HMMT February, 2025

Solution:

Let the internal and external bisectors of $\angle BAC$ meet $BC$ at $D$ and $E$. Similarly, let the internal and external bisectors of $\angle BPC$ meet $BC$ at $D'$ and $E'$. The angle condition implies that $AD\parallel PD'$ and $AE\parallel PE'$. Thus, triangles $ADE$ and $PD'E'$ are homothetic. Hence, the requested ratio is $\frac{DE}{D'E'}$. Repeated applications of the angle bisector theorem yield

$$BD = 7 \cdot \frac{6}{6 + 8} = 3,$$
$$BE = 7 \cdot \frac{6}{8 - 6} = 21,$$
$$BD' = 7 \cdot \frac{1}{1 + 2} = \frac{7}{3},$$
$$BE' = 7 \cdot \frac{1}{2 - 1} = 7,$$
so $DE = 24$ and $D'E' = 28 / 3$. Hence, the answer is $\frac{28 / 3}{24} = \left[\frac{7}{18}\right]$.

Solution 2:

Let $D = AP\cap BC$, $E = BP\cap AC$ and $F = CP\cap AB$. Then, $BCEF$ is cyclic, so by Power of a Point, $\frac{AE}{AF} = \frac{AB}{AC} = \frac{3}{4}$. Let $AE = 3x$ and $AF = 4x$. Then,

$$\triangle PFB \sim \triangle PEC \Longrightarrow \frac{BF}{CE} = \frac{BP}{CP} \Longrightarrow \frac{6 - 4x}{8 - 3x} = \frac{1}{2}.$$
Solving for $x$ gives $x = \frac{4}{5}$, so we get that $AF = \frac{16}{5}$, $FB = \frac{14}{5}$, $AE = \frac{12}{5}$, and $EC = \frac{40}{5}$. By Ceva's theorem on $\triangle ABC$ and point $P$, we have

$$\frac{BD}{DC} = \frac{AE}{EC} \cdot \frac{BF}{FA} = \frac{AE}{AF} \cdot \frac{BF}{CE} = \frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}.$$
Finally, by Menelaus's theorem on $\triangle ADC$ and line $BE$, we get that

$$\frac{AP}{PD} = \frac{CB}{BD} \cdot \frac{AE}{EC} = \frac{11}{3} \cdot \frac{12}{28} = \frac{11}{7},$$
which implies that $\frac{|BPC|}{|ABC|} = \left[\frac{7}{18}\right]$

Solution 3:

We use barycentric coordinates with respect to $\triangle ABC$. From $\angle ABP = \angle ACP$, we get

$$\frac{|ABP|}{|ACP|} = \frac{AB \cdot BP}{AC \cdot CP} = \frac{3}{8},$$
so $P$ has coordinate $(- :8:3)$. Let $Q$ be the isogonal conjugate of $P$, so $Q$ lies on the perpendicular bisector of $BC$. By Steiner ratio theorem, $Q$ has coordinate $(- :8^2 / 8:6^2 / 3) = (- :8:12) = (- :2:3)$. Let the coordinate be $(t:2:3)$ for some real number $t$. Then, recall the equation for the perpendicular bisector (from Corollary 6 of https://web.evanchem.cc/handouts/bary/bary-full.pdf):

$$a^2(z - y) + x(c^2 - b^2) = 0$$
$$\Rightarrow 7^2(3 - 2) + t(6^2 - 8^2) = 0,$$
so $t = \frac{7^2(3 - 2)}{8^2 - 6^2} = \frac{7}{4}$. Hence, point $Q$ has coordinates $(7:8:12)$, so point $P$ has coordinates $(7^2 / 7:8^2 / 8:6^2 / 12) = (7:8:3)$. Therefore, $|BPC| / |ABC| = \left[\frac{7}{18}\right]$