Problem 24 · 2011 Math Kangaroo
Stretch
Geometry & Measurement
spatial-reasoning
Given is a regular tetrahedron ABCD whose face ABC lies on the plane \(\varepsilon\). The edge BC lies on the straight line s. Another tetrahedron BCDE shares one face with ABCD. Where does the straight line DE intersect the plane \(\varepsilon\)?
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Answer: C — Outside of ABC, not on the same side of s as A.
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Hint 1 of 3
Since ABCD is regular, A is itself an apex over face BCD, so E (the other regular tetrahedron's apex on BCD) is just A reflected across the plane BCD.
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Hint 2 of 3
Set simple coordinates with BC on the x-axis and A on the positive-y side, then find where line DE crosses z = 0.
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Hint 3 of 3
Check the y-sign of that crossing point to decide which side of s it lands on.
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Approach: reflect A across plane BCD to get E, then intersect line DE with the base plane
- All edges of ABCD are equal, so A is an apex of face BCD; the second regular tetrahedron's apex E is therefore the mirror image of A in the plane BCD.
- Placing B and C on line s with A on the positive-y side and D the apex above triangle ABC, reflecting A across plane BCD puts E below, with a negative y-coordinate.
- Extending DE down to the base plane ε gives a point with negative y — outside triangle ABC and on the far side of s from A.
- So DE meets ε outside ABC, not on A's side of s, choice (C).
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