Mathematics VIII: rectangle (not so hard)

[⛄]

In the picture depicted below the rectangle is divided into several segments.
All secant lines go through a corner point of the quadrilateral.
The green segment has an area of 20, the blue segment has an area of 2, and the yellow segment has an area of 3.
How big is the red area?



This problem is very hard. I'm curious if anyone can solve it.

[tik 🪀✨]

can this problem be solved with basic trig and algebra?

also, the areas of the yellow and blue triangles do not make sense to me just looking at them.

[⛄]

can this problem be solved with basic trig and algebra?

Yes.

also, the areas of the yellow and blue triangles do not make sense to me just looking at them.

It's just supposed to be a sketch. But I've changed the numbers.

[True Federalist (진정한 연방 주의자)]

25.
The red area is equal to the sum of the blue, yellow, and green areas.

The area of the four right triangles is equal to that of the rectangle.  Subtract them and you are left with the red area, but you've subtracted the blue, yellow, and green areas twice since they are where the triangles overlap.  Hence, the red area is equal to the sum of the blue, yellow, and green areas.

Letting a, b, c, and d equal the areas of the four white areas in order around the red tetragon, and R, B, G, and Y equal the areas of the red, blue, green, and yellow areas respectively.

Then the area of the four right triangles in the figure are:
a+Y, Y+b+G, G+c+B, and B+d.
As I said, the four right triangles are in sum equal to the area of the rectangle, so:
(a+Y)+(Y+b+G)+(G+c+B)+(B+d)=a+b+c+d+R+B+G+Y
That simplifies to:
Y+G+B=R

[True Federalist (진정한 연방 주의자)]

How did you calculate it?

Calculation given in updated post, which I just updated again to give some algebra to back up my answer.

[⛄]

Good. Did you already know this problem? Or are you just intelligent?

[True Federalist (진정한 연방 주의자)]

Good. Did you already know this problem? Or are you just intelligent?

The latter.  I started out trying this method with the complication of trying to find the four unknown lengths of the line segments of the right and bottom of the rectangle, but realized as I was doing it, that I didn't need to find them.

[○∙◄☻¥tπ[╪AV┼cVê└]

25.

Confirmed

[○∙◄☻¥tπ[╪AV┼cVê└]

25.
The red area is equal to the sum of the blue, yellow, and green areas.

The area of the four right triangles is equal to that of the rectangle.  Subtract them and you are left with the red area, but you've subtracted the blue, yellow, and green areas twice since they are where the triangles overlap.  Hence, the red area is equal to the sum of the blue, yellow, and green areas.

Letting a, b, c, and d equal the areas of the four white areas in order around the red tetragon, and R, B, G, and Y equal the areas of the red, blue, green, and yellow areas respectively.

Then the area of the four right triangles in the figure are:
a+Y, Y+b+G, G+c+B, and B+d.
As I said, the four right triangles are in sum equal to the area of the rectangle, so:
(a+Y)+(Y+b+G)+(G+c+B)+(B+d)=a+b+c+d+R+B+G+Y
That simplifies to:
Y+G+B=R

I solved it a different way.


Using the formula 1/2 * base * height for the area of the triangle, the two triangles b+d+R and a+c+R  must have half the area of the rectangle. Together with the other halves, we have b + d + R = a + c + R = Y + G + B + a + c = Y + G + B + b + d.

Since Y + G + B + a + c = a + c + R, we have R = Y + G + B = 25

[Foucaulf]

I solved it a different way.

Jfern's solution is undeniably nicer, by the way.

If we want to be purely algebraic, then proceed as follows:

Let W denote all non-coloured space and R the red space. We find a unique solution provided we can establish a system of two equations.

Observing that the sum of the two triangles formed between the secant lines and one side of the rectangle is the area (here the corner point assumption is crucial) we have
(1) W + 2R = A.
Brute force summation gives also (2) W + R = A - (Y+G+B). Subtract (2) from (1) to get R = Y+G+B = 25.

Notice the values of the other coloured areas do not matter at all, nor was any trigonometry used, because the slopes of the secant lines were irrelevant. They are relevant insofar as they determine the amount of area covered by the Y+G+B segments.


Extension: would this strategy still hold if we were working with a parallelogram? (Yes.) In fact, the parallelogram is probably the more natural ground for examining this question. Does it work if we were working with a rectangular prism? (Almost certainly not.)

[⛄]

Why could you solve that problem so quickly?
I needed five years?

[angus]

Why could you solve that problem so quickly?
I needed five years?

Amazing that it takes so long to accomplish one's goals when one spends all his free time making asinine threads on internet forums, isn't it?

[⛄]

Why could you solve that problem so quickly?
I needed five years?

Amazing that it takes so long to accomplish one's goals when one spends all his free time making asinine threads on internet forums, isn't it?

What?

[doktorb]

Why could you solve that problem so quickly?
I needed five years?

Amazing that it takes so long to accomplish one's goals when one spends all his free time making asinine threads on internet forums, isn't it?

What?

I think that was an attempt at a burn