Difference between revisions of "1962 IMO Problems/Problem 7"

Revision as of 19:36, 18 November 2023

Problem

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.

Solution

Part (a)

Let points $P_{SA}, P_{SB}, P_{SC}, P_{BC}, P_{CA}, P_{AB}$ be the points where the smallest sphere is tangent to the edges $SA, SB, SC, BC, CA, AB$ respectively.

For each face of the tetrahedron there is a circular cross section of the smallest sphere. Since that circle also needs to be tangent to the edges, then that circle is the incircle of each triangular face.

From the properties of an incircle we know that:

$|AP_{AB}|=|AP_{AC}|$ , $|BP_{BC}|=|BP_{BA}|$ , and $|CP_{CA}|=|CP_{CB}|$ in $\Delta ABC$

There is a larger circle that is tangent to $AB$ and the extensions of $CA$ and $CB$ .

This larger circle is a cross section of the sphere that's also tangent to the extension of $CS$ away from $S$ and the edges of $\Delta ABC$

Therefore, this larger circle and the incircle of $\Delta ABC$ are part of that same sphere.

In order for these two circles to be part of the same sphere and also tangent to line $AB$ , then the point of the tangent of this larger circle needs to be the same as point $P_{AB}$

The only way these to circles can share the same tangent point on edge $AB$ is if $|AP_{AB}|=|BP_{AB}|$

Using the same argument with the larder circle of edge $BC$ then $|BP_{BC}|=|CP_{BC}|$

and with the larger circle of edge $AC$ then $|CP_{AC}|=|AP_{AC}|$

This results in:

$|AP_{AB}|=|AP_{AC}|=|BP_{BC}|=|BP_{BA}|=|CP_{CA}|=|CP_{CB}|$ in $\Delta ABC$

which means that

$|AB|=|BC|=|AC|$ in $\Delta ABC$ , thus $\Delta ABC$ is an equilteral triangle.

Likewise,

$|AB|=|BS|=|AS|$ in $\Delta ABS$ , thus $\Delta ABS$ is an equilteral triangle.

$|AS|=|SC|=|AC|$ in $\Delta ASC$ , thus $\Delta ASC$ is an equilteral triangle.

$|SB|=|BC|=|SC|$ in $\Delta SBC$ , thus $\Delta SBC$ is an equilteral triangle.

Since all four faces are equilateral triangles, then tetrahedron $ABCS$ is a regular tetrahedron.

Part (b)

Let's consider a regular tetrahedron in the cartesian space with center at $(0,0,0)$ with $r_c$ as the circumradius.

We can write the coordinates for the four vertices as:

$V_1=\left( \frac{2\sqrt{2}}{3}r_c ,\;0,\;-\frac{r_c}{3}\right)$

$V_2=\left( -\frac{\sqrt{2}}{3}r_c ,\;\frac{\sqrt{6}}{3}r_c,\;\frac{-r_c}{3}\right)$

$V_3=\left( -\frac{\sqrt{2}}{3}r_c ,\;-\frac{\sqrt{6}}{3}r_c,\;\frac{-r_c}{3}\right)$

$V_4=\left(0 ,\;0,\;r_c\right)$

And the midpoints of all edges as:

$M_{12}=\frac{V_1-V_2}{2}=\left( \frac{\sqrt{2}}{6}r_c ,\;\frac{\sqrt{6}}{6}r_c,\;-\frac{r_c}{3}\right)$

$M_{13}=\frac{V_1-V_3}{2}=\left( \frac{\sqrt{2}}{6}r_c ,\;-\frac{\sqrt{6}}{6}r_c,\;-\frac{r_c}{3}\right)$

$M_{23}=\frac{V_2-V_3}{2}=\left( -\frac{\sqrt{2}}{3}r_c ,\;0,\;-\frac{r_c}{3}\right)$

$M_{14}=\frac{V_1-V_4}{2}=\left( \frac{\sqrt{2}}{3}r_c ,\;0,\;\frac{r_c}{3}\right)$

$M_{24}=\frac{V_2-V_4}{2}=\left( -\frac{\sqrt{2}}{6}r_c ,\;\frac{\sqrt{6}}{6}r_c,\;\frac{r_c}{3}\right)$

$M_{34}=\frac{V_3-V_4}{2}=\left( -\frac{\sqrt{2}}{6}r_c ,\;-\frac{\sqrt{6}}{6}r_c,\;\frac{r_c}{3}\right)$

Now we calculate the following six dot products:

$(V_1-V_2) \bullet M_{12}=\left( -\sqrt{2}r_c \right)\left( \frac{\sqrt{2}}{6}r_c \right)+ \left( \frac{\sqrt{6}}{3}r_c\right)\left( \frac{\sqrt{6}}{6}r_c \right)+ 0=-\frac{1}{3}r_c^2+\frac{1}{3}r_c^2=0$

$(V_1-V_3) \bullet M_{13}=\left( -\sqrt{2}r_c \right)\left( \frac{\sqrt{2}}{6}r_c \right)+ \left(- \frac{\sqrt{6}}{3}r_c\right)\left( -\frac{\sqrt{6}}{6}r_c \right)+ 0=-\frac{1}{3}r_c^2+\frac{1}{3}r_c^2=0$

$(V_2-V_3) \bullet M_{23}=0+\left(- \frac{\sqrt{6}}{6}r_c\right)\left( 0\right)+ 0=0$

$(V_1-V_4) \bullet M_{13}=\left( 2\frac{\sqrt{2}}{3}r_c \right)\left( \frac{\sqrt{2}}{3}r_c \right)+0+ \left(- \frac{4}{3}r_c\right)\left( \frac{1}{3}r_c \right)=\frac{4}{9}r_c^2-\frac{4}{9}r_c^2=0$

$(V_2-V_4) \bullet M_{13}=\left( -\frac{\sqrt{2}}{3}r_c \right)\left( -\frac{\sqrt{2}}{6}r_c \right)+\left( \frac{\sqrt{6}}{3}r_c \right)\left( \frac{\sqrt{6}}{6}r_c \right)+ \left(- \frac{4}{3}r_c\right)\left( \frac{1}{3}r_c \right)=\frac{1}{9}r_c^2+\frac{1}{3}r_c^2-\frac{4}{9}r_c^2=0$

$(V_3-V_4) \bullet M_{13}=\left( -\frac{\sqrt{2}}{3}r_c \right)\left( -\frac{\sqrt{2}}{6}r_c \right)+\left( -\frac{\sqrt{6}}{3}r_c \right)\left( -\frac{\sqrt{6}}{6}r_c \right)+ \left(- \frac{4}{3}r_c\right)\left( \frac{1}{3}r_c \right)=\frac{1}{9}r_c^2+\frac{1}{3}r_c^2-\frac{4}{9}r_c^2=0$

Since all these dot producs equal to $0$ that means that the lines from the center to each of the midpoints are all perpendicular.

Now we calculate the distances from $(0,0,0)$ to all the midpoints:

$|M_{12}|=\sqrt{\left( \frac{\sqrt{2}}{6}r_c\right)^2+\left(\frac{\sqrt{6}}{6}r_c\right)^2+\left(-\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$

$|M_{13}|=\sqrt{\left( \frac{\sqrt{2}}{6}r_c\right)^2+\left(-\frac{\sqrt{6}}{6}r_c\right)^2+\left(-\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$

$|M_{23}|=\sqrt{\left( -\frac{\sqrt{2}}{3}r_c\right)^2+\left(0\right)^2+\left(-\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$

$|M_{14}|=\sqrt{\left( \frac{\sqrt{2}}{3}r_c\right)^2+\left(0\right)^2+\left(\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$

$|M_{24}|=\sqrt{\left( -\frac{\sqrt{2}}{6}r_c\right)^2+\left(\frac{\sqrt{6}}{6}r_c\right)^2+\left(\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$

$|M_{34}|=\sqrt{\left( -\frac{\sqrt{2}}{6}r_c\right)^2+\left(-\frac{\sqrt{6}}{6}r_c\right)^2+\left(\frac{r_c}{3}\right)^2}=\frac{\sqrt{3}}{3}r_c$

Since all distances are all the same and all dot products are $0$ , then we have our first sphere at $(0,0,0)$ with a radius of $\frac{\sqrt{3}}{3}r_c$

Now we will look at the other 4 spheres.

Let $E_{ij}$ be the point of tangent of the larger sphere on the extension of line $V{i}V{j}$ in the direction of $V{i}$ to $V{j}$ and beyond $V{j}$ for $i=1,2,3,4$ ; $j=1,2,3,4$ ; and $i \ne j$

Since $|V_j E_{ij}|=|V_j M_{ij}|=\frac{1}{2} |V_j V_i|$ , then $\frac{1}{2}(V_j-V_i)=E_{ij}-V_j$ , thus $E_{ij}=\frac{3V_j-V_i}{2}$

Using this formula we calculate the following:

$E_{41}=\left( \sqrt{2} r_c ,\;0,\;-r_c \right)$

$E_{42}=\left( -\frac{\sqrt{2}}{2}r_c ,\;\frac{\sqrt{6}}{2}r_c,\;-r_c \right)$

$E_{43}=\left( -\frac{\sqrt{2}}{2}r_c ,\;-\frac{\sqrt{6}}{2}r_c,\;-r_c \right)$

We will start with the sphere below the base of the tetrahedron opposite of vertex $V_{4}$ below $\Delta V{1}V{2}V{3}$

The center of this larger sphere is at $C_1=(0,0,-2r_c)$ and it is tangent at points $M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}$

We calculate the following dot products:

$(V_1-V_2) \bullet (M_{12}-C_1)=\left( -\sqrt{2}r_c \right)\left( \frac{\sqrt{2}}{6}r_c \right)+ \left( \frac{\sqrt{6}}{3}r_c\right)\left( \frac{\sqrt{6}}{6}r_c \right)+ 0=-\frac{1}{3}r_c^2+\frac{1}{3}r_c^2=0$

$(V_1-V_3) \bullet (M_{13}-C_1)=\left( -\sqrt{2}r_c \right)\left( \frac{\sqrt{2}}{6}r_c \right)+ \left(- \frac{\sqrt{6}}{3}r_c\right)\left( -\frac{\sqrt{6}}{6}r_c \right)+ 0=-\frac{1}{3}r_c^2+\frac{1}{3}r_c^2=0$

$(V_2-V_3) \bullet (M_{23}-C_1)=0+\left(- \frac{\sqrt{6}}{6}r_c\right)\left( 0\right)+ 0=0$

$(E_{41}-V_4) \bullet (E_{41}-C_1) = \left( \sqrt{2}r_c \right)^2+0+\left( -2r_c \right)\left( -r_c+2r_c \right)=2r_c^2-2r_c^2=0$

$(E_{42}-V_4) \bullet (E_{42}-C_1) = \left( -\frac{\sqrt{2}}{2}r_c \right)^2+\left( \frac{\sqrt{6}}{2}r_c \right)^2+\left( -2r_c \right)\left( -r_c+2r_c \right)=2r_c^2-2r_c^2=0$

$(E_{43}-V_4) \bullet (E_{43}-C_1) = \left( -\frac{\sqrt{2}}{2}r_c \right)^2+\left( -\frac{\sqrt{6}}{2}r_c \right)^2+\left( -2r_c \right)\left( -r_c+2r_c \right)=2r_c^2-2r_c^2=0$

Since all these dot producs equal to $0$ that means that the lines from the center $C_1$ to each of the point $M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}$ are all perpendicular.

Now we calculate the distances from $C_1=(0,0,-2r_c)$ to points $M_{12}, M_{13}, M_{23}, E_{41}, E_{42}, E_{43}$

$|C_1M_{12}|=\sqrt{\left( \frac{\sqrt{2}}{6}r_c\right)^2+\left(\frac{\sqrt{6}}{6}r_c\right)^2+\left(-\frac{5}{3}r_c\right)^2}=\sqrt{3}r_c$

$|C_1M_{13}|=\sqrt{\left( \frac{\sqrt{2}}{6}r_c\right)^2+\left(-\frac{\sqrt{6}}{6}r_c\right)^2+\left(-\frac{5}{3}r_c\right)^2}=\sqrt{3}r_c$

$|C_1M_{23}|=\sqrt{\left( -\frac{\sqrt{2}}{3}r_c\right)^2+\left(0\right)^2+\left(-\frac{5}{3}r_c\right)^2}=\sqrt{3}r_c$

$|C_1E_{41}|=\sqrt{\left( \sqrt{2}r_c\right)^2+\left(0\right)^2+\left(-r_c\right)^2}=\sqrt{3}r_c$

$|C_1E_{42}|=\sqrt{\left( -\frac{\sqrt{2}}{2}r_c\right)^2+\left(\frac{\sqrt{6}}{2}r_c\right)^2+\left(-r_c\right)^2}=\sqrt{3}r_c$

$|C_1E_{43}|=\sqrt{\left( -\frac{\sqrt{2}}{2}r_c\right)^2+\left(-\frac{\sqrt{6}}{2}r_c\right)^2+\left(-r_c\right)^2}=\sqrt{3}r_c$

Since all distances are all the same and all dot products are $0$ , then we have one of the larger spheres at $(0,0,-2r_c)$ with a radius of $\sqrt{3}r_c$

Then, the other three larger spheres which are the same size as the sphere with center at $C_1$ are congruent and tangent to their respective sides near the other faces of the tetrahedron.

and this proves that a tetrahedron with any circumradius $r_c$ will have these 5 spheres, one with radius $\frac{\sqrt{3}}{2}r_c$ , at the center of the tetrahedron and the other 4 with radius $\sqrt{3}r_c$ at centers that are $3r_c$ away from any of the vertices of the tetrahedron in the direction from that vertex to the center of its opposite face.

Thus these five spheres exist for any regular tetrahedron.

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 50: / Line 50: @@
 Since all four faces are equilateral triangles, then tetrahedron <math>ABCS</math> is a regular tetrahedron.
-part (b)
+Part (b)
 Let's consider a regular tetrahedron in the cartesian space with center at <math>(0,0,0)</math> with <math>r_c</math> as the circumradius.

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Difference between revisions of "1962 IMO Problems/Problem 7"

Revision as of 19:36, 18 November 2023

Problem

Solution

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