In triangle XYZ, the bisector of $\angle XYZ$ 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 $XY=\frac{3}{7}(XY+YZ)$ 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 $3\cdot 42=\color{red}{126}$.

  • $\begingroup$ Thank you very much for your response. Why is it that XY is three-sevenths of itself and YZ? $\endgroup$ – Db3010 Sep 18 '16 at 1:12
  • $\begingroup$ Ah, nevermind I see. The proportion allows for that conclusion. $\endgroup$ – Db3010 Sep 18 '16 at 1:19
  • $\begingroup$ But what if we let $XY=3x$ and $YZ=4x$. Using your inequalities, we get $6 \lt x \lt 42$, which would make the greatest integer value $41*3=123$. Is this incorrect? $\endgroup$ – Db3010 Sep 18 '16 at 1:30

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