*Posted by Alex in : potd , trackback
*
A circle with center $C$ passes through point $B$. Let $A$ be a point outside of the circle and $Q$ be the intersection of the circle and $\overline{AB}$. If $\angle ACB$ is a right angle and $\overline{AB}$, $\overline{AC}$, $\overline{BC}$, and $\overline{BQ}$ have integer lengths, find the minimum possible perimeter of $\triangle CBQ$.

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**Solution:**
Let $a$ be the length of $\overline{BC}$, $b$ be the length of $\overline{AC}$, $c$ be the length of $\overline{AB}$, and $x$ be the length of $\overline{BQ}$. Because $\triangle ABC$ is a right triangle, $a^2 + b^2 = c^2$. By the power of a point theorem, $(b – a)(b + a) = c(c – x)$. By substituting the first equation into the second and simplifying, one obtains: $x = \frac{2a^2}{c}$. Because $a$, $b$, $c$, and $x$ are all integers, $2a^2$ is an integer multiple of $c$. Considering Pythagorean triples that are multiples of $(3, 4, 5)$ leads to $(15, 20, 25)$, where the $b$ and $c$ would share the common factor of $5$ and $c$ is a perfect square. By letting $a = 15, b = 20, c = 25$, $x$ has the value of $18$. The perimeter of $\triangle CBQ$ is $a + a + x = \fbox{048}$. By considering other Pythagorean triples, there is no way to obtain a smaller perimeter.

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