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A. 547. A tetrahedron OA1A2A3 is given. For i=1,2,3 let Bi be a point in the interior of the edge OAi and let Ci be a point on the ray OAi, beyond Ai. Suppose that the polyhedron bounded by the six planes OAi+1Ai+2 and BiAi+1Ai+2 (i=1,2,3) circumscribes a sphere, and the polyhedron bounded by the planes BiAi+1Ai+2 and CiAi+1Ai+2 also circumscribes a sphere. Prove that the polyhedron bounded by the planes OAi+1Ai+2 and CiAi+1Ai+2 also circumscribes a sphere.

(5 points)

Deadline expired on 12 December 2011.


Solution (outline). We will use oriented lines, planes and angles. For arbitrary, non-collinear points X,Y,Z, the orinetation of the line XY the same as the direction of the vector \overrightarrow{XY}, and the vector \overrightarrow{XY}\times\overrightarrow{XZ} points into the positive side of the plane XYZ,

Lemma. Let \ell be an oriented line which is contained by the oriented planes \mathcal{P}_0, \mathcal{P}_1 and \mathcal{P}_2. Let \varphi_i=\angle(\mathcal{P}_0,\mathcal{P}_i) be the angle by which the plane \mathcal{P}_0 can be rotated to \mathcal{P}_i around \ell (i=1,2; these angles are considered modulo 360o). Suppose that some point X has signed distances d,r,-r from the planes \mathcal{P}_0, \mathcal{P}_1, \mathcal{P}_2, respectively. Then we have


\frac{r+d}{r-d} = \frac{\tg\dfrac{\varphi_2}2}{\tg\dfrac{\varphi_1}2}
\quad\text{and}\quad
\frac{d}{r} =
\frac{\tg\dfrac{\varphi_2}2+\tg\dfrac{\varphi_1}2}{
  \tg\dfrac{\varphi_2}2-\tg\dfrac{\varphi_1}2}.

Proof. Denote by \mathbf{n}_i the unit normal vector of the plane \mathcal{P}_i which points to the positive side of the plane and let \mathbf{v} be the direction vector of the line \ell Let \mathbf{x} be a vector from a point of \ell to X. Then


d = \mathbf{n}_0\cdot\mathbf{x}, \qquad
r = \mathbf{n}_1\cdot\mathbf{x}, \qquad
-r = \mathbf{n}_2\cdot\mathbf{x}.

Let \mathbf{m}=\mathbf{n}_0\times\mathbf{v}.Then


  \mathbf{n}_1 = \cos\varphi_1\cdot\mathbf{n}_0-\sin\varphi_1\cdot\mathbf{m}
  \quad\text{and}\quad
  \mathbf{n}_2 = \cos\varphi_2\cdot\mathbf{n}_0-\sin\varphi_2\cdot\mathbf{m},


  \sin\varphi_2\cdot \mathbf{n}_1-\sin\varphi_1 \cdot \mathbf{n}_2 =
  (\sin\varphi_2\cos\varphi_1-\sin\varphi_1\cos\varphi_2) \cdot \mathbf{n}_0 =
  \sin(\varphi_2-\varphi_1)\cdot\mathbf{n}_0.

Taking the dot-product with \mathbf{x},


  \sin\varphi_2 \cdot (\mathbf{n}_1\cdot\mathbf{x}) - \sin\varphi_1 \cdot (\mathbf{n}_2\cdot\mathbf{x})
  = \sin(\varphi_2-\varphi_1)\cdot (\mathbf{n}_0\cdot\mathbf{x})

sin \varphi2.r-sin \varphi1.(-r)=sin (\varphi2-\varphi1).d


  \dfrac{d}{r} =
  \dfrac{\sin\varphi_2+\sin\varphi_1}{\sin(\varphi_2-\varphi_1)} =
  \dfrac{
    2\sin\dfrac{\varphi_2+\varphi_1}2\cos\dfrac{\varphi_2-\varphi_1}2
  }{
    2\sin\dfrac{\varphi_2-\varphi_1}2\cos\dfrac{\varphi_2-\varphi_1}2
  } =
  \dfrac{
    \sin\dfrac{\varphi_2+\varphi_1}2
  }{
    \sin\dfrac{\varphi_2-\varphi_1}2
  } =


  = \dfrac{
    \sin\dfrac{\varphi_2}2\cos\dfrac{\varphi_1}2+
    \sin\dfrac{\varphi_1}2\cos\dfrac{\varphi_2}2
  }{
    \sin\dfrac{\varphi_2}2\cos\dfrac{\varphi_1}2-
    \sin\dfrac{\varphi_1}2\cos\dfrac{\varphi_2}2
  } =
  \dfrac{
    \tg\dfrac{\varphi_2}2 + \tg\dfrac{\varphi_1}2
  }{
    \tg\dfrac{\varphi_2}2 - \tg\dfrac{\varphi_1}2
  }.

From this we get


\dfrac{d+r}{d-r} = 
\dfrac{\dfrac{d}{r}+1}{\dfrac{d}{r}-1} =
\dfrac{
  \dfrac{
    \tg\frac{\varphi_2}2+\tg\frac{\varphi_1}2
  }{
    \tg\frac{\varphi_2}2-\tg\frac{\varphi_1}2
  }+1
}{
  \dfrac{
    \tg\frac{\varphi_2}2+\tg\frac{\varphi_1}2
  }{
    \tg\frac{\varphi_2}2-\tg\frac{\varphi_1}2
  }-1
}
= \dfrac{
  \tg\dfrac{\varphi_2}2
}{
  \tg\dfrac{\varphi_1}2
}.

The lemma is proved.

Let X and rX be the center and radius of the sphere, respectively, which is circumcribed by the planes OAi+1Ai+2 and BiAi+1Ai+2. Let dX be the signed distance between X and the plane A1A2A3 síktól dX. The signed distances between X and the planes BiAi+1Ai+2 are all rX, and the signed distances between X and the planes OAi+1Ai+2 are all -rX.

Analogously, let Y and rY be the center and radius of the sphere which is circumscribed by the planes BiAi+1Ai+2 and CiAi+1Ai+2. let dY by the signed distance between Y and the plane A1A2A3. The signed distances between Y and the planes CiAi+1Ai+2 are all rY, and the signed distances between X and the planes BiAi+1Ai+2 are all -rY.

For every i=1,2,3 take the external angle bisector of the planes Ai+1Ai+2O and Ai+1Ai+2Ci, and let Z be the intersection point of the three angle bisector planes --- Z will be center of the desired sphere. (Now assume that none of the angle bisectors coincides with the plane A1A2A3; then the point Z is unique and does not lie on A1A2A3.)

Let dZ and \varrho be the signed distance between Z and A1A2A3 and A1A2C3, respectively. Since Z lies on the external angle bisector between A1A2C3 and A1A2O, the signed distance between Z and A1A2O is -\varrho.

Let \alpha=\angle(A1A2A3,A1A2O), \beta=\angle(A1A2A3,A1A2B3) and \gamma=\angle(A1A2A3,A1A2C3), where the angles are measured around the oriented line A1A2.

Applying the Lemma for the line A1A2, the planes A1A2A3, A1A2B3, A1A2O and the point X we get


\frac{\tg\dfrac{\alpha}2}{\tg\dfrac{\beta}2} = \frac{r_X+d_X}{r_X-d_X}.
(1)

Similarly, applying the Lemma for the line A1A2, the planes A1A2A3, A1A2C3, A1A2B3 and the point Y we get


\frac{\tg\dfrac{\beta}2}{\tg\dfrac{\gamma}2} = \frac{r_Y+d_Y}{r_Y-d_Y}.
(2)

Finally, Similarly, applying the Lemma for the line A1A2, the planes A1A2A3, A1A2C3, A1A2O and the point Y we obtain


\frac{d_Z}{\varrho} =
\frac{\tg\dfrac{\alpha}2+\tg\dfrac{\gamma}2}{
  \tg\dfrac{\alpha}2-\tg\dfrac{\gamma}2} =
\frac{
  \dfrac{\tg\frac{\alpha}2}{\tg\frac{\gamma}2}+1}{
  \dfrac{\tg\frac{\alpha}2}{\tg\frac{\gamma}2}-1} =
\frac{
  \dfrac{r_X+d_X}{r_X-d_X} \cdot \dfrac{r_Y+d_Y}{r_Y-d_Y} +1}{
  \dfrac{r_X+d_X}{r_X-d_X} \cdot \dfrac{r_Y+d_Y}{r_Y-d_Y} -1}.

As can be seen, due to dZ\ne0 the distances dX,rX,dY,rY,dZ determine \varrho. If we cyclicly shift the indices 1,2,3, we obtain the same value. Hence, the point Z has the same distance from all planes OAi+1Ai2 and CiAi+1Ai2.

In the particular case when some planes CiAi+1Ai+2 and OAi+1Ai+2 are symmetric over A1A2A3, the points C1,C2,C3 can be moved in such a way that the three spheres exist; then the original configuration can be considered as a limit case.


Statistics on problem A. 547.
2 students sent a solution.
0 point:2 students.


  • Problems in Mathematics of KöMaL, November 2011

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