Squares Climbing a Hypotenuse
Difficulty: Medium · Topics: Calculus, Series, Geometric Series
Gipotenuza bo‘ylab ko‘tariluvchi kvadratlar
A right isosceles triangle has the two sides forming the right angle of length 2. Inside it, squares of areas S 1, S 2, S 3, … are drawn without end in the following way.
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First S 1 is the largest square that fits with two of its sides lying along the two legs. Removing it leaves two smaller right isosceles triangles, one at each end of the hypotenuse, and in each of them the largest square is drawn in the same way, giving two squares of area S 2. The construction then continues inside the leftover triangle nearest each acute vertex, giving two squares of area S 3, and so on without end.
If S is the sum of the areas of all these squares, find the value of 6S.
Teng yonli to‘g‘ri burchakli uchburchakning to‘g‘ri burchak hosil qiluvchi ikki tomoni uzunligi 2 ga teng. Uning ichida yuzalari S 1, S 2, S 3, … bo‘lgan kvadratlar quyidagicha cheksiz chiziladi.
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Avval S 1 — ikki tomoni ikkala katet bo‘ylab yotadigan eng katta kvadrat. Uni olib tashlagach, gipotenuzaning har bir uchida ikkita kichikroq teng yonli to‘g‘ri burchakli uchburchak qoladi va ularning har birida xuddi shunday eng katta kvadrat chiziladi — yuzasi S 2 bo‘lgan ikkita kvadrat hosil bo‘ladi. So‘ng qurish har bir o‘tkir uchga eng yaqin qolgan uchburchak ichida davom ettiriladi — yuzasi S 3 bo‘lgan ikkita kvadrat va hokazo, cheksiz.
Agar S — bu kvadratlarning barcha yuzalari yig‘indisi bo‘lsa, 6S ning qiymatini toping.