# In triangle XYZ, the bisector of angle XYZ intersects overline XZ at E if XY/YZ=3/4 an XZ=42, find the greatest integer value of XY. Thus far, I have determined that XE=18 and ZE=24 by the angle bisector theorem, but I am unsure how to find XY.

Chaim Ferguson 2022-10-16 Answered
In triangle XYZ, the bisector of $\mathrm{\angle }XYZ$ intersects $\overline{XZ}$ at E if $\frac{XY}{YZ}=\frac{3}{4}$ an $XZ=42$, find the greatest integer value of XY.
Thus far, I have determined that $XE=18$ and $ZE=24$ by the angle bisector theorem, but I am unsure how to find XY.
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The bisector theorem gives $\frac{XE}{EZ}=\frac{XY}{YZ}=\frac{3}{4}$, hence $XZ=42$ implies $XE=18$ and $EZ=24$.
By the triangle inequality, $XY+YZ$ has to be greater than $XZ=42$, hence $XY=\frac{3}{7}\left(XY+YZ\right)$ has to be greater than 18. On the other hand, $YZ-YX$ has to be smaller than $XZ=42$, hence the length of $XY$ is at most $3\cdot 42=126$