Download Andrew Ryba Math Intel Research Final Paper 6/7/09 (revision 6/17

Andrew Ryba
Math Intel Research Final Paper
(revision
6/7/09
6/17/09)
Euler's formula tells us that for every triangle, the square of the distance between its
circumcenter and incenter is R2 - 2rR, where R is the circumradius and r is the inradius.
Not all quadrilaterals have either a circumcenter or an incenter, let alone both, but if one
does, is there a relationship between the inradius, the circumradius, and the distance
between the circumcenter and incenter?
Though this question is asked about a special type of quadrilateral, in the end it is really a
question about a system of two circles. Such a system is determined by their radii and the
distance between their centers.
The first thing we must think about is how to go about finding such a formula relating the
radii to the distance between the centers of the incircle and circumcircle. We do not know
for certain if such a formula even exists.
We know that a quadrilateral is cyclic if and only if its opposite angles are
supplementary. The condition for a quadrilateral to have an inscribed circle is not so well
known. There is an inscribed circle if the four angle bisectors intersect at a point. (This is
because the center of any inscribed circle must be equidistant from the four sides.
However a point is equidistant from two adjacent sides if it is on the bisector of the angle
formed by the two sides. This shows that an incenter must be a point of intersection of the
four angle bisectors.)
If we examine an arbitrary cyclic quadrilateral whose angle bisectors intersect (at the
incenter of its inscribed circle), we would require a formula relating R (the circumradius),
r (the inradius), and d (the distance between the circumcenter and the incenter). These
quantities would depend on two adjacent angles of the cyclic quadrilateral. We could
plan to express r and d in terms of R and the two adjacent angles and then eliminate the
angle parameters. However, we can simplify the procedure by examining cyclic
quadrilaterals where one of the angles is fixed at 90 degrees. The other angle will still be
available as a parameter to eliminate in order to find a relationship between R, r, and d.
Now we are looking at a kite, with two 90 degree angles, and another pair of angles
which add to 180 degrees. The kite ABCD is shown in the diagram, and has incenter I,
circumcenter O, both of which clearly lie along diagonal BD. O is obviously the midpoint
of BD, and thus BD has length 2R.
C
x
y
a
B
I
D
O
A
The line BD bisects both the angles at B and D. Therefore the incenter I is at the point of
intersection of BD and the bisector of the angle C.
By the angle bisector theorem,
BI / ID = y/x = tan a.
Also BI + ID = 2R
Thus ID = 2R – AI
And BI = (2R tan a) / ( 1 + tan a) = (2R sin a) / ( sin a + cos a)
And BO = R
Let IZ be a radius of the incircle, r, where IZ intersects BC at Z (see the diagram below).
Since IZ is a radius, angle BZI = 90 degrees, and IZ is parallel to CD. So angle ZIB = a .
C
90
Z
90
a
a
B
I
O
D
Thus cos a = r ( 1 + tan a) / (2R tan a)
Therefore r = ( 2Rsin a) / ( 1 + tan a) = (2R sin a cos a) / ( cos a + sin a)
OI = d = abs [ R ( (2 sin a )/ (sin a + cos a) – 1)]
= abs [ R ((sin a – cos a)/ (sin a + cos a))]
Hence d2 = R2 ( ( 1 – 2sin a cos a)/ (1 + 2 sin a cos a))
so d2 / R2 = ( ( 1 – 2sin a cos a)/ ( 1 + 2sin a cos a))
and (R2 – d2) / (R2 + d2) = 2 sin a cos a
So r2 = 4 R2 ( sin 2 a cos 2 a)/( 1 + 2 sin a cos a))
and r2 = (R2 ( (R2 – d2)/ (R2 + d2) ) 2) / ( 1 + (R2 – d2) / ((R2 + d2))
= (R2 (R2 – d2 ) 2) / ( (R2 + d2) 2 + (R2 + d2)( R2 – d2) )
Thus r2 = (R2 (R2 – d2) 2) / (R4 – 2 R2 d2 + d4 + R4 – d4)
= (R2 – d2) 2 / 2 (R2 + d2)
Therefore 2r2 ( R2 + d2) = (R2 – d2) 2
And now we have a relationship between R, r and d for a kite.
What we need to do now is test whether this relationship appears to hold for all cyclic
quadrilaterals that have an incircle. So my plan is to diagram a generic cyclic
quadrilateral that has an incircle with the help of Geometer’s Sketchpad, and test out the
relationship for it.
In such a quadrilateral ABCD, opposite angles must add to 180 degrees (making it
cyclic), and the angle bisectors must meet at a point I (which is the incenter of its
inscribed circle).
To draw this quadrilateral, we can start with a point I as the incenter. Then we draw three
rays through I that will contain the points A, B, and D. The rays IB and ID form angles
with the ray IA that we call θ and φ. We can calculate that the angle of the quadrilateral
at A must be 270 – θ – φ. (This is because the sum of the angles IBA and IDA is half the
sum of a pair of opposite angles in a cyclic quadrilateral which is 90. But the sum of the
angles of quadrilateral IBAD is 360 and is 90 + θ + φ plus the angle BAD.) The point A
can be anywhere on the ray IA but the locations of B and D are then determined by the
condition that the lines BA and DA make angles of 135 – θ / 2 – φ / 2 with the line IA.
r = 2.27 cm
R = 3.65 cm
d = 1.05 cm
B
s
R2-d2 2-2⋅r2
(
)
⋅(R2+d2) = 0.00 cm4
A
t
q
K
I
O
C
D
p
r
In the diagram, three lines IA, IB, and ID are shown as ordinary blue lines. A dotted blue
ray (marked p) that makes an angle of 90 with ID is also shown. The angle between this
ray and the ray IB is 270 – θ – φ. A second dotted blue ray (marked q) bisects this angle.
Copies of the bisected angle are placed at A to form the two bold lines AB and AD. The
condition that IB and ID bisect the angles at B and D determines the lines BC and BD
which are also shown in bold. Dotted red lines (r and s) mark the perpendicular bisectors
of two sides of the quadrilateral ABCD. They meet at the circumcenter O of ABCD. A
third dotted red line (t) is a perpendicular from I to AD. This perpendicular distance
gives the value of r. The distance OC gives the value of R and the distance OI gives the
value of d.
A computer calculation of (R2 – d2) 2 - 2r2 ( R2 + d2) shows a result of 0. The result
remains 0 as the point A and the rays IB and ID are moved in the Geometer’s Sketchpad
diagram. This suggests very strongly that the relationship must hold for any cyclic
quadrilateral that has an incircle.
Online research now led me to a result known as Poncelet’s Porism that can use my
calculation for cyclic kites to prove the general formula. I also found that various forms
of the formula were discovered by Steiner in 1827. (See
http://mathworld.wolfram.com/PonceletsPorism.html ).
The web reference indicates that the problem has been repeatedly considered and other
early solutions were found by Jacobi in 1823 and Fuss in 1792.
Poncelet’s Porism states that if it is possible to find, for a given n > 2, an n-sided polygon
which is simultaneously inscribed in a circle C and circumscribed around a circle D, then
it is possible to find exactly one such a polygon with any desired point belonging to C as
a vertex.
C
Q
D
d
P
O
I
K
Now let us assume that a quadrilateral Q is inscribed in circle C and has an inscribed
circle D. (In the diagram Q is the quadrilateral with bold blue edges.) Let O be the center
of circle C and I be the center of circle D. Let R and r be the radii of the circumcircle C
and the incircle D. Let d be the distance between the centers of C and D.
Extend the line joining O and I to meet the circle C at point P. Poncelet’s Porism tells us
that exactly one new quadrilateral K with P as a vertex, C as its circumcircle and D as its
incircle can be formed. (The quadrilateral K is shown with red sides in the diagram.)
Since there is only such quadrilateral it must be the same as its mirror image across the
line OI. But if a quadrilateral is symmetric across a line through a vertex it is a kite. This
proves that K will have to be a kite. The circumcircle and incircle of K are the same
circles C and D. According to our formula for kites, which we proved earlier,
2r2 ( R2 + d2) = (R2 – d2) 2.
Therefore this formula holds for the quadrilateral Q as well since Q and K have the same
system of two circles, and thus R, r, and d are the same for Q and for the kite K.
This argument shows that the formula is true for all cyclic quadrilaterals with incircles.