Square ratios

Let $x$ be the length of the smallest side of the smaller blue triangle plus the length of one of the sides of the yellow square, let $y$ be the length of the medium side of the larger blue triangle plus the length of one of the sides of the yellow square, and $z$ be the length of one of the sides of the yellow square, as shown below. Also, let the large square be a unit square, so that the red square has an area of $\frac{29}{50}$ and therefore a side length of $\sqrt{\frac{29}{50}}$ .

Then by Pythagorean's Theorem, $x^2 + y^2 = \frac{29}{50}$ , and as a unit square, $x + y = 1$ , Solving the system of equations for $x < y$ gives $x = \frac{3}{10}$ and $y = \frac{7}{10}$ .

By similar triangles, $\frac{x}{y} = \frac{z}{y - z}$ , or $\frac{\frac{3}{10}}{\frac{7}{10}} = \frac{z}{\frac{7}{10} - z}$ , which solves to $z = \frac{21}{100}$ .

This means that the ratio of the area of four yellow squares to the area of the large square is $\frac{4z^2}{1} = \frac{4(\frac{21}{100})^2}{1} = \frac{441}{2500}$ , so $a = 441$ and $b = 2500$ , and $a + b = \boxed{2941}$ .

Let the large square be a unit square, the longer leg of the larger gray triangle be $G$ , the shorter leg of the smaller gray triangle be $g$ , and the side of the red square be $r$ . The area of all the gray triangles is given by $2rG+2rg=2r(G+g)$ . Because the two triangle legs and two red squares make up one edge of the unit square, this is equal to $2r(1-2r)$ . Adding up all the areas inside the unit square gives $2r(1-2r)+4r^2+\frac{29}{50}=1 \implies 2r=\frac{21}{50} \implies 4r^2=\boxed{\frac{441}{2500}}$

Let area of the large square be 1 hence its side length is also 1. The area of the red square is therefore $\frac {29}{50}$ The area of the triangle inscribing the yellow square (two blue triangles and the yellow square) is $\frac 14 \left(1-\frac {29}{50}\right)= \frac {21}{200}$ .

Let the short and long side length of the triangle inscribing the yellow square be $a$ and $b$ respectively. Then we have $a+b=1$ and:

$\begin{aligned} \frac {ab}2 & = \frac {21}{200} \\ ab & = 0.21 & \small \color{#3D99F6} \text{Since }a+b=1 \\ a(1-a) & = 0.21 \\ \implies a^2 - a + 0.21 & = 0 & \small \color{#3D99F6} \text{Solving the quadraic for }a \end{aligned}$

$\implies a, b = \dfrac {1 \pm \sqrt{1-0.84}}2 \implies \begin{cases} a = 0.3 \\ b = 0.7 \end{cases}$

Let the side length of the yellow square be $x$ . We note that the large blue triangle is similar to the triangle inscribing the yellow triangle. Therefore, we have:

$\begin{aligned} \frac x{b-x} & = \frac ab = \frac 37 \\ 7x & = 2.1 - 3x \\ \implies x & = \frac {21}{100}\end{aligned}$

The area of the four yellow squares is $4x^2 = 4 \times \dfrac {21^2}{100^2} = \dfrac {441}{2500}$ . Therefore, $a+b = 441+2500 = \boxed{2941}$ .

This could be easily solved in the mind. You move grey triangles making two grey rectangles, leaving blue rectangle in the centre (without touching red area). Then you move blue rectangle down, and then join two red rectangles together making one red square.

Its size can be easily calculated, cause if big square have side length equals $1$ , then blue rectangle have side lengths $1$ and $\frac{29}{50}$ . So red square have side length $1-\frac{29}{50}=\frac{21}{50}$ .

Area equals $\big(\frac{21}{50}\big)^2$ , answer is $21^2+50^2=\boxed{2941}$

Let the side length of the biggest square be $10$ . So the area of the biggest square is $100$ . The area of the red square is $\dfrac{29}{50} \times 100 = 58$ . The side length of the red square is $a+b=\sqrt{58}$ . By applying similar triangles, we have

$\dfrac{c}{a}=\dfrac{b}{10-2c}$

$ab=10c-2c^2$ (equation 1)

Let's square $a+b=\sqrt{58}$ . It becomes $a^2+2ab+b^2=58$ . This becomes our equation 2. Substituting equation 1 to equation 2, we get $a^2+b^2=58-20c+4c^2$ . This is our equation 3.

Apply pythagorean theorem,

$(10-2c)^2=a^2+b^2$

$100-40c+4c^2=a^2+b^2$ (equation 4)

Substituting equation 3 to equation 4, we get $c=\dfrac{21}{10}$ .

The area of one yellow square is $\left(\dfrac{21}{10}\right)^2=\dfrac{441}{100}$ .

The area of four yellow squares is $4 \times \dfrac{441}{100}=\dfrac{441}{25}$

The fraction required is $\dfrac{\dfrac{441}{25}}{100}=\dfrac{441}{2500}$

The desired answer is $441+2500=\boxed{2941}$

Let $x$ denote the sidelength of a yellow square and let $y$ denote the ratio of the longer leg to the shorter leg in each triangle (they are all similar, so this is well-defined). Then the largest square has area $(x+xy+\frac{x}{y}+x)^2 = 1 \implies \frac{(1+y)^2}{y} = \frac{1}{x}$ and the red square has area $(x+xy)^2 + (x+\frac{x}{y})^2 = \frac{29}{50}$

This second equation can be rewritten as $\frac{29}{50} = x^2\left(\frac{(1+y)^2}{y} - 2\right)\left(\frac{(1+y)^2}{y}\right) = x^2\left(\frac{1}{x} - 2\right)\left(\frac{1}{x}\right) = 1-2x$ which gives $x = \frac{21}{100}$ so the total area of the yellow squares is $4x^2 = \frac{21^2}{50^2} = \frac{441}{2500}$ and since we defined the area of the largest square to be $1$ , this is also the fraction of the largest square that the yellow squares fill. Therefore, the answer is $441 + 2500 = \boxed{2941}$

Let $x$ be the side length of the red square and $y$ be the length of the largest square. Equating the area of the blue regions in two different way, $(y-2x)^2+4(y-2x)x/2 = (29/50)y^2,$ $50y^2-100xy = 29y^2,$ $21y = 100x,$ $4x^2/y^2 = 4(21/100)^2 = 441/2500,$ $a+b = 2941.$

Refer to the above diagram. Let $x$ be the side length of the red square. By similar triangles we get the relation $\frac{a-x}{x} = \frac{a}{b}.$ Thus $x = \frac{ab}{a+b}.$ The side length of the blue square be $c$ . The sides of the blue square forms a right triangle with the sides $a$ and $b$ . Since the side length of the large square is $a+b$ , $\frac{c^2}{(a+b)^2} = \frac{29}{50}$ or $\frac{a^2+b^2}{(a+b)^2} = \frac{29}{50}.$ Let $(a+b)^2 = 50$ (the area of the large square) and $a^2+b^2 = 29$ (the area of the blue square). This means $(a+b)^2-(a^2+b^2) = 50-29$ , which shows that $2ab = 21$ or $ab = \frac{21}{2}$ . The area of red is four times the area of the red square. $4x^2 = 4\left( \frac{ab}{a+b}\right)^2 = 4\cdot\frac{441}{4} \cdot \frac{1}{50} = \frac{441}{50}.$ Since the area of the large square is 50 $\frac{4x^2}{50} = \frac{441}{2500}.$ Just add the numerator and denominator to get $\boxed{2941}$ .

Let 'c' be the side of large square. Let 'b' be the side of middle square and 'a' the side of small four squares. Then, we have the relation b^2=(c-2a)^2+2a(c-2a)=c(c-2a), Assuming c=10 with (b^2/c^2)=(29/50), we get, a=(21/10), which gives, (4a^2/c^2)=(21/50)^2=441/2500, Therefore, Answer=2941

I am sorry, but I do not know how to make the image smaller.

Let the larger square be a unit square.

We note that $x+y=(1-2s)$ . The area of the four triangles is $1-\frac{29}{50}=\frac{21}{50}$ . Therefore, $4(s^{2}+\frac{1}{2}xs+\frac{1}{2}ys)=\frac{21}{50}$

$4[s^{2}+\frac{1}{2}s(x+y)]=4[s^{2}+\frac{1}{2}s(1-2s)]=4(s^{2}+\frac{1}{2}s-s^{2})=2s=\frac{21}{50}$ So $s=\frac{21}{100}$

Finally, $4s^{2}=\frac{441}{2500}$ So the answer is $\boxed{2941}$

Divide up as in the picture, swap the colors as indicated by the arrow, and you end up with a blue rectangle in the middle and a rectangle at the top and bottom both the height of the red squares. Given that the blue rectangle obtained that way is 29/50 of the height, twice the height of the red square is 1 - 29/50 = 21/50 of the height of the full square. Likewise for the width. The 4 squares form a single square of twice the height and width of a single square, so it is $(21/50)^2 = 441/2500$ of the total square, and thus the answer is 441 + 2500 = 2941. Done.

Assuming that the length of the largest square is 10,the area of the bule square is 58 accordingly. The area of the blue square can be presented as (10-2a) a 2+(10-2a)^2 which equals 58.Then we get a( $\frac{7}{5}$ ).

Let the side of the larger square be $1$ .

$\implies 1 = 2(y)(1 - y) + \dfrac{29}{50} \implies 100y^2 - 100y + 21 = 0 \implies y_{1} = \dfrac{7}{10}$ and $y_{2} = \dfrac{3}{10}$ .

Using similar triangles above

$\implies \dfrac{7 - 10x}{10x} = \dfrac{10x}{3 - 10x} \implies x = \dfrac{21}{100} \implies A_{red} = 4(\dfrac{21}{100})^2 = (\dfrac{21}{50})^2 = \dfrac{441}{2500} = \dfrac{a}{b} \implies a + b = \boxed{2941}.$

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This problem reminds me of deriving the pythagorean theorem using areas.

The area of the square $(a + b)^2 = 2ab + c^2 \implies a^2 + b^2 = c^2$ .

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I posted a somewhat similar problem:

Let the diagram above be extended to a regular $n-gon$ and $A_{j}$ , where $(1 \leq j \leq n)$ , be the area of one of the red triangles.

If $\lim_{n \rightarrow \infty} \sum_{j = 1}^{n} A_{j} = \pi(a^2 + 2ab - 12b^2)$ find $\dfrac{a}{b}$ .

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The area of what simple closed curve does $\sum_{j = 1}^{n} A_{j}$ converge to?

Link to above problem

The big square is four congruent right triangles plus the red square. Call the sides of those four congruent right triangles $a,b,c,$ with $c$ the hypotenuse. Then $\frac{c^2}{(a+b)^2} = \frac{29}{50},$ but $c^2 = a^2+b^2$ by Pythagoras, so $\frac{a^2+b^2}{(a+b)^2} = \frac{29}{50}$ and subtracting both sides from $1$ gives $\frac{2ab}{(a+b)^2} = \frac{21}{50}.$ Now the side length $x$ of the four yellow squares is half the harmonic mean of $a$ and $b$ , $x = \frac{ab}{a+b},$ and the ratio of the areas of the four yellow squares to the big square is $\frac{4\left(\frac{ab}{a+b}\right)^2}{(a+b)^2} = \frac{4(ab)^2}{(a+b)^4}.$ But this is just the square of the quantity we looked at above. So the answer is $\left( \frac{21}{50}\right)^2 = \frac{441}{2500},$ and the answer is $\fbox{2941}.$

Another way to get the dimensions of the grey triangles:

Label the diagram as per David Vreken's solution.

Let the large square be a unit square. The area of the blue square is $\frac{29}{50}$ , so the area of the four grey triangles is $\frac{21}{50}$ . Thus, by symmetry, the area of the four blue triangles is $\frac{21}{50}$ . Thus, the area of the small inner blue square is $\frac{8}{50}=\frac{4}{25}$ . Thus its side length is $\sqrt{\frac{4}{25}}=\frac{2}{5}$ . Looking at the sum of vertical measures, the length of the inner blue square is $1-2x$ . Thus $1-2x=\frac{2}{5} \Rightarrow x=\frac{3}{10}$ . Now $y=(x)+(1-2x)=(1-x)=\frac{7}{10}$ .

From here proceed as per David Vreken's solution to find the side length of the red squares and thus their area.

Let S = side of large square, d = side of red square, t = side of yellow square, x = distance from a corner of large square to closest point where red squae touches large square. Then, d^2 = x^2 + S - x)^2 gives a quadratic in x. Substituting d^2 = (29/50)S62, and solving the quadratic for x, x = 3S/10. The Blue area in one corner = (1/2)(t)(x - t) + (1/2)(t)(S - x - t). Then the total Blue area is (multiplying by 4 and simplifying), 2tS - 4t^2.We have Blue area = 2tS - 4t^2, yellow area = 4t^2, red area = (29/50(S^2, and total area = S^2. Equating the sum of area to S^2, we have 2tS =(21/50)S^2, or t = 21S/100. The ratio of interest is 4t^2/S62, which is 4*441/10000, which reduces to 441/2500, and 441 + 2500 = 2941. Ed Gray