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$OABC$ is a regular tetrahedron. Let $E$, $F$, $G$, $H$ be the centroids of the triangles $OBA$, $OCB$, $OAC$, $ABC$, respectively. You are given that EFGH is also a regular tetrahedron. Using $OA=a$, $OB=b$, $OC=c$, find the ratio of the side length of tetrahedron $EFGH$ to the side length of tetrahedron $OABC$.

Show using vectors. Please give full proof to the answer.

Link to the picture of the tetrahedron used in this question: http://imgur.com/SFzGA5E

the correct answer is a 1:3 ratio, however i am unsure as to how this is calculated

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marked as duplicate by Blue, Andrew D. Hwang, user26857, Christopher, Mike Pierce Jun 1 '15 at 14:42

This question was marked as an exact duplicate of an existing question.

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    $\begingroup$ can you please share your ideas as to how you approached this problem $\endgroup$ – happymath Jun 1 '15 at 8:55
  • $\begingroup$ Welcome to Math.SE. Thoughtful questions, even homework-related, are welcome. However, you should not expect others to do your homework for you, and it's inappropriate to post your homework verbatim. (That may not be what you're doing, but all people here can go by is appearances.) Instead, please try to ask questions about specific places where you're stuck, or about particular concepts you don't understand. (Here's a hint for your question: A regular tetrahedron can be inscribed in a cube by "taking alternate vertices".) $\endgroup$ – Andrew D. Hwang Jun 1 '15 at 10:20
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    $\begingroup$ Albert beat you to it. $\endgroup$ – Blue Jun 1 '15 at 11:35
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This proof is very simple through geometrical approach as I explain here.

In general, if a regular tetrahedron has its each edge of length $a$ then the radius ($\color{blue}{R_{inscribed}}$) of inscribed sphere (i.e. the normal distance of each equilateral triangular face from the center of tetrahedron) is given by the following formula (derived in HCR's formula for platonic solids) $$\bbox[4pt, border: 1px solid blue;]{\color{blue}{R_{\text{inscribed}}}=\color{blue}{\frac{a}{2\sqrt{6}}}}$$

And the radius ($\color{red}{R_{circumscribed}}$) of circumscribed sphere (i.e. the distance of each vertex from the center of tetrahedron) is given by the following formula (derived in HCR's formula for platonic solids) $$\bbox[4pt, border: 1px solid blue;]{\color{red}{R_{\text{circumscribed}}}=\color{red}{\frac{a}{2}\sqrt{\frac{3}{2}}}}$$ Now, let the edge length of regular tetrahedron $OABC$ (the larger one) be $L$ & that of small regular tetrahedron $EFGH$ (the smaller one) be $l$.

Since tetrahedron $EFGH$ is obtained by joining the centroids (centers) $E$, $F$, $G$ & $H$ of the triangular faces $OBA$, $OCB$, $OAC$ & $ABC$ of regular tetrahedron $OABC$ Hence, their centers are coincident i.e. both have a common center let it be $I$ then in this case, we have $$IE=IF=IG=IH=\text{inscribed radius of tetrahedron}\space OABC \space \text{with edge length} \space L$$ $$\implies IE=IF=IG=IH=\color{blue}{\frac{L}{2\sqrt{6}}}\tag 1$$ And also
$$IE=IF=IG=IH=\text{circumscribed radius of tetrahedron}\space EFGH \space \text{with edge length} \space l$$ $$\implies IE=IF=IG=IH=\color{red}{\frac{l}{2}\sqrt{\frac{3}{2}}}\tag 2$$ Now, equating (1) & (2), we get $$\frac{L}{2\sqrt{6}}=\frac{l}{2}\sqrt{\frac{3}{2}}$$ $$\frac{l}{L}=\frac{2\sqrt{\frac{2}{3}}}{2\sqrt{6}}$$ $$\implies \color{green}{\frac{l}{L}=\frac{1}{3}} \quad \text{or} \quad \color{green}{l:L=1:3}$$ Hence, the ratio of the side length ($l$) of tetrahedron $EFGH$ to the side length ($L$) of tetrahedron $OABC$ is $\color{blue}{1:3}$

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