SUBIECTUL 3

Let ABCDA'B'C'D' be a cube with an edge length of $10 \mathrm{~cm}$. If M is the midpoint of edge [B'C'], and N is the midpoint of edge [C'D'], find the distance from point C' to the plane (ABM) and the cosine of the angle formed by lines AM and BN.
SUBIECTUL 3 Consider the cube ABCDA'B'C'D' with an edge length of $10 \mathrm{~cm}$. If $M$ is the midpoint of edge [B'C'], and $N$ is the midpoint of edge [C'D'], find the distance from point C' to the plane (ABM) and the cosine of the angle formed by lines AM and BN.

![](https://cdn.mathpix.com/cropped/2024_06_07_2543a31e59990b7d1ba3g-2.jpg?height=543&width=617&top_left_y=1188&top_left_x=180)

| $A B \perp\left(B C C^{\prime}\right)$ and $A B \subset(A B M) \Rightarrow(A B M) \perp\left(B C C^{\prime}\right)$, and then  $d\left(C^{\prime},(A B M)\right)=d\left(C^{\prime}, B M\right)$, where $B M=(A B M) \cap\left(B C C^{\prime}\right)$. | 1p |
| :---: | :---: |
| $C^{\prime} P$ is the altitude in $\triangle C^{\prime} M B, C^{\prime} P=2 \sqrt{5}$ | $2 p$ |
| OBS: If the distance is obtained without showing which one it is, 3p is awarded. |  |
| Let $A Q \\| B N, Q \in D^{\prime} C^{\prime} \Rightarrow m[\varangle(A M, B N)]=m[\varangle(A M, A Q)]=m(\varangle M A Q)$ |  |
| An isosceles triangle $A M Q$ is obtained with $A M=A Q=15 \mathrm{~cm}$ and $M Q=5 \sqrt{10}$ | 2p |
| $\cos (\varangle M A Q)=\frac{4}{9}$ | $2 p$ |