So many Circles

Geometry Level 5

$\triangle A_0B_0C$ has $A_0B_0=13$ , $B_0C=14$ , and $CA_0=15$ .

A circle with diameter $A_0B_0$ is drawn; it intersects $A_0C$ at $A_1$ and $B_0C$ at $B_1$ .

A circle with diameter $A_1B_1$ is drawn; it intersects $A_1C$ at $A_2$ and $B_1C$ at $B_2$ .

This process continues infinitely. A circle with diameter $A_kB_k$ is drawn; it intersects $A_kC$ at $A_{k+1}$ and $B_kC$ at $B_{k+1}$ .

The value of $\sum_{i=0}^{\infty} |A_iB_i|$ can be expressed as $\dfrac{p}{q}$ for positive coprime integers $p,q$ . Find $p+q$ .

Details and Assumptions

$A_k\ne A_{k+1}$ and $B_k\ne B_{k+1}$ for all integers $k \ge 0$ .

The answer is 67.

1 solution

Billy Bob
Jul 27, 2014

Very fun problem! Note that quadrilateral $A_0A_1B_1B_0$ is cyclic, and so $\angle A_1A_0B_0 = \angle CB_1A_1$ . So $\bigtriangleup A_0B_0C$ is similar to $\bigtriangleup B_1A_1C$ . We can see that our answer is simply $13+\frac{B_1C}{A_0C} \cdot 13+\frac{B_1C^{2}}{A_0C^{2}} \cdot 13 +...$ or $\frac{13}{1-\frac{B_1C}{A_0C}}$ , so it suffices to find $B_1C$

Note that $\bigtriangleup A_0B_1B_0$ is right with hypotenuse $A_0B_0$ . So $A_0B_1$ is an altitude of $\bigtriangleup A_0B_0C$ onto side $B_0C$ , and we can easily find from Pythagorean Theorem that $B_1C = 9$ , and so our answer is $\frac{13}{1-\frac{9}{15}} = \frac{65}{2}$

To make the angle sign, instead of using $<$ , use \angle, which displays as $\angle$ . I edited your solution for you to see.

Daniel Liu - 6 years, 10 months ago

Ah thanks :)

billy bob - 6 years, 10 months ago

So many Circles

The answer is 67.

1 solution

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