| Note 1442.0
>Suppose we have a triangle ABC, with base BC, and draw a line l thru A
>parallel to BC. Draw lines from A to BC, dividing BC in n segments,
>not necessarily equal.
>Now incribe a circle in each of the n small triangles. Let h[i] be
>the distance from the top of the i'th circle to line l.
>The problem is to show that the product of the h[i] is a constant.
Note 1442.1
>Since each division produces two new triangles, it suffices to prove
>the theorem for a single division. Induction will carry on from there.
INDUCTIVE BIT
The product of the h[i] does depend on n. So given that we know
the result for all integers up to n >= 1, let's prove it for n+1. Suppose
we have two subdivisions of the line BC, determined by the points
X_0,X_1,...,X_n
Y_0,Y_1,...,Y_n
Let's say that |X_i-B|=x_i, |Y_i-B|=y_i for all i=0,1,...,n, and that
0 < x_0 < x_1 < ... < x_n < |BC|
& 0 < y_0 < y_1 < ... < y_n < |BC|
Then if x_n=y_n, we apply the result for n to the triangle ABX_n, and
are home. Otherwise, x_n <> y_n. Wlog x_n < y_n. Since we have proved
the theorem for n=1, we can consider the triangle AX_(n-1)C, and slide X_n
along X_(n-1)C without affecting the product of the h[i]. So we can make
x_n match y_n, and then we are home as before.
THE CASE N=1
So, it remains to prove the result for n=1.
Let take a triangle with sides a,b,c, altitude perpendicular to a of
length h, perimeter p and radius of incircle, r. Then it's the work of
moments to see that:
rp = ah (= twice the area of the triangle)
h-2r = h(-a+b+c)/(a+b+c).
Now we have two triangles with the same altitude h, one has sides
x,b,d, and the other has sides (a-x),c,d. It's our task to show:
K = h(-x+b+d)/(x+b+d) * h(-(a-x)+c+d)/((a-x)+c+d)
is independent of x. (Note: d is a function of x.)
Application of the cosine rule shows us that:
ad� = xc� + (a-x)b� - ax(a-x)
and when we use this substitution we get:
K = h�(-c�+b�+a�+2ad-2ax)/(2ab-c�+b�+a�) *
(-b�+c�-a�+2ad+2ax)/(2ac-b�+c�+a�)
whence:
K = h�[4b�c� - (-a�+b�+c�)�]/[(2ab-c�+b�+a�)(2ac-b�+c�+a�)]
duly scoured of x and d.
Andrew.
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