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Geometry Difficulty 5.2 AIME, harder Prove it Romania

Let ABCDABCDABCDA'B'C'D', be a cube with side length aa. Let MM and PP be the midpoints of the edges [AB][AB] and [DD][DD'], respectively.
a) Prove that MPACMP \perp A'C.
b) Find the distance between the lines MPMP and ACA'C.

Solution

Let OO be the midpoint of [AC][A'C].

a) Triangle MACMA'C is isosceles, hence MOACMO \perp A'C.
Similarly PACPA'C is isosceles, therefore POACPO \perp A'C, yielding AC(PMO)A'C \perp (PMO), hence ACMPA'C \perp MP.

Figure 1

b) Let SS be the midpoint of [MP][MP]. The triangles MACMA'C and PACPA'C are congruent, yielding [PO]=[MO][PO] = [MO]. Then OSMPOS \perp MP, hence OSOS is the distance between the lines MPMP and ACA'C.
We have MC=a52MC = \frac{a\sqrt{5}}{2}, therefore MO=a22MO = \frac{a\sqrt{2}}{2}. Next, MP=a62MP = \frac{a\sqrt{6}}{2}, and from the right triangle OSMOSM, we obtain OS=a24OS = \frac{a\sqrt{2}}{4}.

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