Problem

As shown, there is a triangle among three equal circles that are tangent to each other.

The vertex of the triangle is in the center of the upper circle, the left edge of the triangle base is in the center of the lower left circle, and the right edge is on the circumference of the lower right circle. What is the diameter of the equal circles when the area of the triangle is 259.8 square inches?

---

$$ $$

$$ $$

$\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

$$ $$

---

Solution

Let $S$ be the area of the triangle and $x$ be the diameter of the equal circles.

 

 

First, the following can be seen from the figure:

$$AB=x, \qquad BD=\frac{x}{2}, \qquad BC=\frac{3}{2} x.$$

For $⊿ABD$, from the Pythagorean theorem, we have

$$BD^2+DA^2=AB^2,$$

$$(\frac{x}{2})^2+DA^2=x^2,$$

$$DA^2=\frac{3}{4} x^2,$$

$$∴ \ DA=\frac{\sqrt{3}}{2} x.$$

For $∆ABC$, we have

$$S=\frac{BC×AD}{2},$$

$$S=\frac{1}{2}×\frac{3}{2} x×\frac{\sqrt{3}}{2} x,$$

$$\therefore \ S=\frac{3\sqrt{3}}{8} x^2.$$

Since $S=259.8$ from the problem statement, we know that

$$\frac{3\sqrt{3}}{8} x^2=259.8,$$

$$x^2=\frac{259.8×8}{3\sqrt{3}}=\frac{2078.4×\sqrt{3}}{9},$$

$$\therefore x=\frac{\sqrt{2078.4×\sqrt{3}}}{3}=19.9997…$$

(Answer) approximately 20 inches.

---

Reference

Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.56; p.290.