find the radius of the circle circumscribed around triangle $BDC$.

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-071.jpg?height=554&width=773&top_left_y=942&top_left_x=356)
Solution. Based on problem 35 for $\triangle A D C$, we write:

$A D^{2}=A C^{2}=A W \cdot A H=$ $=2 R_{2} \cdot 2 R_{1}=4 R_{1} R_{2}$.

Thus,

$$
A D=A C=2 \sqrt{R_{1} R_{2}}
$$

But, according to Archimedes' problem, $A B$ is also equal to $2 \sqrt{R_{1} R_{2}}$, therefore, point $A$ is the center of the circumcircle of triangle $B D C$. Hence, the answer is: $2 \sqrt{R_{1} R_{2}}$.

## Chapter 14. ARCHIMEDES' PROBLEM ON THE ANGLE BISECTOR. A REMARKABLE PROPERTY OF THE INCENTER OF A TRIANGLE

In the works of Archimedes that have come down to us, we find only one problem related to the property of the angle bisector. In the interpretation of Sabit ibn Qurra [4] and in accordance with Fig. 94, it looks like this:

$$
\frac{b+c}{a}=\frac{c}{y}(*)
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-072.jpg?height=457&width=649&top_left_y=877&top_left_x=1069)

In modern elementary geometry, the formula is known:

$$
\frac{A I}{I L_{1}}=\frac{b+c}{a} \text { (where } I-\text { the incenter of the triangle), }
$$

which, as one can easily notice, essentially duplicates Archimedes' problem. Indeed, from $\triangle A B L_{1}$ (Fig. 94) by the property of the angle bisector: $\frac{A I}{I L_{1}}=\frac{c}{y}$.

The formula (*) helped Archimedes to determine with a high degree of accuracy the boundaries of the number $\pi$.

Let's see how Archimedes found the upper limit of the number $\pi^{1}$. Let $A C=a_{6}$ be the side of the circumscribed regular hexagon (Fig. 95). Let $O B=R=1 . \angle A O C=60^{\circ}$ and $\angle A O B=30^{\circ}$. Then $A B=\frac{1}{2} a_{6}=\frac{1}{\sqrt{3}} ; A O=\frac{2}{\sqrt{3}}$. Let also $O D$ be the bisector of angle $A O B$. Then, by Archimedes' problem, $\quad \frac{A O+O B}{A B}=\frac{O B}{B D} . \quad$ Or

$$
\frac{1+\frac{2}{\sqrt{3}}}{\frac{1}{\sqrt{3}}}=\frac{1}{B D}, \text { from which } \quad B D=\frac{1}{\sqrt{3}+2}
$$

Note that $B D$ is half the side of the circumscribed dodecagon, i.e.,

$$
\frac{1}{2} a_{12}=\frac{1}{\sqrt{3}+2}
$$

By the Pythagorean theorem for $\triangle B O D$, we find $O D$. We draw the next bisector, then another one, and thus find $\frac{1}{2} a_{24} ; \frac{1}{2} a_{48} ; \frac{1}{2} a_{96}$. The circumscribed 96-sided polygon in Archimedes' time provided a fairly good approximation of the number $\pi$ :

$$
2 \pi R \approx I_{96}=a_{96} \cdot 96, \text { from which } \pi \approx \frac{a_{96} \cdot 96}{2 R}
$$

Thus, unexpectedly, beautifully, and effectively, Archimedes applied the property of the angle bisector of a triangle that he derived.

As we have already noted, in modern problems, the formula $\frac{A I}{I L_{1}}=\frac{b+c}{a}$ finds wider application. By proving this formula, we will also solve Archimedes' problem.

Theorem-Problem. Prove that the incenter $I$ divides the bisector $A L_{1}$ of triangle $A B C$ in the ratio

$$
\frac{A I}{I L_{1}}=\frac{b+c}{a}
$$

First Solution. Draw the bisector $C L_{3}$ in triangle $A B C$ (Fig. 96). By the known property of the bisector for triangle $A L_{1} C$, we have:

72

$$
\frac{A I}{I L_{1}}=\frac{b}{x}
$$

Draw a line $k$ through vertex $B$ parallel to $A L_{1}$. Line $C A$ intersects $k$ at point $D$. It is not difficult to show that triangle $B A D$ is isosceles:

$$
B A=A D=c .
$$

Then from the fact that

$$
\triangle A L_{1} C \sim \triangle D B C, \frac{b}{x}=\frac{b+c}{a},
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-074.jpg?height=574&width=603&top_left_y=375&top_left_x=1109)
it follows that $\frac{A I}{I L_{1}}=\frac{b+c}{a}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-074.jpg?height=474&width=634&top_left_y=1062&top_left_x=380)

Second Solution. By the property of the bisector for triangle $A B C$ (Fig. 97):

$$
\frac{b}{c}=\frac{x}{a-x}, \text { from which } x=\frac{a b}{b+c} \text {. }
$$

Then by this same property for triangle $A L_{1} C$:

$$
\frac{A I}{I L_{1}}=\frac{b}{x}=\frac{b(b+c)}{a b}=\frac{b+c}{a}
$$

Third Solution. Draw a line $t$ through vertex $A$ of triangle $A B C$ parallel to $B C$ (Fig. 98). Draw $T K \perp B C$ through the incenter $I$. $I K=r, I T=h_{a}-r$. $\Delta A I T \sim \Delta L_{1} I K$, from which $\frac{A I}{I L_{1}}=\frac{I T}{I K}$, or

$$
\frac{A I}{I L_{1}}=\frac{h_{a}-r}{r}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-074.jpg?height=509&width=631&top_left_y=1573&top_left_x=1089)

Since $h_{a}=\frac{2 S}{a}, r=\frac{S}{p}$, then $\frac{A I}{I L_{1}}=\frac{\frac{2 S}{a}-\frac{S}{p}}{\frac{S}{p}}=\frac{S(2 p-a)}{S a}=\frac{b+c}{a}$.