In the diagram, $\triangle P Q R$ is right-angled at $P$ and has $P Q=2$ and $P R=2 \sqrt{3}$. Altitude $P L$ intersects median $R M$ at $F$. What is the length of $P F$ ?
(A) $\frac{\sqrt{3}}{2}$
(B) $\frac{3 \sqrt{3}}{7}$
(C) $\frac{4 \sqrt{3}}{7}$
(D) $\frac{5 \sqrt{3}}{9}$
(E) $\frac{3 \sqrt{3}}{5}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-095.jpg?height=475&width=331&top_left_y=218&top_left_x=1293)
Since $P Q=2$ and $M$ is the midpoint of $P Q$, then $P M=M Q=\frac{1}{2}(2)=1$.

Since $\triangle P Q R$ is right-angled at $P$, then by the Pythagorean Theorem,

$$
R Q=\sqrt{P Q^{2}+P R^{2}}=\sqrt{2^{2}+(2 \sqrt{3})^{2}}=\sqrt{4+12}=\sqrt{16}=4
$$

(Note that we could say that $\triangle P Q R$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, but we do not actually need this fact.)

Since $P L$ is an altitude, then $\angle P L R=90^{\circ}$, so $\triangle R L P$ is similar to $\triangle R P Q$ (these triangles have right angles at $L$ and $P$ respectively, and a common angle at $R$ ).

Therefore, $\frac{P L}{Q P}=\frac{R P}{R Q}$ or $P L=\frac{(Q P)(R P)}{R Q}=\frac{2(2 \sqrt{3})}{4}=\sqrt{3}$.

Similarly, $\frac{R L}{R P}=\frac{R P}{R Q}$ so $R L=\frac{(R P)(R P)}{R Q}=\frac{(2 \sqrt{3})(2 \sqrt{3})}{4}=3$.

Therefore, $L Q=R Q-R L=4-3=1$ and $P F=P L-F L=\sqrt{3}-F L$.

So we need to determine the length of $F L$.

Drop a perpendicular from $M$ to $X$ on $R Q$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-161.jpg?height=474&width=325&top_left_y=1823&top_left_x=954)

Then $\triangle M X Q$ is similar to $\triangle P L Q$, since these triangles are each right-angled and they share a common angle at $Q$. Since $M Q=\frac{1}{2} P Q$, then the corresponding sides of $\triangle M X Q$ are half as long as those of $\triangle P L Q$.

Therefore, $Q X=\frac{1}{2} Q L=\frac{1}{2}(1)=\frac{1}{2}$ and $M X=\frac{1}{2} P L=\frac{1}{2}(\sqrt{3})=\frac{\sqrt{3}}{2}$.

Since $Q X=\frac{1}{2}$, then $R X=R Q-Q X=4-\frac{1}{2}=\frac{7}{2}$.

Now $\triangle R L F$ is similar to $\triangle R X M$ (they are each right-angled and share a common angle at $R$ ).

Therefore, $\frac{F L}{M X}=\frac{R L}{R X}$ so $F L=\frac{(M X)(R L)}{R X}=\frac{\frac{\sqrt{3}}{2}(3)}{\frac{7}{2}}=\frac{3 \sqrt{3}}{7}$.

Thus, $P F=\sqrt{3}-\frac{3 \sqrt{3}}{7}=\frac{4 \sqrt{3}}{7}$.