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Geometry Difficulty 7.5 National olympiad, round 2 Prove it Hong Kong

In $\triangle ABC$ , $AB < AC$ . The internal bisector of $\angle BAC$ meets $BC$ at $D$ , while the external bisector of $\angle BAC$ meets $CB$ produced at $E$ . If $EB = 2022$ and the lengths of $BD$ and $DC$ are integers, how many possible lengths of $BD$ are there?

Solution

Let the lengths of $BD$ and $DC$ be $x$ and $y$ respectively. By the angle bisector theorem, we have $\frac{x}{y} = \frac{AB}{AC} = \frac{EB}{EC}$ , i.e.
$\frac{x}{y} = \frac{2022}{2022 + x + y}$

$y = \frac{x^2 + 2022x}{2022 - x} = -x - 4044 + \frac{2022 \cdot 4044}{2022 - x}$
Hence we must choose positive integers $x$ for which $2022 - x$ is a positive integer dividing $2022 \cdot 4044$ (note that every such $x$ will result in a positive integer value of $y$ and the existence of such a figure satisfying all the conditions). As $2022 \cdot 4044 = 2^3 \cdot 3^2 \cdot 337^2$ , its only positive factors that are less than $2022$ are the 12 positive factors of $2^3 \cdot 3^2$ as well as $337, 337 \cdot 2, 337 \cdot 3$ and $337 \cdot 4$ . Hence there are altogether 16 such values of $2022 - x$ , and they correspond to 16 possible values of $x$ .

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