# Plane cutting a pyramid (drawing)

We have a triangular pyramid (see the picture) in which the side of the base and the height are equal to $$a$$ and the side walls are isosceles triangles.

The pyramid was cut it with a plane passing through one of the edges of the base such that we obtain two solids of the same volume. Determine the tangent of the angle of inclination of this plane to the base plane.

I made a drawing. Since both solids have the same volume and the base has not changed, the yellow line $$h = \frac{H}{2}=\frac{a}{2}$$. I do not know what to do next. The blue line in the base is equal to $$\frac{a\sqrt{3}}{2}$$ because it is a hight of the equilateral triangle of side $$a$$. I believe that to find the tangent I need to calculate the parts in which the height $$h$$ of the smaller pyramid divides the height of the base triangle.

I would be grateful for any help.

• It seems to me that the intercept theorem applies to H and h which means that the top vertex of the red triangle should be at the middle of the pyramid's edge. Dec 19, 2019 at 8:51
• You mean the yellow line should be on $H$? Dec 19, 2019 at 8:56
• I mean that the yellow line is parallel to the vertical black line and its length is half the length of that line. Then use the intercept theorem en.wikipedia.org/wiki/Intercept_theorem Dec 19, 2019 at 8:58
• But we do not know what length is this part of $H$ which is inside the blue triangle. Dec 19, 2019 at 9:03
• This can also be found by the intercept theorem because the bottom vertex of H is the centroid of the base triangle, which position is known and the base vertex of the yellow line is the middle point between the centroid and the pyramid's vertex. Dec 19, 2019 at 9:10

## 1 Answer

Give names to the pyramid's vertices. $$T$$ is the top, the other are $$X, Y, Z$$ where $$Y$$ is the rightmost and $$Z$$ the leftmost. Let $$C$$ be the centroid of the triangle $$X Y Z$$, let $$M$$ and $$P$$ be the ends of the yellow segment, with $$P$$ in the base plane, finally let $$Q$$ be the middle of $$X$$ and $$Z$$.

We know that $$TC = 2 MP = a$$ The intercept theorem tells us that $$T Y = 2 M Y$$ and that $$C Y = 2 P Y =2(CY-CP)$$. Hence $$2CP = C Y = 2 Q C$$ It follows that $$QC = C P$$, hence $$\tan(\alpha) = \frac{a/2}{2 a \sqrt{3}/6} = \frac{\sqrt{3}}{2}$$