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CMJ Problem 650 Solution
by Peter Y. Woo, 3/27/99
Problem. Let D be a point on the side BC of an
equilateral triangle ABC of length s. Let the inradii of triangles ABD, ACD
be r1, r2. Find s in terms of r1 and r2.
Solution. Let t = AD, x1 = BD, x2 = CD. Assume
x1 ³ x2 without loss of generality.
From law of cosines on DABD and DACD, we get x1 and x2 as
roots of
t2 = s2 + x2 - s x, or x2 - s x + (s2 - t2) = 0. . . . (1)
Quadratics theory then gives x1 + x2 = s and x1 x2 = s2 - t2.
and (x1 - x2) = Ö[(x1 + x2)2 - 4 x1 x2]
= Ö[4 t2 - 3 s2]. . . . (2)
Now r1 = 2 × area(DABD) / perimeter(DABD), giving
r1 = q s x1 / (s + t + x1) . . . (3) where q = sin 60° = Ö3 / 2.
or 1/r1 = (1/q) [1/s + (s + t) / s x1]
Similarly for r2. Now Let pi = q/ri
for i = 1,2, then
pi = (s + t) / s xi + 1/s. . . . (4)
Hence p1+ p2 = 2/s + (s + t)(x1+x2) / s x1 x2
= 2/s + (s + t)/(s2 - t2), i.e.,
p1+ p2 = 2/s + 1/(s - t) . . . .(5a), also
= (3 s - 2 t) / s (s - t). . . . (5b)
and p2 - p1 = (1 + t/s) (1/x2 - 1/x1)
= (1 + t/s) (x1 - x2) / x1 x2
= Ö(4 t2 - 3 s2) / s (s - t). . . . (6).
We want to eliminate t from (5b) and (6). Try with
p1 p2 = [(p1+ p2)2 - (p1-p2)2]/4
= [9 s2 + 4 t2 - 12 s t - 4 t2 + 3 s2] / [4 s2 (s - t)2],
= 3 / s (s - t).
From (5a) we get 1/(s - t) = p1+ p2 - 2/s,
\ p1 p2 = (3 / s) (p1+ p2 - 2/s), free of t.
\ p1 p2 s2 - 3(p1+ p2) s + 6 = 0. . . . (7)
Quadratic formula then gives
s = (1/2p1p2) [3(p1+ p2) + Ö[9(p1+ p2)2 - 24p1p2]]
= (3/2)(1/p1+ 1/p2) + Ö[(9 p12 + 9 p22 - 61 p2)
/ (4 p12 p22)] giving
s = Ö3 (r1 + r2) + Ö[3 r12 + 3 r22 - 2 r1 r2].
QED
