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?)
Update:Addition to above detail: I meant to add, "what's the maximum area possible in such a case of 2 different triangles with equal areas?"
Update 3:Jeffery, very careful reasoning, but you quit a bit too soon.
Update 5:Ah, scratch that last comment I made, let me figure out why you can't get to the correct conclusion.
Update 7:How did I figure out that the triangles have sides (1,√2,√3), and (1,√3,√6), and Sin(θ) = 1/√3? First, I figured out which 2 out of the 4 possible triangles in the general case would be candidates for the maximum areas. Then I equated the two to figure out what y has to be in terms of Sin(θ). Then I used this y to figured out the area, and found the maximum by conventional calculus. I was kind of suprised with the answer.
Answers & Comments
Verified answer
First, note that every argument for either triangle in this problem can be applied symmetrically, since there is no difference between switching z and y.
Now, we have the formula A=ab*sin(θ)/2.
We can let theta be in any of the three places for both triangles.
Setting the areas equal, we have:
ab*sin(θ)/2 = cd*sin(θ)/2
ab = cd, where a and b are side lengths from one triangle, and c and d are from the other. Theta is placed inbetween these lengths.
Say we let a and b be 1 and x. Theta could be placed in three places on the second triangle.
If c and d are also 1 and x in the other triangle, y=z, which creates the same triangle, causing a contradiction.
If c and d are 1 and z, then z=x, causing the same triangle once again.
If c and d are x and z, then z=1, causing again, the same triangle.
Thus, a and b cannot be 1 and x, and by symmetry, cannot be 1 and x in the other triangle.
Now, say a and b are 1 and y.
If c and d are 1 and z, then z=y. This results in the same triangle, another contradiction.
So, we know that we must place the 2 equal angles between 1 and y, and x and z. Or, 1 and z, and x and y, but this is just symmetry so we ignore it.
Now we have, y=xz.
Our first triangle is with the angle between 1 and xz with opposite side x. Our second triangle is with the angle between x and z and opposite side 1. At this point, it may help to draw a picture.
Now use cosine formula with theta twice.
cos(θ) = (a^2 + b^2 - c^2)/2ab.
(1 + x^2 z^2 - x^2) / (2xz) = cos(θ) = (x^2 + z^2 - 1) / (2xz)
1 + x^2 z^2 - x^2 = x^2 + z^2 - 1
x^2 z^2 - 2x^2 - z^2 + 2 = 0
x^2 (z^2 - 2) = z^2 - 2
Thus either x = 1, -1, or z = root(2), -root(2)
Obviously negative values don't work. and x=1 creates the same triangle.
Thus, z=root(2), and cos(θ) = (x^2+1) / (2root(2)x)
Let's get sin(θ)
sin^2(θ) = 1 - cos^2(θ)
= 1 - (x^4+2x^2+1) / (8x^2)
= (8x^2 - x^4 - 2x^2 - 1) / (8x^2)
= (- x^4 + 6x^2 - 1) / 8x^2
Now, we have
A = 1/2 ab sin(θ)
= xz/2 * sin(θ)
= x root(2)/2 * root((- x^4 + 6x^2 - 1) / 8x^2)
= x * root(2) / 2 * root(- x^4 + 6x^2 - 1) / (2 root(2) x)
= root(- x^4 + 6x^2 - 1) / 4
Now we need to maximize this with domain restriction root(2)-1 < x < root(2)+1 and x is not 1.
Clearly, we must simply maximize -x^4+6x^2-1.
So, -(x^2-3)^2 + 8 = 0
x^2 = 3
x = root(3)
A = root(8) / 4 = root(2) / 2
x = root(3)
y and z = root(6), root(2)
θ = arccos(root(6)/6) = 1.150
Note: So pretty much, scythian, I had been retarded.
Law of sines.... why not law of cosines. It was yelling to me the whole time. But I'm interested to know how you approached this.
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.