A cube with vertices $F G H J K L M N$ has edge length 200. Point $P$ is on $H G$, as shown. The shortest distance from $G$ to a point inside $\triangle P F M$ is 100 . Which of the following is closest to the length of $H P$ ?
(A) 53
(B) 55
(C) 57
(D) 59
(E) 61

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-023.jpg?height=404&width=439&top_left_y=747&top_left_x=1301)
Suppose that $G P=x$.

Since the edge length of the cube is 200 , then $H P=200-x$.

We consider tetrahedron (that is, triangle-based pyramid) $F G M P$ and calculate its volume in two different ways.

The volume of a tetrahedron is equal to one-third times the area of its triangular base times the length of its perpendicular height.

First, we consider tetrahedron $F G M P$ as having base $\triangle F G M$ and height $G P$, which is perpendicuar to the base.

$\triangle F G M$ is right-angled at $G$ and has $F G=G M=200$, so its area is $\frac{1}{2} \times F G \times G M$ which equals $\frac{1}{2} \times 200 \times 200$ which equals 20000 .

Thus, the volume of $F G M P$ is $\frac{1}{3} \times 20000 \times x$.

Next, we consider tetrahedron $F G M P$ as having base $\triangle P F M$.

From the given information, the shortest distance from $G$ to a point inside this triangle is 100 . This means that the height of tetrahedron $F G M P$ when considered to have base $\triangle P F M$ is 100 .

We need to calcualte the area of $\triangle P F M$.

Since $\triangle F G M$ is right-angled at $G$ and $F M>0$, then by the Pythagorean Theorem,

$$
F M=\sqrt{F G^{2}+G M^{2}}=\sqrt{200^{2}+200^{2}}=\sqrt{200^{2} \times 2}=200 \sqrt{2}
$$

Since $\triangle F G P$ is right-angled at $G$ and $F P>0$, then by the Pythagorean Theorem,

$$
F P=\sqrt{F G^{2}+G P^{2}}=\sqrt{200^{2}+x^{2}}=\sqrt{x^{2}+40000}
$$

Similarly, $M P=\sqrt{x^{2}+40000}$.

This means that $\triangle P F M$ is isosceles with $F P=M P$.

Let $T$ be the midpoint of $F M$.

Then $F T=T M=100 \sqrt{2}$.

Since $\triangle P F M$ is isosceles, then $P T$ is perpendicular to $F M$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-047.jpg?height=347&width=379&top_left_y=1670&top_left_x=927)

By the Pythagorean Theorem,

$P T=\sqrt{F P^{2}-F T^{2}}=\sqrt{\left(\sqrt{x^{2}+40000}\right)^{2}-(100 \sqrt{2})^{2}}=\sqrt{x^{2}+40000-20000}=\sqrt{x^{2}+20000}$

Therefore, the area of $\triangle P F M$ is $\frac{1}{2} \times F M \times P T$ which equals $\frac{1}{2} \times 200 \sqrt{2} \times \sqrt{x^{2}+20000}$. This means that the volume of tetrahedron $F G M P$ is equal to

$$
\frac{1}{3} \times\left(\frac{1}{2} \times 200 \sqrt{2} \times \sqrt{x^{2}+20000}\right) \times 100
$$

We can now equate the two expressions for the volume of $F G M P$ to solve for $x$ :

$$
\begin{aligned}
\frac{1}{3} \times 20000 \times x & =\frac{1}{3} \times\left(\frac{1}{2} \times 200 \sqrt{2} \times \sqrt{x^{2}+20000}\right) \times 100 \\
20000 \times x & =\left(\frac{1}{2} \times 200 \sqrt{2} \times \sqrt{x^{2}+20000}\right) \times 100 \\
x & =\frac{1}{2} \times \sqrt{2} \times \sqrt{x^{2}+20000} \\
2 x & =\sqrt{2} \times \sqrt{x^{2}+20000} \\
4 x^{2} & =2\left(x^{2}+20000\right) \\
2 x^{2} & =40000 \\
x^{2} & =20000
\end{aligned}
$$

Since $x>0$, then $x=\sqrt{20000}=\sqrt{10000 \times 2}=\sqrt{100^{2} \times 2}=100 \sqrt{2}$.

This means that $H P=200-x=200-100 \sqrt{2} \approx 58.58$.

Of the given answers, this is closest to 59 , which is (D).

ANSWER: (D)