Three squares of different areas are side by side. The middle square’s area is 16, and its opposite sides are each collinear with an adjacent square’s side.

What is the area of the red quadrilateral?

12
15
16
18

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Good explanation brother

Ravi Bendi
- 2 years, 10 months ago

The question seems random to me but once saw the solution it makes sense!

Francvien Paraphantakul
- 2 years, 10 months ago

can someone please explain to me what seems to be intermediate steps that are not included to get from the top line to the middle and then to the 3rd. I can't reconcile the signs and multiplication to end up like it is presented here. Where did the denominator of 2 go? Thank you.

James Price
- 2 years, 10 months ago

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is the formule to calcul an are of a triangle

adem outlaw
- 2 years, 10 months ago

Triangles are half rectangles and look close at the formula, it describes which parts are which triangle areas.

Olaf Doschke
- 2 years, 10 months ago

O started punching letters a, b as well, but thought it would not help me. So I have not had the idea of calculating bits and pieces like you did. Now I realized that in the end all literal calculations cancelled out. Nice solution

Rogerio De Souza
- 2 years, 10 months ago

Why did you subtract the pink area? It wasn't part of the 3 squares.

Denise Haskins
- 2 years, 10 months ago

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? Pink area is added after being calculated.

Laz Valdes
- 2 years, 10 months ago

The pink area is supposed to be added, because it is part of the original red quadrilateral.

Gonçalo Freitas
- 2 years, 10 months ago

It's not subtracted, it's added

Olaf Doschke
- 2 years, 10 months ago

Beautiful presentation. Congratulations.

Akin Kosetorunu
- 2 years, 10 months ago

Why do you change the - sign between a-4 and 4-b? Maybe not why, but how?

Erwin Turpault-Petibon
- 2 years, 10 months ago

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$a>4$ but $b<4$ .

Chew-Seong Cheong
- 2 years, 10 months ago

- the side lengths of pink triangle are b and 4-b (with b<4)
- all other triangles are subtracted from the sum of all square areas, but the pink triangle is added.

Olaf Doschke
- 2 years, 10 months ago

wow, this is epic! very beautiful and clear!

Vladimir Fekete
- 2 years, 10 months ago

Implied that the answer does not depend on a or b, let's set a=4 and b=0. (Technically, a=4+e and b=e where e is infinitesimally small.) The blue, green and pink areas vanish and the yellow and red areas become equal. So the answer is half of the area of a 4x8 rectangle.

Tapani Lindgren
- 2 years, 7 months ago

Label the vertices and connect
$AC, GE, FI$
.

Notice that
$\overline{AC}, \overline{GE}$
and
$\overline{FI}$
are parallel to each other. Then,
$\Delta AEG$
and
$\Delta CEG$
have the same areas (equal base, same height),
$\Delta EIG$
and
$\Delta EFG$
have the same areas (ditto), which concludes that the area of quadrilateral
$AEIG$
is the area of the square
$CEFG$
.

連AC、GE、FI，可知AC平行GE平行FI，所以三角形AEG面積=三角形CEG面積，三角形EIG面積=三角形EFG面積，則四邊形AEIG面積=正方形CEFG面積。

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For those who can't read Traditional Chinese, here is the translation that is worded slightly different from the literal one:

Notice that $\overline{AC}, \overline{GE}$ and $\overline{FI}$ are parallel to each other. Then, $\Delta AEG$ and $\Delta CEG$ have the same areas; $\Delta EIG$ and $\Delta EFG$ have the same areas. In this case, the area of quadrilateral $AEIG$ is the area of the square $CEFG$ .

Michael Huang
- 2 years, 11 months ago

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Thanks for translating. That's a great solution.

Jeremy Galvagni
- 2 years, 10 months ago

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Not a problem. It is really my first time, doing this here. :)

Fun fact: I work out with various of languages, not only in Chinese! I can understand and use French and German, which are both useful in modern mathematical articles. :)

Michael Huang
- 2 years, 10 months ago

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@Michael Huang – Amazing but how did you learn this?

James Bacon
- 2 years, 10 months ago

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@James Bacon – I was influenced by my math college years when I was studying various of technical terms coined from European, Russian and Asian mathematicians. Some modern articles were written in the languages I already mentioned, which is why I became more interested in understanding how statements are formed, much like how patterns are formulated. :)

Michael Huang
- 2 years, 10 months ago

Yeah me too.

James Bacon
- 2 years, 10 months ago

Besides the upvotes (which simply push the solutions to the top), I consider this solution to be the best so far due to more geometry involved.

I like solving with various of methods. However, I prefer to post the most elegant method at any level. Seems like I am not the only one who prefers to avoid textbook-like methods not taught here!

Michael Huang
- 2 years, 10 months ago

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@Michael Huang – I agree. Do you know if there's a name for this technique/theorem?

Garoh Seven
- 2 years, 10 months ago

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@Garoh Seven – Robert mentioned the following theorem already:

The area of a triangle is 1/2 base x height. The base of all the of triangles is GE. Look carefully and you should be able to convince yourself that the heights are all the same.

Michael Huang
- 2 years, 10 months ago

Isn't math beautiful in all languages? Thank you for the translation too.

John Church
- 2 years, 10 months ago

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Yes, that is certainly true. Though, it is popular in some countries. Besides Wikipedia (which has a lot of translated articles into separate languages), it is difficult to find scholarly articles of rare languages, such as Urdu or Hindi. While I was in my non-math job, I independently pursue my interest in different subjects for more than 4 years (since my college graduation).

Michael Huang
- 2 years, 10 months ago

Is this a theorem? How are those areas equal?

Aishwary Omkar
- 2 years, 10 months ago

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The area of a triangle is 1/2 base x height. The base of all the of triangles is GE. Look carefully and you should be able to convince yourself that the heights are all the same.

Robert Kelleher
- 2 years, 10 months ago

Just to clarify, EG is a common base for the 2 pairs of triangles, and the common heights (SQRT 32)/2 are the perpendicular distances from EG to FI and EG to AC. Well, I cheated, just estimated (guessed) the dimensions of the various squares and calculated the answer as more than 16 but less than 17, therefore not 15 nor 18. .

Omar Jette
- 2 years, 10 months ago

Thanks a lot!

Gonçalo Freitas
- 2 years, 10 months ago

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Perfect! Missed the small part of the original text. :)

Michael Huang
- 2 years, 10 months ago

That's awesome! I used algebra, but I like this solution better!

Mark Lama
- 2 years, 10 months ago

Very beautiful solution!

Gonçalo Freitas
- 2 years, 10 months ago

Ah, that's beautiful!

Neville Reid
- 2 years, 10 months ago

can some one explain why those lines are parallel?

unite perry
- 2 years, 10 months ago

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There are all diagonals of squares. So all are at 45 degree angles to the sides.

Jeremy Galvagni
- 2 years, 10 months ago

nice!!!!!!!!!!!!!

Shubham Singh
- 2 years, 10 months ago

$AB=a$ and $HE=b$ , then $HI=IJ=4-b$ and $DG=a-4$ .

Let$area[AEIG]=area[ABEIJGD]-area[ABE]-area[GJI]-area[ADG]$

$area[AEIG]=a^2+4^2+(4-b)^2+\dfrac{1}{2}(b)(4-b)-\dfrac{1}{2}(a)(a+4)-\dfrac{1}{2}(a)(a-4)-\dfrac{1}{2}(4-b)(4+4-b)$

$area[AEIG]=a^2+32-8b+b^2+2b-\dfrac{b^2}{2}-a^2-16+6b-\dfrac{b^2}{2}$

$area[AEIG]=32-16=\boxed{16}$

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Using the limiting case and logic:

In the limiting case where the side squares’ sides go to zero, we just end up with the middle square as the red quadrilateral. The answer must be independent of the size of the side square, so the answer must be 16.

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Same reasoning here, except I took the limiting case where the leftmost square has the same size as the middle one (since I was not sure what the red quadrilateral should be if the leftmost square were to be a point.

Antoine G
- 2 years, 10 months ago

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You can also take the case where all square have the same size (as pointed below by Ben Mitchell).

Antoine G
- 2 years, 10 months ago

This. Assuming an answer exists can lead to much simpler solutions.

Matthew Sandy
- 2 years, 10 months ago

The assumption that they are different squares is a red herring. The answer appears independent of the relative sizes, and taking the limit as each square approaches that of the middle settles that.

So, assume all squares are the same size. Once done the red section becomes a parallelogram with base 4 and height 4. Trivial isn't it?

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That was exactly my reasoning. Took all of two minutes, including redrawing the figure and considering the limit.

John Pabrinkis
- 2 years, 9 months ago

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Let the bottom left hand corner of the middle sized square be at the origin of the Cartesian plane. Let the big square have side length b and the small square have side length s. Then it is easy to see that the coordinates of the required quadrilateral are in clockwise order, (-b,b),(0,4),(4+s,4-s) and (4,0).

Apply the marvellously simple shoelace formula to these coordinates to see the side lengths of the big and the small square vanish in a puff of logic leaving the answer 16.

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Given that the question has an answer independent of the sizes of the larger or smaller squares, it suffices to calculate a numerical solution to one special case e.g. when all three squares are equal.

It's plain to see that in such a case the quadrilateral is a parallelogram with area equal to the center square. Thus answer must be 16.

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The problem statement learns us nothing about the size of the external squares. The possible solutions are all integer. Therefore I conclude that the solution is independent of the area of the external squares. (Which indeed has been proven in earlier solutions - it all cancels out). So let us reconsider the problem with three equal squares:

The red figure then simply becomes a parallellogram. Its area is Base (4) * Height (4) = 16
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Let the side length of the large square be $a$ and that of the small square be $b$ . We have no information about $a$ and $b$ , they can have any value, yet the problem asks for a fixed solution.

So, the answer is independent of the values of $a$ and $b$ . Let's look at the border case $a=0$ and $b=0$ . In this case, all we have left is the center square, it coincides with the red area, which is 16.

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What we know:

```
1) Triangles that have the same base and height are equal in area
2) The middle square (quadrilateral BCEF) = 16
```

According to 1),
**
triangle ABF is equivalent to triangle AEF
**

Both triangle ABF and triangle AEF include triangle AFG within themselves, therefore
**
triangle ABG has an area equivalent to triangle EFG
**

Similarly,
**
triangle BCH has an area equivalent to triangle DEH
**

Since

```
triangle BCH and triangle EFG, and quadrilateral BHEG = quadrilateral BCEF and
quadrilateral BCEF = 16 according to 2)
```

we can deduce that triangle ABG, triangle DEH, and quadrilateral BHEG (what we are looking for) has the same area as the Square in the middle; 16.

Sincere apologies to the color blind...

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This is the proof I discovered! I was going to post, but you beat me to it.

Ken Haley
- 2 years, 10 months ago

I used the way of the cheater to quickly find the solution, lol. I'll tell you how. The middle square is the only one that has fixed dimensions, so your conclusion must be indifferent to the sizes of the squares on the edges. I know that in this situation the red area is clearly 16.

So, it must be still 16 with a bigger left square and a smaller right square (plus the corner triangle, that becomes a line when all the squares have the same size). Either this is true or the question has no answer. They should have added "Depends on the sizes of the other squares" as an alternative among the answers to prevent this cheat.
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The solutions provided are only valid if the squares have common CORNERS. The text of the question only stipulates common sides. If a configuration such as:

is made then the solution is not the same as the area of the central square.
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Good point!

Ken Haley
- 2 years, 10 months ago

I multiplied 8 by 2 and got 16.

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......revolutionary.

Simon The Great
- 2 years, 10 months ago

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Yes, but the real challenge would be to prove the answer holds regardless of the relative sizes. I have discovered a purely geometrical proof of this, which I'll post if no one else already has.

Ken Haley
- 2 years, 10 months ago

If all square are the same, the answer is 16. So...

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**
This is not a rigorous solution!
**

Let the sides of the outer squares be $a$ and $b$ , $a \geq 4 \geq b$ . The statement doesn't provide enough information to determine the value of $a$ and $b$ and the answers suggest the result does not depend on $a$ or $b$ .

If we choose convenient values for them (f.e. $a=4$ and $b=4$ or $b=0$ ), the red quadrilateral turns into a shape with known sides and whose area can be easily computed. For $a=4$ and $b=0$ the red shape is a right triangle with sides $8$ and $4$ ; its area is $\frac{8 \times 4}{2} = 16$

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You can complete the middle square by using the red pieces

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This may be cheating, somewhat, but based on the way the problem was worded, I inferred that it didn’t matter how large the other squares were, and the answer would be the same no matter what. Therefore, we can set the side lengths of both of the other squares to 4, giving us a parallelogram of height 4 and base 4, and an area of

16 |

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The sizes of the larger and smaller squares are not given, the result is therefore invariant on these values.

Let the larger and smaller squares be of side length 4, In this case the required area constitutes a parallelogram of parallel side length 4 and height 4

Thus area = 4 x 4 = 16

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The centre square is made up of a red band and two white triangles. The big square on the left contains a red triangle.

Statement 1: The area of the red triangle in the big square is equal to the bottom white triangle in the centre square.

Statement 2: The red area to the right of the centre square is equal to the top white triangle in the centre square.

It follows that the complete red area in the diagram is equal to the area of the centre square.

Proof of Statement 1:

Let the side of the centre square be c, and the side of the left hand (big) square be b. Consider the line from the top left corner of the big square to the bottom right corner of the centre square. This line intersects the vertical line common to the two squares. Let the distance from the bottom of the diagram to the intersection point be x.

Considering similar triangles: x = bc/(b + c).

Area of the bottom white triangle = ½ cx = ½ bc^2/(b + c)

Area of left red triangle = ½ b(c – x) = ½ b(c – bc/(b + c)) = ½ b (c2^/(b + c)).

Statement 2 can be proved in a similar way.

Therefore the red area is equal to the area of the centre square which is given as 16.

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Somewhat cheating solution (assuming that there is an invariant answer instead of proving that it is invariant):

The areas of the larger and smaller squares are not given, therefore we can assume that the answer does not depend on those sizes, and the area will remain the same if those sizes vary. So we can set those areas to be 16 also, giving a vastly simplified parallellogram as the red area with base and height equal to the middle square. Thus the area is 16.

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Let the side length of the large square be $a$ and that of the small square be $b$ . Then the area of the red region is:

$\begin{aligned} A_{\text{quad}} & = \underbrace{a^2 + 16 + b^2}_{\text{Area of 3 squares}} - \underbrace{\frac {a(a+4)}2}_{\triangle_{\text{yellow}}} - \underbrace{\frac {a(a-4)}2}_{\triangle_{\text{blue}}} - \underbrace{\frac {b(b+4)}2}_{\triangle_{\text{green}}} + \underbrace{\frac {b(4-b)}2}_{\triangle_{\text{pink}}} \\ & = a^2 + 16 + b^2 - \frac {a^2+4a+a^2-4a}2 - \frac {b^2+4b-4b+b^2}2 \\ & = a^2 + 16 + b^2 - a^2 - b^2 \\ & = \boxed{16} \end{aligned}$