[Math Lair] Pillow-Problems: Problem #22

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

Problem:

22.

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

[4/6/89

Answer:

22.

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

2α²β²γ² · (&alpha² + &beta² + &gamma²) − (β4γ4 + γ4α4 + α4β4)
4β4γ4
k²’,

  (1) α =
1
kα
, &c.;

  (2) sin A = kβγ, &c.;

  (3) area =
1
k²
.

Solution:

22.
[diagram for problem 22]

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,

4γ4 + &c.) − 2α²β²γ² · (α² &c.) + 4k²α4β4γ4 = 0;

[diagram for problem 22]

k² =
2α²β²γ² · (&alpha² + &beta² + &gamma²) − (β4γ4 + γ4α4 + α4β4)
4β4γ4
.

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.