Pillow-Problems: Problem #22

Problem #21 | Problem #23
For more information on this collection, see Pillow-Problems by Charles L. Dodgson (Lewis Carroll).

Problem:

Given the 3 'altitudes' of a Triangle: find its (1) sides, (2) angles, (3) area.

Answer:

Calling the given altitudes ‘α,β,γ’ and the fraction

Solution:

Call given altitudes ‘α,β,γ’.

Now aα = bβ = cγ

∴ α sin A = β sin B = γ sin C;

∴

sin A

βγ

=

sin B

γα

=

sin C

αβ

= k (say);

∴ sin A = kβγ, sin B = kγα, sin C = kαβ.

Now sin (A + B) = sin C;

∴ sin A cos B + cos A sin B = sin C;

∴ sin A cos B = sin C − cos A sin B;

∴ sin² A(1 − sin²B) = sin² C + sin² B(1 − sin² A) − 2 sin C cos A sin B;

∴ sin² A − sin² A sin²B = sin² C + sin² B − sin² A sin² B − 2 sin B sin C cos A;

∴ sin² A − sin² B − sin² C = −2 sin B sin C cos A;

∴, squaring, (sin ⁴ A + &c.) − 2 sin² A sin² B − 2 sin² A sin² C + 2 sin² B sin² C = 4 sin² B sin² C(1 − sin² A);

(sin ⁴ A + &c.) − 2(sin² B sin² C &c.) + 4 sin² A sin² B sin² C = 0;

∴, substituting for sin A, &c., and dividing by k⁴,

(β⁴γ⁴ + &c.) − 2α²β²γ² · (α² &c.) + 4k²α⁴β⁴γ⁴ = 0;

∴ k² =

2α²β²γ² · (α² + β² + γ²) − (β4γ4 + γ4α4 + α4β4)

4α4β4γ4

.

Now sin A = kβγ, &c.; which answers (2).

Also α = b sin C; and similarly γ = α sin B;

∴ α =

γ

sin B

=

γ

kγα

=

1

kα

, &c.; which answers (1).

Also area =

bc sin A

2

= ½ ·

1

kβ

·

1

kγ

· kβγ =

1

2k

; which answers (3).

Q.E.