Reflect on this

[bonanova]

A triangle with vertices (6,5), (8,-3) and (9,1) is reflected about the line x = 8

to create a second, overlapping triangle with vertices at (10,5), (8,-3) and (7,1).

What is their combined area?

[nana77]

spoiler alert

Flipping about line x creates 2 more points for a 5 point polygon (since we are only interested in total area and not the overlapping area). The (8,-3) point remains the same after the flip. The two new points are (7,1) and (10,5). We can simplify this 5 point polygon into a normal triangle by realizing that the points (7,1) and (9,1) lie exactly along the lines created by the points along y=5 and y=-3. Now just solve for that area.

Simplify further by splitting that triangle into 2 right angle triangles with a new point at (8,5)

a=1/2 bh, and since there are two, the total area is base times height. b=4, h=8, area = 32.

[bonanova]

That is the area of a rectangle that contains all the vertices: x in [6, 10] and y in [-3, 5].

The area we need is less than that.

[nana77]

Did you only want the area of overlap?

[bonanova]

No. It should be included, but only once.

The two triangles (the original and its mirror image) partially overlap,

making a four-sided figure with vertices at (6,5) (8,-3) (10,5) and a point

that is the intersection of lines segments from (6,5) to (9,1) and from (10,5) to (7,1).

We want the area of that four-sided figure.

[nana77]

Oh. I had a scatter plot to see it and was giving the area of everything in the plot. But there is a piece to subtract out of that.

[DeGe]

Lets say the area of the triangle is A

The area of reflected triangle in also A

Lets say area of overlap is O

The area we are looking for is 2A - O

Area O is a kite with diagonals 2 and 5 (haven't drawn it but only pictured it... so not sure if this is correct)

Therefore area (d1d2/2) of O = 5

The area we want is 2A - 5

Now for the aha moment:

2A - 5 = A + 1/4A = 5/4A

3A/4 = 5

A = 20/3

So the area we want (2A - O) = 25/3

[witzar]

To get from (7,1) to (10,5) you need to take 3 steps right and 4 steps up,

so a straight line through these points croses line x=8 4/3 units above (7,1) (since it's 1 step right).
Therefore the coordinates of 4th vertex of the quadrilateral are (8, 1+4/3).
Given this caclulation of area of the quadrilateral is elementary: 4*(8-(2+2/3))/2 = 2*(5+1/3) = 10+2/3.

[k-man]

32/3

The interesting point of intersection is on x=8 line. The y coordinate of that point is 7/3. Due to the symmetry along the x=8 line we can only consider the are of the left side and multiply it by 2. The area of that triangle is half of the area of a rectangle 16/3 x 2. So the area we're looking for is the area of that rectangle and is 32/3

[k-man]

witzar beat me to it

[bonanova]

NaNa claimed enough real estate to include everything.

DeGe's kite was slightly under-sized.

k-man had it, but was still typing when witzar hit the Post button

witzar by a nose!

Good job everyone.