HMMT November, 2024

Solution:

Let $X=P B \cap A D$ and $Y=P C \cap A D$. Let $A X=p$, $X Y=q$, and $Y D=r$. From $A B \| C D$, we get that $A B=C D$, and so $\angle A P X=\angle D P Y$. Thus, we may apply Steiner ratio theorem on $\triangle P A D$ and $\triangle P X Y$ to get that
$$\frac{p(p+q)}{r(q+r)}=\frac{3^{2}}{4^{2}}, \quad \frac{p(q+r)}{r(p+q)}=\frac{5^{2}}{6^{2}} .$$
Multiplying these two equations gives $p: r=5: 8$, and using each individual equations gives $p: q : r=5: 22: 8$. Thus, $p=10$, $q=44$, and $r=16$.

Now, from $X Y \| B C$, we have $P X: X B=4: 1$, so set $P X=4 t$ and $X B=t$. However, $4 t^{2}=P Y \cdot Y C=10 \cdot 60=600$. Solving this gives $t=\sqrt{150}=5 \sqrt{6}$, hence $P B=5 t=25 \sqrt{6}$.

Solution:

Let $A B=C D=a$, $A C=B D=b$, $\frac{A P}{3}=\frac{D P}{4}=x$, and $\frac{B P}{5}=\frac{C P}{6}=y$. Applying Ptolemy's theorem for the quadrilaterals $A B C P$, $B C D P$, and $A B C D$ yields:
$$\begin{aligned} b \cdot 5 y & =55 \cdot 3 x+a \cdot 6 y \\ b \cdot 6 y & =55 \cdot 4 x+a \cdot 5 y \\ b^{2} & =55 \cdot 70+a^{2} \end{aligned}$$
Equating the left-hand sides of (1) and (2) leads to
$$6 \cdot(165 x+6 a y)=5 \cdot(220 x+5 a y) \Longrightarrow 110 x=11 a y \Longrightarrow 10 x=a y$$
Substituting $220 x=22 a y$ into (2) implies $27 a y=6 b y$, or $b=\frac{9}{2} a$. Plugging this into (3), we find $a^{2}=200$, so $a=10 \sqrt{2}$, and therefore $b=45 \sqrt{2}$. Furthermore, $x=y \sqrt{2}$ after replacing $a$ with $10 \sqrt{2}$ in (4). We now apply Law of Cosines for $\triangle A B C$ and $\triangle A P C$ :
$$\begin{aligned} 55^{2}+(10 \sqrt{2})^{2}-2 \cdot(10 \sqrt{2}) \cdot 55 \cos \theta & =(45 \sqrt{2})^{2} \\ (3 \sqrt{2} y)^{2}+(6 y)^{2}+2 \cdot(3 \sqrt{2} y) \cdot(6 y) \cos \theta & =(45 \sqrt{2})^{2} \end{aligned}$$
where $\theta=\angle A B C$. Solving (5) yields
$$\cos \theta=\frac{55^{2}+(10 \sqrt{2})^{2}-(45 \sqrt{2})^{2}}{2 \cdot(10 \sqrt{2}) \cdot 55}=-\frac{3 \sqrt{2}}{8}$$
Plugging this into (6), we can compute:
$$y^{2}=\frac{(45 \sqrt{2})^{2}}{(3 \sqrt{2})^{2}+6^{2}-27}=\frac{4050}{27}=150$$
Therefore, $B P=5 y=5 \sqrt{150}=25 \sqrt{6}$.