In $\triangle XYZ$, $XY = 4$, $YZ = 7$, and $XZ = 9$. Let $M$ be the midpoint of $\overline{XZ}$, and let $A$ be the point on $\overline{XZ}$ such that $\overline{YA}$ bisects angle $XYZ$. Let $B$ be the point on $\overline{YZ}$ such that $\overline{YA} \perp \overline{AB}$. Let $\overline{AB}$ meet $\overline{YM}$ at $C$. Find $AC: CB$.


We know that $XM$ = $MZ$ = $\dfrac{9}{2}$, and by the Angle-Bisector theorem, $$\dfrac{4}{x} = \dfrac{7}{9-x}$$ $$\implies 36-4x = 7x$$ $$\implies x= \dfrac{36}{11}=XA$$ Therefore, $AZ$ = $\dfrac{63}{11}$, and $AM$ = $\dfrac{9}{2} - \dfrac{36}{11}= \dfrac{27}{22}.$ Also, by the Menelaus theorem, $$\dfrac{ZM}{MA} \times \dfrac{AC}{CB} \times \dfrac{YB}{YZ} = 1$$ $$\implies \dfrac{11}{3} \times \dfrac{AC}{CB} \times \dfrac{YB}{7} = 1$$ From here, I got stuck. I'm wondering if one could use mass points, since the problem wants to find the ratio of lengths, and not specific side lengths. Any help is appreciated!


You have already got $$\dfrac{11}{3} \times \dfrac{AC}{CB} \times \dfrac{YB}{7} = 1\tag1$$ Also, we have $$YB=\frac{AY}{\cos \frac Y2}\tag2$$ So, we want to find $\cos\frac Y2$ and $AY$.

By the law of cosines, $$9^2=4^2+7^2-2\cdot 4\cdot 7\cos Y\implies \cos Y=-\frac 27$$ from which we have $$\cos\frac Y2=\sqrt{\frac{1-\frac 27}{2}}=\sqrt{\frac{5}{14}}$$ Also, the length of $AY$ is given by $$AY=\sqrt{YX\cdot YZ\left(1-\frac{YZ^2}{(YX+YZ)^2}\right)}=\frac{4\sqrt{70}}{11}$$ (see here or here for this formula)

So, from $(1)$ and $(2)$, we get $$\color{red}{AC:CB=3:8}$$


(Wrong!!! This is the case for $YB\perp AB.$) Following your attempts, we get $\frac{11}{3}\times\frac{AC}{CB}\times\frac{YB}{7} = 1.$ Hence it remains to find the length of $YB.$ Let $K$ be the point on $YZ$ such that $XK\perp YZ.$ Then we can use the area of the triangle to find $XK$: $$\sqrt{10(10-9)(10-7)(10-4)}=\frac{7\times XK}{2}.$$ Therefore $XK=\frac{12\sqrt{5}}{7}.$ Then since $\triangle XKZ\sim\triangle ABZ,$ we can get $$AB = XK \times\frac{63/11}{9}=\frac{12\sqrt{5}}{11}$$ and obtain $BZ = \sqrt{(\frac{63}{11})^2-(\frac{12\sqrt{5}}{11})^2}=\frac{57}{11}.$ Then our $YB=7- \frac{57}{11} = \frac{20}{11},$ so the answer should be $$\frac{AC}{CB} = \frac{3}{11}\times\frac{7}{YB} = \frac{21}{20}.$$

  • $\begingroup$ @tommy xu3. I'm sorry, but I don't get why △XKZ∼△ABZ. Could you elaborate? $\endgroup$ – Albus Dumbledore Jan 1 '17 at 6:55
  • $\begingroup$ $\triangle{XKZ}$ and $\triangle{ABZ}$ should not be similar since $\angle{XKZ}=90^\circ$ and $\angle{ABZ}$ is obtuse. $\endgroup$ – mathlove Jan 1 '17 at 7:52
  • $\begingroup$ You are right. I misunderstood the problem statement as $YB\perp AB$ Sorry. $\endgroup$ – User Jan 1 '17 at 8:00

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