Maximize it
The maximum area of an equilateral triangle that can be inscribed in a unit square can be expressed in the form
a b − c . Where a and c are positive integers and b is a square-free positive integer. Find a + b + c .
The answer is 8.
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2 solutions
The proof of the 1 5 ∘ part is:
Let the other angles be α and θ .
We know that α + θ = 3 0 ∘ .
And since it is an equilateral triangle,
sec α = x and
sec θ = x .
Thus, α = θ = 1 5 ∘ .
Question: How do you type the degree symbol in LaTeX? Update: Thanks Mr. Calvin!
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For degree symbol, I use ^ \circ, which gives ∘ , e.g. 1 5 ∘ .
Some people just use ^0, which gives 0 , e.g 1 5 0 .
Yes, that's a good explanation of my "By symmetry" comment. :) As for the degree symbol, it would appear that you have figured it out. I was trying to figure that out myself, (I'm new at LaTeX, as you can probably tell.)
There needs to be a slight explanation of why "maximum area is achieved when vertex coincides". That is not immediately obvious.
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Yes, I was sort of just waving my hands around there. I just felt that it seemed intuitively clear that the equilateral triangle of maximum area would have to be symmetric about a diagonal of the square. This then would necessitate that one of its vertices would coincide with one of the corners of the square. The rest then follows from my and Sean's analyses. I'm not sure yet how to formalize my initial "intuition", but I'll keep thinking about it; suggestions are both encouraged and welcome. :)
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Let ABCD be the unit square and CDE an equilateral triangle on CD. E is far away from BA. So we have room for expansion, till E is on BA. This new CD'E' will be the max. area triangle.
D' is on DA and CD' is one side of the triangle.
I may add:-
3 1 = t a n ( 3 0 ) = 1 − t a n 2 ( 1 5 ) 2 t a n ( 1 5 ) g i v e s t a n ( 1 5 ) = 2 − 3
C D ′ 2 = D ′ D 2 + C D 2 . . . . . . D ′ D = tan ( 1 5 ) ∗ C D ∴ C D ′ 2 = tan 2 ( 1 5 ) + 1 . . . A r e a = 4 3 ∗ { 1 + ( 2 − 3 ) 2 } = 4 3 ∗ ( 8 − 4 ∗ 3 ) = 3 ( 2 − 3 ) . . . = 2 ∗ 3 − 3 = a b − c .
I solved this problem in two ways, using the property that the equilateral triangle has equal length sides and by using complex numbers and rotation. For my first method, its just an application of the Pythagorean Theorem twice and setting the two lengths equal to get the coordinates of the triangle then further finding the length and using s^2sqrt(3)/4 for the area. For the second method you can set the coordinates of the triangle at (0,0), (1,b) and (b,1). we rotate point (1,b) 60 degrees counter clockwise to (b,1). (1,b) is also defined as 1+bi in the complex plane, so (b,1) is b+i. Solve the equation to find b and carry out the rest of calculations.
i guess there is nothing like maximum or minimum ... the equilateral triangle inscribed itself is the only one .. !
The maximum area is achieved when one vertex of the equilateral triangle coincides with one of the corners of the square, and the other two vertices lie on the sides opposite the aforementioned corner. By symmetry, this means that the "base" of the triangle makes an angle of 15 degrees with the base of the square, which in turn means that the sides of the triangle have length sec(15).
The area of the triangle will then be ( 3 / 4) * ( s e c ( 1 5 ) ) 2 .
Now sec(15) = 2 2 − 3 , so we end up with an area of 2 3 - 3, leaving us with
a + b + c = 2 + 3 + 3 = 8 .