Teaser 2438: [Nested quadrilaterals]
From The Sunday Times, 14th June 2009 [link]
Quadrilateral A has its vertices on a circle, and no two angles are the same. Another circle is drawn inside A that touches its sides at four points, and these points form the corners of Quadrilateral B. Then Quadrilateral C is formed inside B in the same way, and so on.
Quadrilateral K’s angles are whole numbers of degrees.
What (in increasing order) are Quadrilateral A’s angles?
This puzzle was originally published with no title.
[teaser2438]
Frits, 
GeoffR and 
Hugh Casement are discussing. 
S2T2 
Jim Randell 10:42 am on 8 April 2026 Permalink |
A quadrilateral that has a circumcircle (i.e. a circle that passes through each of the vertices) is called a cyclic quadrilateral.
A quadrilateral is cyclic if (and only if) opposite internal angles are supplementary (i.e. if the angles at the vertices are (𝛂, 𝛃, 𝛄, 𝛅) we have: 𝛂 + 𝛄 = 𝛃 + 𝛅 = 180°).
A quadrilateral that has a incircle (i.e. a circle inscribed inside the quadrilateral that touches all four sides) is called a tangential quadrilateral.
A quadrilateral is tangential if (and only if) opposite sides sum to the same total length (i.e. if the length of the sides are (a, b, c, d) we have: a + c = b + d).
Note that the condition for a tangential quadrilateral is more restrictive than the condition for a cyclic quadrilateral. For example, all rectangles (all internal angles = 90°) are cyclic, but only squares are tangential.
A quadrilateral that is both cyclic and tangential is called bicentric.
This puzzle concerns a series of nested bicentric quadrilaterals.
Consider a bicentric quadrilateral ABCD. The tangential points on the incircle form the contact quadrilateral WXYZ. (Such that W is on AB, X on BC, Y on CD, Z on DA).
Considering the internal angles of the ABCD quadrilateral (as a, b, c, d), the tangent lengths from a vertex to the vertices of the the contact quadrilateral are equal (see: Pitot Theorem), and so the angle subtended at the incentre of the contact quadrilateral (I) is supplementary to the internal angle of the original quadrilateral. (And so it is the same as the opposite internal angle of the original quadrilateral):
(Note that knowing the angles subtended at the centre of the incircle means we can fully determine the shape of the contact quadrilateral, and we can then draw the tangents to the vertices of the contact quadrilateral to find the original quadrilateral (which has opposite angles supplementary, and so is necessarily cyclic). [*]
And so, each internal angle of the contact quadrilateral is the mean of the internal angles at the vertices on the original quadrilateral at either end of the tangent side:
So, if the original angles are (a, b, c, d), then the final angles (a′, b′, c′, d′) after 10 applications of this transformation are:
And remembering a + c = 180°, and b + d = 180°, we get:
And these are integers.
So, if we assume a and b are angles where 0° < a < b < 90°, we get:
Solution: The angles in the original quadrilateral were: 26°, 58°, 122°, 154°.
Calculating successive internal angles we get:
Note that the angles are approaching 90°, and a square forms a stable scenario, which can be extended in either direction indefinitely.
The sequence would continue:
by which time the angles are all within 0.2° of 90°.
However if we attempt to actually construct the series of nested quadrilaterals we run into a problem.
From [*] we know that if the internal angles of quadrilateral A are: (26°, 58°, 154°, 122°), then the angles subtended at the centre of the contact quadrilateral (= quadrilateral B) are: (122°, 26°, 58°, 154°), and this is enough information to draw quadrilateral B inside a circle, and then construct quadrilateral A outside the circle.
If we do so we get this:
(quadrilateral A is red; quadrilateral B is black).
Similarly, knowing the internal angles of quadrilateral B are: (42°, 106°, 138°, 74°), then the angles subtended at the centre of quadrilateral C are: (74°, 42°, 106°, 138°), and drawing quadrilateral’s C and B we get this:
(quadrilateral B is red; quadrilateral C is black).
And, although the two quadrilaterals B in the diagrams look superficially similar (as they have the same angles), they are not mathematically similar, so the second cannot be uniformly scaled to fit onto the first. (The easy way to see this is that the distance between the 106° and 138° angles in the second diagram is longer than if the first diagram was scaled so as to fit the opposite side over the corresponding side in the first diagram. It is as if a strip parallel to that side has been removed (keeping the angles the same)).
Here are the two quadrilaterals B superimposed:
(Note the extra red line where one side of the red (smaller) quad B does not overlap the black quad B, and the inscribed circle does not touch all four sides of the black quad B).
So, practically it is not possible to construct such a sequence of quadrilaterals.
Hence the puzzle itself is flawed.
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Jim Randell 2:44 pm on 8 April 2026 Permalink |
Although I have not found a correction printed in The Sunday Times, this puzzle did prompt an article by Victor Bryant (editor of Teaser puzzles at the time) and John Duncan, showing that if a nest of 3 bicentric quadrilaterals can be made, then they are necessarily all squares.
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