III gimnazjum

Policzmy to razem 3

Rozwiązanie 1

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31637/h7PEkVJMVCp69yxrRHrkJg%3D%3D_8b9f749ed77f8dea922be610b1cee765995c3409443d584f2133a509fd0910c0.png" width="515" height="331"> Oznaczmy sobie wierzchołki tego czworokąta jako A,B,C,D, a środki jego boków jako E,F,G,H ∣AE∣=∣EB∣ ∣BF∣=∣FC∣ a zatem zachodzi ∣EB∣∣AE∣​=∣BF∣∣CF∣​ czyli z tw. Talesa odcinki AC i EF są równoległe. <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31638/0HqBAFU9dEwe4AVXzqVFYA%3D%3D_4186ce00fbacc77ea17ffeaab7d901d29854346ddf04ad68ec740edcf2e10502.png" width="520" height="334"> Z innej zależności tw. Talesa mamy: ∣AB∣∣EB∣​=∣AC∣∣EF∣​ ∣AB∣=∣AE∣+∣EB∣=2⋅∣EB∣ 2∣EB∣∣EB∣​=∣AC∣∣EF∣​ 21​=∣AC∣∣EF∣​

Oznaczamy sobie promień średniego okręgu jako R, a promień najmniejszego okręgu jako r.   <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31651/wOSHKR7ynRgkWrIrV5wDJA%3D%3D_7bfede070b9cf908b57f11082495e15d5a331c69dd6e26e20611196275670d11.jpg"> Zgodnie z twierdzeniem Talesa układamy proporcje: 12∣AD∣​=R∣AC∣​ 1262<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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M834 80h400000v40h-400000z'/></svg>​​=R62<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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