AMC 10 Daily Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 22 Hard

Consider four spheres in the same space with radii 2, 2, 3, 3, respectively. Each of the four spheres are externally tangent to the other three spheres. There is a fifth sphere such that is externally tangent to all four original spheres. Determine the radius of the fifth sphere.

  • A.

    \frac 14

  • B.

    \frac 13

  • C.

    \frac 12

  • D.

    \frac{6}{11}

  • E.

    \frac{7}{12}

Answer:D

Let the centers of the spheres with radii 3 be A and B, respectively. Similarly, let the centers of the spheres with radii 2 be C and D, respectively.

Then we can readily get AB=6 and CD=4.

In addition, AC=AD=BC=BD=5.

Let the center of the fifth sphere be O and radius r. So O is inside the tetrahedron ABCD. AO=BO=3+r and CO=D0=2+r. Take the midpoint of AB and denote as E. Connect CE and DE, so CE\bot AB and DE\bot AB.

∴ Plane CDE is the perpendicular bisector plane of AB and we denote as a,

∴ O is in the plane CDE. Also, by OC=OD=2+r, we have O is in the perpendicular bisector plane of CDE,

∴ O is on the altitude of the base CD of \triangle CED, which is EF. F is the midpoint of CD. Then, we calculate that ED=EC=\sqrt{{{5}^{2}}-{{3}^{2}}}=4,

∴ \triangle ECD is an equilateral triangle adn EF=\frac{\sqrt{3}}{2}ED=2\sqrt{3}, OF=\sqrt{O{{C}^{2}}-C{{F}^{2}}}=\sqrt{{{(2+r)}^{2}}-{{2}^{2}}}=\sqrt{r(4+r)}, OE=\sqrt{O{{A}^{2}}-A{{E}^{2}}}=\sqrt{{{(3+r)}^{2}}-{{3}^{2}}}=\sqrt{r\left( 6+r \right)},

Plug in, OE+OF=EF=2\sqrt{3}.

The final equation is \sqrt{r(4+r)}+\sqrt{r(6+r)}=2\sqrt{3}. Solve to get r=\frac{6}{11}.

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