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Source: 4th Iranian Geometry Olympiad (Advanced) P5

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#1 h Sep 15, 2017, 6:11 AM • 1 Y

Y by Adventure10

Sphere $S$ touches a plane. Let $A,B,C,D$ be four points on the plane such that no three of them are collinear. Consider the point $A'$ such that $S$ in tangent to the faces of tetrahedron $A'BCD$ . Points $B',C',D'$ are defined similarly. Prove that $A',B',C',D'$ are coplanar and the plane $A'B'C'D'$ touches $S$ .

Proposed by Alexey Zaslavsky (Russia)

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#3 h Feb 10, 2018, 6:53 PM • 2 Y

Y by Adventure10, Mango247

Nice problem!

My solution:

Invert at the contact point of $S$ and plane $ABCD$ , say $X$ . From equal tangents from each of the points $A,B,C,D$ separately, we have that $P$ is on a sphere associated with each point (say $S_a$ ). In fact, since the sphere intersects at a circle with another sphere, the points lie on a circle. Denote by $X_{LMN}$ , the image of the tangency point of face $LMN$ with $S$ . The points $X_{ABC'} \equiv X_{ABD'}, X_{ACB'} \equiv X_{ACD'}, X_{ADB'} \equiv X_{ADC'}$ are collinear. Thus we have 4 such lines. Let's dropped the primed points from the name, as they are redundant.
The circle through $X_{BC}, X_{BD}, X_{CD}$ is the image of the circle intercepted by the tangent cone from $A'$ to $S$ , by $S$ . Consider the Miquel point of the four lines mentioned. The image of this point is on the sphere, and the line $A'P$ is tangent to the sphere $S$ . By symmetry, the lines $B'P, C'P, D'P$ are also tangent to $S$ , and thus we have the conclusion from the fact that all tangents at a point on the sphere lie on a plane.

This post has been edited 1 time. Last edited by WizardMath, Feb 10, 2018, 6:54 PM

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#4 h Sep 13, 2022, 11:42 AM • 1 Y

Y by guptaamitu1

Take the polarity of the problem wrt. $S$ . Then the four complanar point became $4$ planes that cut the the sphere in $4$ circles ( $w_a,w_b,w_c,w_d$ ) passing through the tangency point P of the sphere, then the point $A'$ become the plane passing throught the intersection of $w_b,w_c,w_d$ different from $P$ and call the circle in which cuts the sphere $w_{a'}$ ; do the same construction for the other 3 point $B', C', D'$ . Now the condition becomes $w_{a'}, w_{b'},w_{c'},w_{d'}$ all intersect in a common point. Now invert in P with arbitrarius radius, the the problem becomes given 4 lines in a plane $l_a,l_b,l_c,l_d$ , inversion of the four circle passing through $P$ , then taken circumcircle of the 3 intersection of $l_a,l_b,l_c$ and and simlarly for the other 3 choises of 3 lines, they pass through a single point. But this is just the definition of the miquel point of the quadrilateral, hence we are done. Q. E. D.

This post has been edited 1 time. Last edited by FairyBlade, Sep 13, 2022, 11:42 AM

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