A right triangle is divided into two smaller right triangles by the altitude $CD$ to its hypotenuse $AB$ as shown in the diagram. Circle $O$ with radius $r$ is inscribed in the $\unicode {0x25FA} BCD$. The $\unicode {0x25FA} CAD$ contains 3 circles which are tangent to each other and also to the sides of the triangle as depicted in the diagram. All four circles are congruent. What is the ratio of the sum of the areas of the four circles to the area of the original $\unicode {0x25FA} ABC$?
My solution:
Say, $BC = a, CA = b, AB = c$.
As $\unicode {0x25FA} BCD \sim \unicode {0x25FA} CAD$, it is easy to see that $\angle OBC = \angle ACE$ and $\angle OCB = \angle CAG$. Given that the circles are congruent,
$BH = CI, CH = AJ, IJ = 4r$.
So we get, $b-a = 4r$.
Given $\unicode {0x25FA} BCD \sim \unicode {0x25FA} ABC$, the ratio of their inradius will be the ratio of their hypotenuse (or other sides). So,
$r \times c = a \times \dfrac{a+b-c}{2}$ or $c(b-a) = 2a(a+b-c)$ or $(a+b)(2a-c) = 0$.
So, $c = 2a$ and hence $\angle A = 30^0, \angle B = 60^0$.
$BC = 2BD = 2r(1+cot30^0) = 2r(\sqrt3+1)$
$AC = BC+4r = 2r(3+\sqrt3)$
Area of $\unicode {0x25FA} ABC = \dfrac{1}{2} \times AC \times BC = 4r^2(2\sqrt3 + 3)$
So, the ratio of the sum of the areas of 4 circles to the triangle ABC
$= \dfrac{4 \pi r^2}{4r^2(2\sqrt3+3)} = \dfrac{\pi}{3} (2\sqrt3-3)$
Coming to the purpose of posting the question here -
- With the arrangement of the congruent circles and division of the triangle $ABC$, we come to the conclusion that $c = 2a$. I am getting to it with some calculation. Is there a more obvious way to get to the conclusion or is there any theorem which establishes it?
- Is there a better and faster solution to the problem than what I already have?
