It's All Squares

Level 2

The problem below is a brillant problem of the week that I slightly modified.

The blue square touches each side of the largest square and each of the 4 red squares at the corners of the largest square touch the blue square and the largest square is a unit square.

The unit square is the square in the Cartesian plane with corners at the four points ( 0 , 0 ) , ( 1 , 0 ) , ( 0 , 1 ) (0, 0), (1, 0), (0, 1) , and ( 1 , 1 ) (1, 1) .

Let ( 1 2 < c < 1 ) (\dfrac{1}{\sqrt{2}} < c < 1) .

If c c is a side of the blue square and the sum of the areas of the red squares are A = ∑ j = 1 4 A j = c 4 + 1 − ( 2 + 1 2 ) 2 A = \sum_{j = 1}^{4} A_{j} = c^4 + 1 - (\dfrac{\sqrt{2} + 1}{2})^2

(1): Find the value of c c .

Let ( 1 ≤ j ≤ 4 ) (1 \leq j \leq 4) .

(2): Find the coordinates ( x j , y j ) (x_{j},y_{j}) , where each of the 4 red squares at the corners of the largest square touch the blue square.

Find ∏ j = 1 4 x j + ∏ j = 1 4 y j \prod_{j = 1}^{4} x_{j} + \prod_{j = 1}^{4} y_{j} to eight decimal places.


The answer is 0.02751975.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Jul 24, 2018

Let the side of the largest square have a length of 1 1 .

⟹ 1 = 2 y ( 1 − y ) + c 2 ⟹ 2 y 2 − 2 y + 1 − c 2 = 0 ⟹ y = 1 ± 2 c 2 − 1 2 \implies 1 = 2y(1 - y) + c^2 \implies 2y^2 - 2y + 1 - c^2 = 0 \implies y = \dfrac{1 \pm \sqrt{2c^2 - 1}}{2} where y 1 = 1 + 2 c 2 − 1 2 y_{1} = \dfrac{1 + \sqrt{2c^2 - 1}}{2} and y 2 = 1 − 2 c 2 − 1 2 y_{2} = \dfrac{1 - \sqrt{2c^2 - 1}}{2} and ( 1 2 < c < 1 ) (\dfrac{1}{\sqrt{2}} < c < 1) .

Using similar right triangles below we obtain:

y 1 − x x = x y 2 − x ⟹ y 1 y 2 − ( y 1 + y 2 ) x + x 2 = x 2 ⟹ \dfrac{y_{1} - x}{x} = \dfrac{x}{y_{2} - x} \implies y_{1}y_{2} - (y_{1} + y_{2})x + x^2 = x^2 \implies

x = y 1 y 2 y 1 + y 2 = 1 − c 2 2 ⟹ A = 4 ( 1 − c 2 2 ) 2 = ( 1 − c 2 ) 2 = c 4 + 1 − ( 2 + 1 2 ) 2 \boxed{x = \dfrac{y_{1}y_{2}}{y_{1} + y_{2}} = \dfrac{1 - c^2}{2}} \implies A = 4(\dfrac{1 - c^2}{2})^2 = (1 - c^2)^2 = c^4 + 1 - (\dfrac{\sqrt{2} + 1}{2})^2

⟹ 1 − 2 c 2 + c 4 = c 4 + 1 − ( 2 + 1 2 ) 2 ⟹ 2 c 2 = ( 2 + 1 2 ) 2 ⟹ c = 2 + 1 2 2 \implies 1 - 2c^2 + c^4 = c^4 + 1 - (\dfrac{\sqrt{2} + 1}{2})^2 \implies 2c^2 = (\dfrac{\sqrt{2} + 1}{2})^2 \implies \boxed{c = \dfrac{\sqrt{2} + 1}{2\sqrt{2}}} .

For each ( x j , y j ) (x_{j},y_{j}) , where ( 1 ≤ j ≤ 4 ) (1 \leq j \leq 4) :

The coordinates of the four points are: ( x , x ) , ( 1 − x , x ) , ( x , 1 − x ) , ( 1 − x , 1 − x ) ⟹ ∏ j = 1 4 x j = ( x ( 1 − x ) ) 2 = ( ( 1 − c 2 2 ) ( 1 + c 2 2 ) ) 2 = 1 16 ( 47 − 12 2 64 ) 2 (x,x), (1 - x, x), (x,1 - x),(1 - x,1 - x) \implies \prod_{j = 1}^{4} x_{j} = (x(1 - x))^2 = ((\dfrac{1 - c^2}{2})(\dfrac{1 + c^2}{2}))^2 = \dfrac{1}{16}(\dfrac{47 - 12\sqrt{2}}{64})^2

= ∏ j = 1 4 y j ⟹ ∏ j = 1 4 x j + ∏ j = 1 4 y j = 1 8 ( 47 − 12 2 64 ) 2 ≈ 0.02751975 = \prod_{j = 1}^{4} y_{j} \implies \prod_{j = 1}^{4} x_{j} + \prod_{j = 1}^{4} y_{j} = \dfrac{1}{8}(\dfrac{47 - 12\sqrt{2}}{64})^2 \approx \boxed{0.02751975} .

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