Circles on a Line

  1. Consider three identical equilateral triangles, such that one triangle shares a side with each of the other two. A circle is drawn inside each triangle, where all three circles are identical and as large as possible, while keeping the centers of the three circles collinear (all on the same line.) If the height of the triangles is 1, what are the radii of the circles? How far along the side of one triangle is the point of tangency with its internal circle?
  2. Consider three identical squares, where one shares a side with each of the other two (in a right-angle configuration.) As above, identical circles are drawn inside each square such that their centers are collinear and the circles are as large as possible. If the side of the squares is 1, what are the radii of the circles? How far along the side of one square is the point of tangency with its internal circle?
  3. In a 4x4 grid of squares, identical circles are placed in the outer 12 squares, such that they are as large as possible and their centers are all on the same large circle (with center at the middle of the grid.) If the side of the squares is 1, what are the radii of the circles? How far along the side of one square is the point of tangency with its internal circle?
  4. Consider three identical hexagons, where each shares a side with each of the other two. As above, identical circles are drawn inside each hexagon such that their centers are collinear and the circles are as large as possible. If the height of the hexagons is 1 (between opposite sides), what are the radii of the circles? How far along the side of one hexagon is the point of tangency with its internal circle?
  5. Consider three identical circles, where each is tangent to each of the other two. As above, identical circles are drawn inside each circle such that their centers are collinear and the internal circles are as large as possible. If the radii of a large circle is 1, what are the radii of the smaller circles?

Source: Original. The last was suggested by Phillippe Fondanaiche.


Solution
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