# Prismatoid plane section with straightedge and compass

I came up with a task I'm out of ideas how to do a solution. Perhaps I'm not paying attention to an obvious thing, but still. All I came up with is to build orthogonal projections but that's not a good position of a solid to make these clear.

In this image, there is a prismatoid with two bases: bottom one is a pentagon $$ABCDE$$, and the top one is a triangle $$A_1B_1C_1$$. Its lateral faces include two triangles ($$AEA_1$$ and $$BCB_1$$) and three trapezoids ($$ABB_1A_1$$, $$DEA_1C_1$$, and $$CDC_1B_1$$). Point $$M$$ belongs to a plane $$A_1B_1C_1$$, point $$N$$ belongs to a plane $$ABA_1$$, point $$K$$ belongs to a plane $$AEB_1$$. The task is to dissect a prismatoid $$ABCDEA_1B_1C_1$$ with a plane $$MNK$$ using straightedge and compass, i.e. to construct each line where a plane $$MNK$$ intersects a prismatoid $$ABCDEA_1B_1C_1$$.

tl,dr: $$ABCDE\parallel A_1B_1C_1, M\in(A_1B_1C_1), N\in(ABA_1), K\in(AEB_1)$$. Use SE&C to construct lines $$\{ \alpha_1,\alpha_2,\dots,\alpha_n \} \ni (MNK)\ \cap ABCDEA_1B_1C_1$$.

Any tips are highly appreciated.

• Can you take as given some basic constructions? Such as, for instance, the intersection between a line and a plane? May 25, 2021 at 13:34
• @Intelligentipauca Certainly! May 25, 2021 at 13:59

Let $$G$$ be the intersection between line $$KN$$ and plane $$A_1B_1C_1$$: line $$MG$$ is then the intersection between planes $$KMN$$ and $$A_1B_1C_1$$. Its two intersections $$P$$, $$Q$$ with triangle $$A_1B_1C_1$$ are two vertices of the section to be constructed. This doesn't work if $$K$$ lies on $$ABCDE$$: construct in that case the midpoint $$K'$$ of $$KM$$ and do the construction with $$K'$$ instead of $$K$$.
Let $$H$$ be the intersection between line $$KM$$ and plane $$ABCDE$$: line $$NH$$ is then the intersection between planes $$KMN$$ and $$ABCDE$$. Its two intersections $$R$$, $$S$$ with pentagon $$ABCDE$$ are two more vertices of the section to be constructed. This doesn't work if $$K$$ lies on $$A_1B_1C_1$$: construct in that case the midpoint $$K'$$ of $$KN$$ and do the construction with $$K'$$ instead of $$K$$.
To construct the remaining vertices, consider again line $$KN$$: it must intersect one of the lateral faces, for instance $$BCB_1$$, at some point $$I$$. On face $$BCB_1$$ also lies a point already found ($$R$$, for instance): the intersection of line $$RI$$ with triangle $$BCB_1$$ will then give another vertex. Do the same with line $$KM$$, if needed, to complete the construction.
• @Rusurano My figure is not exactly the same as yours, but I made it with GeoGebra and point $K$ does lie on plane $AEB_1$. No other constraints on that point are mentioned in the question. May 25, 2021 at 18:26