Tetrahedron properties

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Source: IMO 1962, Day 2, Problem 7

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DPopov

#1 h Oct 8, 2005, 5:56 PM • 1 Y

Y by Adventure10

The tetrahedron $SABC$ has the following property: there exist five spheres, each tangent to the edges $SA, SB, SC, BC, CA, AB,$ or to their extensions.

a) Prove that the tetrahedron $SABC$ is regular.

b) Prove conversely that for every regular tetrahedron five such spheres exist.

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#2 h Apr 23, 2014, 5:33 PM • 2 Y

Y by Adventure10, Mango247

a) Let $\mathcal{S}$ be the sphere tangent to the edges from the inside, and let $\mathcal{S}_X$ be the sphere tangent to the edges of the face opposite to vertex $X$ , and the extensions of the edges from $X$ .
Consider say $\mathcal{S}$ and $\mathcal{S}_S$ . If we cut these spheres by plane $ABC$ , in both cases we get a circle on plane $ABC$ tangent to the segments $AB,BC,CA$ - this is the incircle of $\triangle ABC$ . Hence $\mathcal{S}$ and $\mathcal{S}_S$ touches the edges $AB,BC,CA$ in the same points.
Now cut the two spheres by plane $SAB$ . We get the incircle and $S$ -excircle of $\triangle SAB$ . We have proven that $\mathcal{S}$ and $\mathcal{S}_S$ touch $AB$ in the same point, so the incircle and $S$ -excircle of $\triangle SAB$ touch $AB$ at the same point. This implies that $SA=SB$ .
Similarly, we get that any two edges are the same length, therefore the tetrahedron is indeed regular.

b) Look at the regular tetrahedron $SABC$ .
By symmetry, the center $O$ of the regular tetrahedron $SABC$ is of equal distance $d$ from all its edges, so taking $\mathcal{S}(O,d)$ works.
Let the incenter of say $\triangle SAB$ be $I$ , the $S$ -excenter of $\triangle SAB$ be $J$ . Consider the homothety with center $S$ that maps $I$ to $J$ . This homothety maps the incircle $\omega$ of $\triangle SAB$ to the $S$ -excircle of $\triangle SAB$ ; since $\omega\subset \mathcal{S}$ , we have $\omega'\subset\mathcal{S}'$ and therefore $\mathcal{S}'$ also touches edge $AB$ . By rotational symmetry, $\mathcal{S}'$ also touches edges $BC$ and $CA$ . Moreover, because $\mathcal{S}$ touched edges $SA,SB,SC$ , $\mathcal{S}'$ will touch the extensions of $SA,SB,SC$ . This proves that $\mathcal{S}_S=\mathcal{S}'$ is up for the job.
By symmetry, so do $\mathcal{S}_A,\mathcal{S}_B,\mathcal{S}_C$ exist.

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