An ugly light fixture is hanging from point $O$ on the ceiling. Wires $O X M, O Y N$ and $O Z P$ pass through the vertices of a very thin wooden equilateral triangle $X Y Z$ of side $60 \mathrm{~cm}$. (A small bulb is attached to the end of each wire.) The plane of the wooden triangle is parallel to the ceiling. If each wire is $100 \mathrm{~cm}$ long and the lower end of each wire is $90 \mathrm{~cm}$ from the ceiling, what is the vertical distance between the wooden triangle and the ceiling?
(A) $40 \mathrm{~cm}$
(B) $45 \mathrm{~cm}$
(C) $50 \mathrm{~cm}$
(D) $55 \mathrm{~cm}$
(E) $60 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-101.jpg?height=469&width=352&top_left_y=264&top_left_x=1296)
Since the vertical distance from $O$ to each of $M, N$ and $P$ is the same, then $O X=O Y=O Z$ since the length of each wire is equal.

Let $x=O X=O Y=O Z$.

Since each wire has total length 100, then $X M=Y N=Z P=100-x$.

Therefore, $M$ is a distance $100-x$ below $X$. Since the total distance of $M$ below the ceiling is 90 , then the vertical distance from the ceiling to $X$ (and thus to the plane of the triangle) is $90-(100-x)=x-10$.

Let $C$ be the centre of $\triangle X Y Z$.

By symmetry, $O$ is directly above $C$ with $O C=x-10$.

Also, $O X=x$.

By the Pythagorean Theorem, $O X^{2}=O C^{2}+X C^{2}$.

Therefore, we need to find the length of $X C$.

Draw altitudes from $X, Y$ and $Z$ to points $F, G$ and $H$ on the opposite sides. Note that $C$ is the point of intersection of $X F, Y G$ and $Z H$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-201.jpg?height=377&width=415&top_left_y=516&top_left_x=909)

Since $\triangle X Y Z$ is equilateral, $H$ is the midpoint of $X Y$ and $X F$ is the angle bisector of $\angle Z X Y$. Therefore, $X H=30$ (since $X Y=60)$ and $\angle C X H=30^{\circ}$.

$\triangle C X H$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, so $C X=\frac{2}{\sqrt{3}} X H=\frac{2}{\sqrt{3}}(30)=\frac{60 \sqrt{3}}{3}=20 \sqrt{3}$.

Since $O X^{2}=O C^{2}+X C^{2}$, then

$$
\begin{aligned}
x^{2} & =(x-10)^{2}+(20 \sqrt{3})^{2} \\
x^{2} & =x^{2}-20 x+100+1200 \\
20 x & =1300 \\
x & =65
\end{aligned}
$$

Therefore, the distance between the triangle and the ceiling is $x-10=55 \mathrm{~cm}$.

ANSWER: (D)