Question 1198962
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<pre>
The area of the k-th triangle is  {{{(k/10)^2}}}  of the area of the greatest triangle, A.


Therefore, the area of k-th individual single shape is  

    {{{(k/10)^2}}} - {{{((k-1)/10)^2}}}  of the area of greatest triangle, A,    (1)

or

    {{{a[k]}}} = {{{(1/100)*(k^2 - (k^2-2k+1))*A}}} = {{{(1/100)(2k-1)*A}}}.    (2)



From here, the area of the grey part is

    A(grey) = {{{(1/100)*(1 + 5 + 9 + 13 + 17)*A}}} = {{{(45/100)*A}}}.      (3)

(summing all (2k-1) over k = 1, 3, 5, 7, 9)



The area of the unshaded (white) part is

    A(white) = {{{(1/100)*(3 + 7 + 11 + 15 + 19)}}} = {{{(55/100)*A}}}.    (4)

(summing all (2k-1) over k = 2, 4, 6, 8, 10)



Therefore,  the ratio under the problem's question is

    {{{A(grey)/A(white)}}} = {{{45/55}}} = {{{9/11}}}.    <U>ANSWER</U>
</pre>

Solved.


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Surely, one can avoid calculating unshaded area (4), when the area of grey part (3) is just calculated.


I did it only for the purpose of completeness.