Repost: Maximum area of triangles with sides 1, x, and angle θ?

Question

This repost is for other contributing answers.

Two different triangles with sides 1, x, and angle θ have the same area. What is the maximum area possible?

(Problem restated: Triangle 1 has sides 1, x, y, and triangle 2 has sides 1, x, z. Both share same angle θ, and both have the same area. For what values x, y, z, θ is this area maximum?)

Answer 1

I've rewritten this and compacted the initial logic:
If θ is adjacent to both 1 and x in one triangle, then it is adjacent to either x or 1 in the second triangle (since there are only 3 sides) so that it must also be adjacent to the other to get the same area (by the law of sines). But this would yield congruent triangles, contradicting the assumption.
Similar logic shows θ is not opposite the same length side in both triangles.



Thus, θ is opposite 1 in one triangle (say the 1, x, y triangle), and opposite x in the other (1, x, z).

Equal areas imply:
xy sin(θ) / 2 = z sin(θ) / 2
z = xy

Law of cosines for triangles 1 and 2, respectively, says:
2xy * cos(θ) = y² + x² - 1
2z * cos(θ) = z² + 1 - x²

Since z=xy the left sides are equal, so that:
z² + 1 - x² = y² + x² - 1
z² - y² = 2(x² - 1)
(x² - 1)y² = 2(x² - 1)
y = √2



We would like to maximize
A = x√2 sin(θ) / 2, subject to, from the law of cosines for triangle 1 (with y = √2):
(1) 2 + x² - 1 - 2x√2 * cos(θ) = 0

So introduce λ and maximize:
x sin(θ) - λ (x² + 1 - 2x√2 cos(θ))
By taking partials with respect to λ we get (1) above.
By taking partials with respect to x and θ, respectively, we get:
(2) sin(θ) - λ (2x - 2√2 cos(θ)) and
(3) x cos(θ) - λ (2x√2 sin(θ)) = 0

The latter yields: λ = cos(θ) / (2√2 sin(θ))
which we plug into (2) to get:
sin (θ) = (cos(θ) / (2√2 sin(θ))) (2x - 2√2 cos(θ))
Clear the denominator:
2√2 sin²(θ) = 2x cos(θ) - 2√2 cos²(θ)
2√2 = 2x cos(θ) so
x * cos(θ) = √2

Plug this back into (1) to find x:
x² - 2√2 * √2 + 1 = 0
x = √3

Thus our two Δs of equal area are:
(1, √2, √3) and (1, √3, √6) where the common angle θ is given by:
cos(θ) = √(2/3) => sin(θ) = 1/√3
for an area of 1/√2

Answer 2

Is θ the angle between 1 and x? If so, then y=z, and there are not two different triangles. Then the problem is restated as what is the maximum area for a triangle with sides 1, x, and angle θ? I suspect the answer is equilateral triangle: θ=60, x=y=1.

Answer 3

What is the position of angle θ.
Is it the included angle. Without knowing the position of the angle, the problem can't be solved.