Discs with radii of 3 cm and 4 cm are placed in the smallest equilateral triangle that can contain them, similar to
Discs in a Triangle.
Then two more discs of equal radii are placed along with the 3 cm and 4 cm discs inside the same triangle. What is the largest radius that the new discs can have?
Let AB be the edge of the triangle that the 4 cm circle doesn't touch. Without loss of generality, let AB be horizontal. Let r be the radii of the small circles.
Place the two small circles so that:
they are both tangent to AB,
one of them touches the 4cm circle (call it C1),
one touches a second edge of the triangle (call it C2),
and they touch each other.
Draw a line segment from the centre of the 4 cm circle to the centre of C1. This line has length (4+r) and makes an angle x with the vertical.
Vertically, height is 33/2 so
4+4+(4+r)cosx + r = 33/2
Horizontally, half the width is 11/2 sqrt3 so
rsqrt3+2r+(4+r)sinx=11/2 sqrt3
rearrange each to get sinx, cosx, then square and add to eliminate (sin^2 + cos^2 =1).
This gives quadratic:
(7+4sqrt3)r^2 + (-58-22sqrt3)r + 147 =0
Which has solution r = 2.2887
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Posted by Pablo
on 2020-08-01 11:17:08 |
An end
