Very interesting math puzzle...?

Question

There is a blue equilateral triangle with one of its vertices at one corner of the surrounding square and the others on two of the sides of the square. Is the black triangle's area equal to the area of the two white triangles?

Please go here to see a diagram:

http://img246.imageshack.us/img246/7084/puzzlept4.png

(remember, the blue triangle is equliateral)

A speedy response is required! Have fun...

Answer 1

let's go....

let the side of the square is 1 unit

the angle at the corner (of white triangles) = 15 deg

so base of the white triangle = tan(15)

so the area of white triangles = 1/2 . 1. tan(15) = t/2 ( t= tan(15))

side of blue triangle = 1 / cos(15)

hence area of blue triangle = sqrt(3) / 4 * 1/(cos(15)^2)

the values are calculated as:

area of white triangle = 0.134 unit^2 each
total for 2 white triangles = 0.268 unit^2

area of blue triangle = 0.464 unit^2

so the area of the black triangle = 1-0.268-0.464=1-0.732=0.268

hence equals the area of the white triangles...


note:

this question can be answered intuitively in a simple way....think!

Answer 2

My first instinct was no, but I was wrong.

What you have are two white triangles (15-75-90). You can figure this out by subtracting 60 from 90 (the corner) and cutting it in half. So that angle is 15 degrees. Then it's obvious the other angle is 75 degrees. The hypotenuse of each is the side of the equilateral triangle.

And the black triangle is a 45-45-90 triangle, also with a hypotenuse equal to the side of the equilateral triangle.

Using the ASA rule for the 15-75-90 triangles, and picking an arbitrary length of 1 unit for the equilateral triangle, each came out to an area of 1/8 sq. units.

Here's the details of that:
The legs of a white triangle will be:
a = ( sqrt(6) + sqrt(2) ) / 4
b = ( sqrt(6) - sqrt(2) ) / 4

a is the leg opposite the 15 degree angle
b is the leg opposite the 75 degree angle

So the area of a white triangle is: ab/2
(sqrt(6) + sqrt(2))(sqrt(6) - sqrt(2)) / (4 * 4 * 2)

= (6 + sqrt(12) - sqrt(12) - 2) / 32
= (6 - 2) / 32
= 4 / 32
= 1 / 8

Note: to double check these values, use the pythagorean theorem and confirm the hypotenuse is 1.

Squaring each:
a² = ( 6 + 2 sqrt(2) + 2 ) / 16
b² = ( 6 - 2 sqrt(2) + 2 ) / 16

So summing we get:
c² = ( 8 + 2 sqrt(2) + 8 - sqrt(2) ) / 16
c² = 1
c = 1

And for the 45-45-90 triangle, call each leg x.
For a 45-45-90 triangle x = 1/sqrt(2).
So the area is x²/2
= (1/sqrt(2))^2 / 2
= (1/2) / 2
= 1/4 sq. units.

So actually the two white triangles (2 * 1/8) do add up to the same as the black triangle (1/4).

Answer 3

Yes, they are the same area, but I can't come up with a formula to prove it. I drew the example in autocad, but I couldn't come up with a way to draw the equilateral triangle so that it touched the edges like in the diagram. So I drew and scaled and rotated and scaled again and rotated and got it close, then I computed the areas. They were very close, so I'm saying yes... but now I need to know how and why. Will you tell me?

Answer 4

not sure but i think the two white triangles would have a bigger area

Answer 5

yes...i think...duno =S is this meant to be similar to the pythagoras thereom? haha i have no idea..