Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.2 National olympiad Prove it Austria

Let $ABCDEF$ be a regular octahedron with lower vertex $E$ , upper vertex $F$ , middle slice plane $ABCD$ , center $M$ and circumsphere $k$ . Furthermore let $X$ be an arbitrary point within side $ABF$ . The line $EX$ intersects $k$ in $E$ and $Z$ and the plane $ABCD$ in $Y$ .
$\text{Show that } \langle EMZ \rangle = \langle EYF \rangle.$

Solutions — 2

Solution 1

We intersect the entire figure with the plane through the points $X$ , $E$ and $F$ . Since $M$ is on line $EF$ , it is part of that plane. Also points $Y$ and $Z$ are part of that plane because they are on line $EX$ . The intersection of the circumsphere $k$ and the plane results in a circle $k'$ , which also has $M$ as its center. The intersection of $ABCD$ with the plane is the perpendicular bisector of line $EF$ , and $Y$ is on that bisector.

We denote $\angle ZEF = \alpha$ .
Since $ZM$ and $EM$ are both radii of $k$ (and $k'$ ), the triangle $ZME$ is isosceles, therefore
$\angle EZM = \angle ZEM = \alpha$ .
Since $Y$ is on the bisector of $EF$ , also triangle $EYF$ is isosceles, therefore $\angle YFE = \angle YEF = \alpha$ .
Therefore the triangles $ZME$ and $EYF$ are similar, and their corresponding angles $\angle EMZ$ and $\angle EYF$ are identical. $\square$

Solution 2

We intersect the figure with the plane $XEF$ as we did in solution 1. Since $\angle YMF = 90^\circ$ , and by Thales also $\angle FZY = 90^\circ$ , the quadrilateral $YMFZ$ has a circumcircle. Therefore $\angle YZM = \angle YFM$ , and we again get that the triangles $ZME$ and $EYF$ are similar. $\square$

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