SPOILER ALERT:
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Graphical proof & solution
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Another spoiler is a Youtube video with same title as this thread subject. Probably a better solution than mine there (I have refrained myself from watching it, but I'll do it presently).

I'm looking forward to your approaches (in case you find this simple problem worthwhile solving).

Have fun!

Code:

solve with coordinate geometry triangle area formula:

Spoiler Alert ...

let points A = (0,2), B = (2,0), C = (4,2), D = (2,4), P=(x,y)

triangle area = (ax(by - cy) + bx(cy - ay) + cx(ay - by)) / 2, points counter-clockwise

right bottom = 2 + area(A,B,P) = x + y

left bottom = 2 + area(B,C,P) = 4 - x + y

right top = 2 + area(A,P,D) = 4 + x - y

right top = 4*4 - (4 + x - y) - (x + y) - (4 - x - y) = 8 - x - y = 8 - right bottom

right top + left bottom = 8 = 16 cm^2 + 32 cm^2 = 48 cm^2

sx area = 8 - right bottom = 48 cm^2 - 20 cm^2 = 28 cm^2

Same answer, different approach - I considered the half-side segments (marked with the two short dashes) as of length "h" each and labeled all 9 vertices by their coordinates [0,0], [h,0], [2h,0], etc. Then found the areas of each of the four sections using the Shoelace Formula mentioned earlier here (thanks!). That gave me four equations in four unknowns. A bit of quick algebra gave h is the square root of 24 and the missing area as 28 for an overall area of 96. Fun - and thank you!

(08-06-2018 09:14 PM)Jim Horn Wrote: [ -> ]Same answer, different approach - I considered the half-side segments (marked with the two short dashes) as of length "h" each and labeled all 9 vertices by their coordinates [0,0], [h,0], [2h,0], etc. Then found the areas of each of the four sections using the Shoelace Formula mentioned earlier here (thanks!). That gave me four equations in four unknowns. A bit of quick algebra gave h is the square root of 24 and the missing area as 28 for an overall area of 96. Fun - and thank you!

That's the way I did it too, although the method shown in the youtube video is simpler!

A similar method to the shoelace method is to split or extend each quadrilateral into a trapazoid/trapezium and a right triangle (which is either added to or subtracted from the trapezoid/trapezium to get the overall area of the quadrilateral). This gives the same four equations as the shoelace formula.

Another way, still using coordinate geometry, but without Shoelace formula:

Again, let A=(0,2), P=(x,y), B=(2,0)

just sum the area like integration with trapezoid rule:

bottom right area (3 points A, P, B)

= (2+y)/2 * (x-0) + (y+0)/2 * (2-x)

= 1/2 * (2x + x y + 2y - x y)

= x + y

it match the cordinate triangle area formula, without the magic.

it is super easy messing up the "shoelace"

Edit: just realized ijabott just said that in previous post. Sorry for the noise ...

I have seen the geometric solution ! (no, i did not peek)

Since all points A=(0,2), B=(2,0), C=(4,2), D=(2,4) is the mid-point of square side,

Join them all, and it is still a square (no coordinates needed, but I can't draw with words)

Within this smaller square, the opposite triangles must sum to half of square.

(same base, 2 height sum to base, thus 2 triangles sum to base*base/2)

so, back to big square ...

opposite quadrilaterals also sum the same, to half of big square.

20 cm^2 + sx = 16 cm^2 + 32 cm^2

-> sx = 28 cm^2

(08-06-2018 11:54 PM)rprosperi Wrote: [ -> ] (08-06-2018 10:52 PM)Albert Chan Wrote: [ -> ]20 cm^2 + sx = 16 cm^2 + 32 cm^2

-> sx = 28 cm^2

I don't follow the explanation at all (don't worry, it's me, not you), but it doesn't seem like the final steps shown above are correct. They seem to imply sx = SQRT(880) = 29.66, don't they?

These are area units (cm^2 = square centimeters). Perhaps cm2 (without the caret symbol) might have avoided the confusion.

All solutions and comments welcome! (Linear Algebra, Analytic Geometry, Geometry...). Thank you all for your interest and contributions.

Gerson.

(08-06-2018 10:52 PM)Albert Chan Wrote: [ -> ]20 cm^2 + sx = 16 cm^2 + 32 cm^2

That was my reasoning as well except that I was reading it as: \(20+5x=16+32\)

Which would lead me to: \(x=5.6\)

Well, I guess I muffed this one.

But seriously: Why would you use an index \(x\) in \(s_x\)?

Thanks for the challenge and the chuckle.

Thomas

PS: It reminded me of

Viviani's theorem.

(08-07-2018 12:41 AM)Thomas Klemm Wrote: [ -> ] (08-06-2018 10:52 PM)Albert Chan Wrote: [ -> ]20 cm^2 + sx = 16 cm^2 + 32 cm^2

That was my reasoning as well except that I was reading it as: \(20+5x=16+32\)

Which would lead me to: \(x=5.6\)

Well, I guess I muffed this one.

But seriously: Why would you use an index \(x\) in \(s_x\)?

I don't get it either, but that sx is what the picture ask to solve ...

Ideally, cm^2 should be left out, and put back at the end.

However, my previous attempt use a dimensionless unit (square side = 4), so units need to be carried to avoid confusion.

(08-07-2018 12:41 AM)Thomas Klemm Wrote: [ -> ]But seriously: Why would you use an index \(x\) in \(s_x\)?

Really a bad choice. \(S_4\) or simply \(S\) would’ve been better. Too late now…

Gerson.

(08-07-2018 12:57 AM)Albert Chan Wrote: [ -> ]Ideally, cm^2 should be left out, and put back at the end.

As a former Physics student (eons ago, for one and a half year), I usually don’t leave them out either. But I did this time as there was no way to make a mistake on this one.

For some reason (perhaps a forgotten knowledge) I had a hunch the sums of the areas of opposite regions would equal each other. I tried two or three examples in scale and it proved right. Hopefully my graphical proof us valid and clear, despite mistakes like S1’ + S2” instead of S1’ + S2’.

Once I had a hard time trying to impose my point

without a proper proof, so I’ve been avoiding participating in these kind of discussions, but I thought this problem was worth being presented here. I’m glad you and others have liked it.

Gerson.

Hi all,

An image to visualize the approach previously presented by some other contributors. ;-)

[

attachment=6193]

Not a Picasso but hopefuly clear enough.

Cheers.

Diego.

Looking back my first post, I had proved opposite quadrilaterals sum to half of big square, without realizing it.

With assumed square side = 4, I had this: right top + left bottom = 8

Only one more step, saying area of square = 4 * 4 = 16 = 2 * 8, and the prove is done.

Oh, so close ...

Hello,

Follow me in my complicated way:

A=16

B=20

C=32

X=?

If d is Big Square side:

=> d*d=A+B+C+X

Surface X=Rectangle+UpTriangle+DownTriangle

Rectangle=(d/2-b)*d/2 (b is base of DownTriangle X)

UpTriangle=a*(d/2-b)/2 (a is base of UpTriangle X)

DownTriangle=b*(d/2+a)/2

=> X=d/4*(d+a-b)

Surface A=Rectangle+UpTriangle+DownTriangle

Rectangle=(d/2-a)*(d/2-b)

UpTriangle=UpTriangle X

DownTriangle=b*(d/2-a)/2

=> A=d/4*(d-a-b)

Surface B=Rectangle+UpTriangle+DownTriangle

Rectangle=(d/2-a)*d/2

UpTriangle=DownTriangle A

DownTriangle=a*(d/2+b)/2

=> B=d/4*(d-a+b)

Surface C=Rectangle-DownTriangle B-DownTriangle X

Rectangle=(d/2+a)*(d/2+b)

=> C=d/4*(d+a+b)

Sanity check of A+B+C+X:

d/4*(d-a-b)+d/4*(d-a+b)+d/4*(d+a+b)+d/4*(d+a-b) simplifies as d*d, without relation to a and b, as expected

Now,

C=2*A => d=3*(a+b)

4*B=5*A => d=a+9*b

5*C=8*B => d=13*a/3-b

Solve system of 3 equations with 3 unknowns

=> a is free, b=a/3, d=4*a

Pushing this back into earlier formula A=d/4*(d-a-b)

=> a=sqrt(6)

=> b=sqrt(6)/3, d=4*sqrt(6)

As X=d*d-A-B-C

=> X=(4*sqrt(6))^2-16-20-32-=28

Delightfully complicated

(08-07-2018 08:59 AM)Pekis Wrote: [ -> ]Follow me in my complicated way:

A=16

B=20

C=32

X=?

...

=> A=d/4*(d-a-b)

=> C=d/4*(d+a+b)

It is pretty good.

But, had you spotted A + C is independent of a,b, it will cut down a lot of details.

A + C = d/4*(2d) = d^2/2

B + X = d^2 - (A + C) = d^2/2

A + C = B + X

X = A + C - B = 28 cm^2

(08-06-2018 10:52 PM)Albert Chan Wrote: [ -> ]I have seen the geometric solution ! (no, i did not peek)

Same solution as Albert in post #6

[

attachment=6195]

I started out by joining the midpoints of the original square hoping to simplify the problem: all triangles. Returning to the matter after an hour or so I noticed de sum of the heights of a pair of opposite triangles being the same as for the other pair.

ha + hc = hb + hx

Bases of the triangles being the same and the red surfaces also the same...

SX + SB = SA + SC

SX = SA + SC - SB = 28

Delightfully simple.

(Lucky day!)

Thanks for posting the problem.

Worth a try if 'I should be able to solve this'.

Robert

Hi,

After attacking this with analytic geometry (?), and simplifying several times, I wanted to solve it using just geometry. Tough to wrap my head around, but I rotated the square to a diamond to set up x-y axes vertical and horizontal, and ended up proving that opposite quads are equal...

(see attachment)

Tom

There is another way to solve top right area without inner square:

Let P = (x,y), inside the square with side = 2

With P, square splitted into 4 triangles, all with base=2, thus triangle area = height

Triangle Area: Left, Bottom, Right, Top = x, y, 2-x, 2-y

SB = 1/2 (Left + Bottom) = (x + y)/2

SX = 1/2 (Right + Top) = 2 - (x + y)/2

SB + SX = 2

SA + SC = 2*2 - (SB + SX) = 4 - 2 = 2

SX = 2 - SB = SA+ SC - SB = (16 + 32 - 20) cm^2 = 28 cm^2

Area of quadrangle by coordinates:

S = ((x1*y2 - y1*x2) + (x2*y3 - y2*x3) + (x3*y4 - y2*x4) + (x4*y1 - y4*x1))/2

[

attachment=6198]

We have four quadrangles:

16 = ((0*y - a*x) + (x*2a - y*a) + (a*2a - 2a*0) + (0*a - 2a*0))/2

20 = ((0*0 - 0*a) + (a*y - 0*x) + (x*a - y*0) + (0*0 - a*0))/2

32 = ((a*0 - 0*2a) + (2a*a - 0*2a) + (2a*y - a*x) + (x*0 - y*a))/2

S = ((x*a - y*2a) + (2a*2a - a*2a) + (2a*2a - 2a*a) + (a*y - 2a*x))/2

After simplification:

16 = (a*x - a*y +2a^2)/2

20 = (a*x + a*y)/2

32 = (a*y - a*x + 2a^2)/2

S = (-a*x - a*y + 4a^2)/2

Take Prime:

[

attachment=6199]

[

attachment=6201]