2 circles

To construct two equal circles, mutually tangent, one tangent to two sides of a rectangle, the other one tangent to the two other sides.

Sorry, your web browser is not Java compliant, or Java not installed or disabled Circles tangent to opposite sides have as diameter the width and length of the rectangle. Hence can be equal only if the rectangle is a square. There are then infinitely many solutions.
Point I is dragable/animated in the applet to chose a pair of circles.
The "rectangle" ABCD aspect ratio is defined by the dragable opposite corners A and C (initially a square)

We then consider only circles, each tangent to adjacent sides !
There are several solutions, depending on the circles being "inside" or "outside" the rectangle.
At symmetries, there are 4 families of circles, depending on the respective angles which "contain" them.
Without loss of generality, we may consider the opposite angles A and C.
Both angles outside is impossible : the circles couldn't touch !

Angle C is then an inside angle, and the 4 choices for angle A give the 4 families.
By symmetries, this results into xx solution pairs of circles.
The problem is to construct these circles, at least one "representant" for each family.

An applet showing a representant in each family.
The rectangle is defined by its opposite dragable corners A and C.

Sorry, your web browser is not Java compliant, or Java not installed or disabled

The other circles in a family are not shown, but result from symmetries and are then circles equal to this representant.
We may note that the first family has two different representants : red and magenta.
The red circles are the only case when both circles are inside the rectangle, if the aspect ratio of the rectangle is in  1/2 ≤ BC/AC ≤ 2 

Constructions

Your turn to play...

Solutions

 

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