Given isosceles triangle $ABC$ below with $\angle{B}=\angle{C}$ and $BC=1$. Let point $P$ be the intersection of the line bisecting $\angle{B}$ and $AC$. Determine the length of $BP$ as the length of $AM$ approaches 0.
My answer was $\frac34$ but the answer was $\frac23$.
My attempt: Since the problem didn't really state what was $AM$, I set $AM$ to be the non-trivial perpendicular bisector. Because we are finding the median, I noted that this triangle is also half of a parallelogram where $AC$ is one diagonal and $BF$ is the other where point $F$ is on the parallelogram. Using the law of cosines, you can find that $$(AC)^2+(BF)^2=2(AB)^2+2(BC)^2$$ Since $BF$ is twice the length of $BP$, divide $BF$ by 2 to get $BP$.
If we let $AM$ by y, then by the Pythagorean theorem we can say that $AC = \sqrt{y^2+\frac14}$
We can then plug into the previous formula to get $$(2BP)^2+(y^2+\frac14)=2(1)+2(y^2+1)$$ So $$BP=\frac{\sqrt{y^2+\frac94}}{2}$$
As $AM$ approaches $0$, then $y$ approaches $0$ as well. This means that $BP=\frac{\sqrt{\frac94}}{2}=\frac34$
Where did I go wrong?
Your line of reasoning is flawed.
$BP$ is not a median, it is a bisector. And in a generic parallelogram, the diagonals are not bisectors of its angles. The intersection of $BF$ and $AC$ would be the midpoint $Q$ of $AC$, not $P$. So, the length of $BF$ is not twice the length of $BP$ - it is twice the length of $BQ$ instead. What you proved is that the length of $BQ$ tends to $\frac{3}{4}$ as $A \to M$, which is correct but not what the problem asked.
Back to the question now, consider that $\frac{PA}{PC} = \frac{BA}{BC}$ by the bisector theorem.
When $A \to M$ we have $PA \to PM$ and $BA \to BM$ so $\frac{PA}{PC} \to \frac{PM}{PC} = \frac{BM}{BC} = \frac{1}{2}$.
Thus at the limit $\frac{PM}{MC} = \frac{1}{3}$ so $PB = BM + MP = \frac{BC}{2} + \frac{MC}{3} = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$ .