STEYNING & DISTRICT U3A
Discovering Mathematics
Session 28
Heron’s Formula - Proof
&
Mathematics & Music
An Algebraic Proof of Heron’s Formula.
You will recall that we recently used Heron’s formula for the area of a triangle;
A = √s(s-a)(s-b)(s-c), where s is the semi-perimeter of a triangle with sides a,b & c.
a
h
x
b
y
c
Area, A = ch/2
&
c=x+y
h 2 = a 2 – x2
& h2 = b 2 – y2
s = (a+b+c)/2
or 2s = (a+b+c)
Therefore; 2(s-a) = (-a+b+ c),
2(s-b) = (a-b+c) & 2(s-c) = (a+b-c)
As y = c – x, then y2 = (c-x)2 & expanding; y2 = c2 – 2cx + x2
Adding h2 to each side gives. h2 +y2 = h2 + c2 – 2cx + x2
But h2 + y2 = b2 & h2 + x2 = a2, therefore; b2 = a2 + c2 – 2cx
Solving for x, gives; x = (a2 + c2 - b2)/2c
But h2 = a2 – x2 = (a + x)(a - x) = [a + (a2 + c2 - b2)/2c][a + (a2 + c2 - b2)/2c]
Heron’s Formula (Cont.)
So; h2 = [2ac + a2 + c2 - b2][2ac -a2 - c2 + b2]/4c2
If we factorize this, we get; h2 = [(a+c)2 – b2][b2 - (a - c)2]/4c2
This can be further factorized to give, [(a + b +c)(-a + b +c)][(-b + a + c)(a + b - c)]
But, 2s = (a + b + c), so h2 = 4s(s –a)(s – b)(s – c)/c2
& h = 2√s(s –a)(s – b)(s – c)/c
Area A = ch/2 = c/2[2√s s(s –a)(s – b)(s – c)/c]
ie. A = √s(s –a)(s – b)(s – c)
The Longest Ladder
A ladder of length l has to be carried along a corridor of width w.
The corridor has a right angled corner. What is the longest possible
ladder(l) which can be carried horizontally?
w
y
Applying Pythagoras; l2 = (x +w)2 + (y + w)2 --Eqn. (a)
w
x
The 2 yellow triangles are similar. So; y/w = w/x or y = w2/x
Substitute y in Eqn. (a).
l2 = (x + w)2 + (w2/x + w)2 , Expanding; l2 =( x2 + 2xw +w2) + (w4/x2 + 2w3/x + w2)
Simplify; l2 = x2 + 2xw + 2w2 + w4x-2 + 2w3x-1
We can differentiate ; d(l2)/dx = 0 for max or min values of l2
d(l2)/dx = 2x + 2w – 2w4/x3 -2w3/x2 = 0
Rationalize by multiplying by x3/2.
x4 + wx3 - w4 – w3x = 0
By examination, an obvious solution is x = w
Then l2 = (w + w)2 + (w2/w + w)2 = (2w)2 + (2w)2 = 8w2
Therefore; l = 2√2 *w and for a 2m wide corridor, the longest ladder is 5.65 m.
Any suggestions for manoeuvering a slightly longer ladder.
The Longest Ladder (Cont’d)
If one end of the ladder is lifted to the roof of the corridor, with the other end at
floor level, a slightly longer ladder can be manoeuvered around the corner.
Assume corridor height is 3m.
The solution can be achieved without further use of Calculus.
The maximum length of the horizontal is l, as before.
L
3
Using Pythagoras and with l = 4√2;
l
L2 = 32 + (4√2)2 = 9 + 32
L = √41 = 6.40 m
Music & Mathematics
by
Professor Marcus du Sautoy
The Mathematics of Music
by Jack H David, Jr.
http://jackhdavid.thehouseofdavid.com/papers/math.html
Suggested Future Topics
Algorithms
Topology
That’s it Folks!