Download Cool Math Essay_April 2014_On Cyclic Quadrilaterals

CURRICULUM INSPIRATIONS: www.maa.org/ci
Math for America DC: www.mathforamerica.org/DC
Innovative Online Courses: www.gdaymath.com
Tanton Tidbits:
www.jamestanton.com
Wockin’ COOL MATH!
CURIOUS MATHEMATICS FOR FUN AND JOY
APRIL 2014
PROMOTIONAL CORNER: Have
you an event, a workshop, a website,
some materials you would like to share
with the world? Let me know! If the work
is about deep and joyous and real
mathematical doing I would be delighted
to mention it here.
***
Looking for really cool ways to teach
problem-solving and connect it to the
curriculum? And/or do you want to learn
how to solve problems and also think
about really innovative and curious
mathematics? Check out MAA
Curriculum Inspirations: www.maa.org/ci.
PUZZLER: Quadrilateral ABCD has
AB = 2 , BC = 5 and CD = 7 . What
must AD be so that the quadrilateral
contains a circle just touching all four
sides? What is the largest possible radius
of such an incircle?
© James Tanton 2014
FLOPPY QUADRILATERALS
WITH INCIRCLES
And by expanding that circle, we can
grow it to a size that makes it touch a
third side.
The two tangents theorem from circle
geometry (see this month’s
CURRICULUM ESSAY) shows that if a
quadrilateral circumscribes a circle, then
opposite sides of that quadrilateral have
lengths that sum to the same value.
Must this circle touch the fourth side as
well? Is there anything mathematically
amiss if it does not?
Since AB + CD = 2 + 7 = 9 this suggests
we need AD = 4 . But this is assuming
first that the quadrilateral has an
inscribed circle! Is setting AD = 4
enough to ensure that an inscribed
circle exists?
We can certainly draw within this figure
a circle that is tangent to two of its
sides.
Question: It seems we are assuming
that the quadrilateral is convex. Can a
concave quadrilateral circumscribe a
circle?
If the circle fails to touch the fourth
side, draw a line as indicated to make a
triangle with one side tangent to the
circle. Call the lengths of the two sides
of that triangle x and y .
We have within this picture a
quadrilateral with all four sides tangent
to a circle, so opposite pairs of its sidelengths must sum to the same value:
5+ y = 2+7− x.
It follows that x + y = 4 . But now look
at the triangle. As two sides of a triangle
must always sum to a value greater than
the length of the third side, something is
www.jamestanton.com and www.gdaymath.com
© James Tanton 2014
indeed mathematically amiss: we should
have x + y > 4 . It must be the case that
the circle touches all four sides of the
original quadrilateral after all.
This work can be generalized to
establish:
Theorem: A convex quadrilateral
ABCD possesses an inscribed circle if,
and only if, opposite pairs of sides have
the same sum of lengths:
AB + CD = BC + AD .
I find this result really quite astounding!
Imagine four rods of lengths 2 , 5 , 7 ,
and 4 hinged together to make a floppy
model of a quadrilateral. Each and every
convex figure one makes with this
model is sure to possess an inscribed
circle!
THE INCIRCLE RADIUS
Part two of the puzzle wants us to
compute the radius of the largest
possible incircle that could appear.
For any quadrilateral (or any polygon for
that matter) possessing an inscribed
circle, the radius r of the inscribed circle
can be computed from the area A and
the perimeter P of the quadrilateral as
follows:
If the sides of the quadrilateral have
lengths a , b , c , and d , then dividing
the quadrilateral into four triangles with
heights given by the radius r of the
incircle gives:
1
1
1
1
A = ar + br + cr + dr
2
2
2
2
1
= (a + b + c + d ) r
2
1
= P⋅r
2
Wow!
In any case, setting AD = 4 certainly
does the trick in answering the first part
of the opening puzzle.
2A
. For our quadrilateral we
P
have P = 2 + 5 + 7 + 4 = 18 and so
1
r = A . We see that the incircle of
9
largest radius sits within the convex
quadrilateral of largest possible area.
Thus r =
So how do we get a quadrilateral of
largest possible area?
www.jamestanton.com and www.gdaymath.com
© James Tanton 2014
THE AREA OF A
QUADRILATERAL
I know three formulas for the area A of
a quadrilateral with one side of length
zero. (The quadrilateral is usually then
called a triangle!) The first two are
basically the same formula:
1
A = × base × height
2
1
A = ab sin θ
2
Here a and b are two sides of the
triangle and θ is the angle between
them. The second formula is really the
first formula with the sophistication of
trigonometry thrown in: base = a and
height = b sin θ .
If the third side of the triangle is length
c then the law of cosines says:
Heron’s formula for the area of a
triangle.
So what if the fourth side of the
quadrilateral is not of zero length? Do
any of these formulas generalize?
DREAM: It seems that Heron’s formula
is easiest to generalize. Wouldn’t it be
lovely if the area A of a quadrilateral
with sides of lengths a , b , c , and d is
just:
A=
( s − a )( s − b )( s − c )( s − d )
a+b+c+d
? This is Heron’s
2
formula when d = 0 .
where s =
Let’s follow the trigonometric approach
used to derive Heron’s formula and see
where get. (What follows look very
scary, but each step forward is
straightforward in concept.)
Consider a quadrilateral with four sides
of lengths a , b , c , and d and two
interior angles as shown. We’ve also
drawn a diagonal of length p .
c 2 = a 2 + b 2 − 2 ab cos θ .
So we have:
2A
ab
c2 − a 2 − b2
cos θ =
2ab
Substituting into sin 2 θ + cos 2 θ = 1
gives a formula involving only the
symbols A , a , b and c .
sin θ =
A page of algebraic manipulation, if you
are up for it, then gives:
A = s ( s − a )( s − b )( s − c )
a+b+c
is the semi2
perimeter of the triangle. This is
where s =
The area A of the quadrilateral is:
1
1
A = ab sin θ1 + cd sin θ 2 .
2
2
The law of cosines gives:
a 2 + b 2 − 2ab cos θ1 = c 2 + d 2 − 2cd cos θ 2
(both sides are p 2 ).
As before, can we use the famous
relation sin 2 θ + cos 2 θ = 1 to get a
www.jamestanton.com and www.gdaymath.com
© James Tanton 2014
formula involving only A and the four
side lengths? Let’s try!
Think of this as:
16 A2 = ( ( c + d ) − ( a − b ) ) ( ( c + d ) + ( a − b ) )
× (( a + b ) − ( c − d )) (( a + b) + ( c − d ))
Squaring the first equation gives:
to get:
4 A2 = a 2 b 2 sin 2 θ1 + c 2 d 2 sin 2 θ 2
(
2
16A2 = ( c + d ) − ( a − b )
+ 2abcd sin θ1 sin θ 2
2
) (( a + b)
2
− (c − d )
× ( 2ab + 2cd − c 2 − d 2 + a 2 + b 2 )
2
= ( 2ab + 2cd ) − ( c 2 + d 2 − a 2 − b 2 )
2
which is:
16 A2 = 4a 2 b 2 + 4c 2 d 2 − ( c 2 + d 2 − a 2 − b 2 )
− 2abcd cos θ1 cos θ 2
Adding these two yields:
Our dream formula matches our actual
formula only if cos (θ1 + θ 2 ) = −1 , that
2
1 2
a + b2 − c 2 − d 2 )
(
4
2 2
= a b + c2 d 2
is, if θ1 + θ 2 = 180o . Hmm.
But this work has revealed to us that:
− 2abcd ( cos θ1 cos θ 2 − sin θ1 sin θ 2 )
16 ( s − a )( s − b )( s − c )( s − d )
= a 2 b 2 + c 2 d 2 − 2abcd cos (θ1 + θ 2 )
= 4 a 2 b 2 + 4c 2 d 2 − ( a 2 + b 2 − c 2 − d 2 )
This can be rewritten:
How does this compare with the “dream
formula”
2
+ 8abcd
Let’s use this observation to rewrite our
actual, non-dream formula:
16 A2 = 4a 2 b 2 + 4c 2 d 2 − ( a 2 + b 2 − c 2 − d 2 )
2
− 8abcd cos (θ1 + θ 2 )
A=
( s − a )( s − b )( s − c )( s − d ) ?
It is hard to tell. Let’s square both sides
and rewrite this formula without the
semi-perimeter s :
A2 =
2
+ 8abcd
4 A2 +
− 8abcd cos (θ1 + θ 2 )
( −a + b + c + d )( a − b + c + d )( a + b − c + d )( a + b + c − d )
16
We can expand this by noting lots of
differences of two squares!
)
This is:
16 A2 = ( 2ab + 2cd + c 2 + d 2 − a 2 − b 2 )
Rewriting the second equation as:
a2 + b2 − c2 − d 2
= ab cos θ1 − cd cos θ 2
2
and squaring gives:
1 2
a + b2 − c2 − d 2 )
(
4
= a 2 b 2 cos 2 θ1 + c 2 d 2 cos 2 θ 2
16 A2 = 4a 2 b 2 + 4c 2 d 2 − ( a 2 + b 2 − c 2 − d 2 )
2
= 16 ( s − a )( s − b )( s − c )( s − d ) − 8abcd
− 8abcd cos (θ1 + θ 2 )
= 16 ( s − a )( s − b )( s − c )( s − d )
− 8abcd ( cos (θ1 + θ 2 ) + 1)
Noting that cos ( 2 x ) = 2 cos 2 ( x ) − 1 this
can be rewritten:
www.jamestanton.com and www.gdaymath.com
2
© James Tanton 2014
16 A2 = 16 ( s − a )( s − b )( s − c )( s − d )
− 16abcd cos 2 (θ )
where θ =
A=
θ1 + θ 2
2
A=
. We see:
( s − a )( s − b )( s − c )( s − d ) − abcd cos2 (θ )
We have established:
Bretschneider’s Formula: The area A of
a quadrilateral with side lengths a , b ,
c , and d is given by:
A=
Brahmagupta’s Formula: If a cyclic
quadrilateral has side lengths a , b , c ,
and d , then its area A of is given by:
( s − a )( s − b )( s − c )( s − d ) .
So we see now that our dream formula
applies only to cyclic quadrilaterals.
Note that every triangle is cyclic. (Again
see this month’s CURRICULUM ESSAY.
That essay has it all!)
FINISHING THE PUZZLER
2
( s − a )( s − b )( s − c )( s − d ) − abcd cos (θ )
a+b+c+d
is its semi2
perimeter and θ is the average of the
measures of a pair of opposite angles in
the quadrilateral.
where s =
To remind us… We have a quadrilateral
with sides of lengths 2 , 5 , 7 and 4
possessing an incircle. The radius of that
1
incircle is r = A where the area A is
9
given by Bretscneider’s formula:
( 9 − 2 )( 9 − 5)( 9 − 7 )( 9 − 4 ) − 2 ⋅ 5 ⋅ 7 ⋅ 9 cos2 (θ )
Comment: Since
cos 2 (180o − x ) = cos 2 ( x ) it does not
A=
matter which pair of opposite interior
angles one chooses to work with.
with θ the average of a pair of opposite
angles. The value of θ can change,
depending on how we “flop” those four
side lengths about to create
quadrilaterals.
CYCLIC QUADRILATERALS
= 280 − 630 cos 2 (θ )
A quadrilateral is called cyclic if all four
of its vertices lie on a common circle.
We learned in this month’s
CURRICULUM ESSAY their complete
characterization:
Theorem: A quadrilateral is cyclic if, and
only if, pairs of opposite angles in the
quadrilateral sum to 180o .
Since the average measure of opposite
angles in a cyclic quadrilateral is 90o and
cos ( 90o ) = 0 we have:
The puzzle asks for the largest possible
value of r . To find it we need the
largest possible value of A .
This maximal area occurs if we can
arrange matters so that cos (θ ) = 0 ,
that is, if we can arrange that opposite
angles x and y in the diagram satisfy
www.jamestanton.com and www.gdaymath.com
© James Tanton 2014
x + y = 180o . That is, if we can make the
quadrilateral cyclic! Then A = 280
1
and r =
280 is the largest possible
9
inradius.
Definition: A quadrilateral is bicyclic if it
both circumscribes a circle (that is, has
an incircle) and is circumscribed by a
circle (that is, is cyclic).
Our solution to the puzzle is bicyclic!
So only one issue remains to complete
this puzzle:
Is it possible to maneuver the rods of a
floppy quadrilateral with sides 2 , 5 , 7
and 4 , so that the opposite angles x
and y sum to 180o to actually obtain
this alleged quadrilateral with largest
possible incircle?
The answer is yes! Start by pivoting the
rod of length 2 to turn the quadrilateral
into triangle as shown, with angle x
becoming 180o . In this extreme,
x + y = 180o + y > 180o .
Question: A bicyclic quadrilateral has
sides of lengths a , b , c , and d . Show
that its area is A = abcd .
RESEARCH CORNER:
Consider a (very degenerate) polygon
with two sides, each necessarily the
same length. We can think of this as a
cyclic quadrilateral with a = b , c = 0 ,
d = 0 . Brahmangupta’s formula gives
the correct value for the area of this
bigon. (Can you see why?)
Brahmagupta’s formula is also correct
for all triangles, which are cyclic
polygons with d = 0 .
Next pivot the rod of length 2 in the
opposite direction to make a second
triangl. As all three angles in a triangle
sum to 180o , we have that x + y < 180o .
The obvious generalization of
Brahmagupta’s formula to cyclic
pentagons, namely,
A=
( s − a )( s − b )( s − c )( s − d )( s − e )
alas, cannot hold: the left side is a
quantity in units squared but the right
side is units to the 2.5 th power,
whatever that means.
Can anything of interest be said about
the areas of cyclic pentagons?
There must be some intermediate pivot
position for that rod of length 2 that
has x + y = 180o , finally proving that the
maximal inradius we identified actually
does happen!
© 2014 James Tanton
[email protected]
www.jamestanton.com and www.gdaymath.com