In the diagram, cube $A B C D E F G H$ has edges of length $a$. Points $M, N$ and $P$ are on edges $G C, F B$, and $E A$, respectively, so that $M C=\frac{1}{2} a, N B=\frac{1}{3} a$ and $P A=\frac{1}{4} a$. If the plane defined by points $M, N$ and $P$ intersects $H D$ at point $T$, determine the length $D T$.

![](https://cdn.mathpix.com/cropped/2024_04_30_790b6557ee0304df033eg-4.jpg?height=461&width=472&top_left_y=1491&top_left_x=881)
We draw the plane $M N P T$ and join $M P$ and $N T$ :

![](https://cdn.mathpix.com/cropped/2024_04_30_264fdb419065898dfdf2g-08.jpg?height=469&width=482&top_left_y=503&top_left_x=884)

Since $M P$ and $N T$ both lie in the plane, then these line segments intersect.

Since $M P$ joins $G C$ to $E A$, it lies above the diagonal $C A$ of the base.

Since $N T$ joins $F B$ to $H D$, it lies above the diagonal $B D$ of the base.

Therefore, the point of intersection $X$ of $M P$ and $N T$ lies above the point of intersection of $C A$ and $B D$, which is the centre of the base.

Thus, $X$ is the midpoint of $M P$ and the midpoint of $N T$.

This means that the height of $X$ will be the average of the heights of $M$ and $P$ above the base, and will be the average of the heights of $N$ and $T$ above the base. (This is because the height changes linearly along each line segment $M P$ and $N T$.)

Therefore, $\frac{1}{2}(M C+P A)=\frac{1}{2}(N B+T D)$.

Thus, $M C+P A=N B+T D$ or $\frac{1}{2} a+\frac{1}{4} a=\frac{1}{3} a+D T$, and so $D T=\frac{6}{12} a+\frac{3}{12} a-\frac{4}{12} a=\frac{5}{12} a$. ANSWER: $\frac{5}{12} a$