# Pillow-Problems: Problem #28

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Problem #27 | Problem #29
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

## Problem:

28.

If the sides of a given Triangle, taken cyclically, be divided in extreme and mean ratio; and if the points be joined: find the ratio which the area of the Triangle, so formed, has to the area of a given Triangle.

[12/78

28.

7 − 3√5.

## Solution:

28.

Let ABC be the given Triangle; and let its sides be divided internally at A′, B′, C′, in extreme and mean ratio.

And let M be the area of ABC.

Let BA′ = x; then x² = a · (ax);

i.e. x² + axa² = 0;

x =
 (−a ± a√5)/ (2)
=
 (a)/ (2)
· (√5 − 1), the other sign being excluded by the terms of the question.

Then area of Triangle ABC

= ½ ·
 (c)/ (2)
· (√5 − 1) · {b
 (b)/ (2)
· (√5 − 1)} · &sinz; A,
 (1)/ (8)
· (√5 − 1)(3 − √5)bc · &sinz; A,
= ¼ · (4√5 − 8) · M = (√5 − 2) · M.

Similarly for BCA′ and CAB′.

Hence the sume of these 3 Triangles = 3 · (√5 − 2) · M, and area of Triangle ABC′ = (7 − 3√5) · M.

Q.E.F.