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Let \(A_1 A_2 A_3 A_4\) be a regular tetrahedron. Let \(P_1\) be the center of face \(A_2 A_3 A_4,\) and define vertices \(P_2, P_3,\) and \(P_4\) the same way. Find the ratio of the volume of tetrahedron \(A_1 A_2 A_3 A_4\) to the volume of tetrahedron \(P_1 P_2 P_3 P_4.\)

Note: A tetrahedron is called regular if all the edges lengths are equal, so all the faces are equilateral triangles.

Thank you!

Apr 17, 2020

#1
+20810
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Since each of the points Pis the centroid of the triangle that includes it, it is found one-third from the distance from the base to the vertext of this triangle.

Therefore, the height of the inner tetrahedron is 1/3rd the height of the outer tetrahedron.

Since these shapes are similar, the volume of the inner tetrahedron is (1/3)3, or one-twenty-seventh the volume of the outer tetrahedron.

This makes the volume of the outer tetrahedron 27 times the volume of the inner tetrahedron.

e

Apr 17, 2020
#2
+1

How did I not see that ?? I really need to get better at thinking outside the box.

Apr 17, 2020
#3
+109501
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Nice work Gino.

I would have to some work before I found the 1/3 part.

But the rest all seems reasonable ... I think.

Melody  Apr 17, 2020
edited by Melody  Apr 17, 2020