A geometry problem by Paola Ramírez

$\large \angle{CAB} = 17^\circ \text{(Triangle 1 )}$

$\large \text{Now let's find }\angle{CBD} \text{(Triangle 2)}$

$\large \space \space \space \angle{CBD} = 2\angle{CAB} \\ \large \space \space \space \angle{CBD} = 2( 17^\circ) \\ \large \space \space \space \angle{CBD} = 34^\circ$

$\large \text{Now find} \space \angle{EDF} \text{(Triangle 3)}$

$\large \space \space \space \angle{EDF} = 2\angle { CBD} \\ \large \space \space \space \angle { EDF} = 2(34^\circ) \\ \large \space \space \space \angle{EDF} = 68^\circ$

$\large \text{Between} \space \triangle{BCD} \space \text{and} \space \triangle {EDF} \text{, there must be another triangle ( } \triangle {CDE} \text{ (Triangle 4))}$

$\large \text{There is no way Triangle 6 existed because one of the base angle of the triangle} = 68^\circ \times 2 = 136^\circ$

$\large\text{We cant draw such a triangle because then the third vertex would not be formed because the equal sides would never intersect as they move apart}$

$\large \text{Since Triangle 6, 7, 8, 9, ... is no longer existed, let's find out about Triangle 5 }$

$\large \space \space \space \angle {DEC} = \frac{180^\circ-(180^\circ - 68^\circ - 34^\circ)}{2} = 51^\circ \\ \large\space \space \space \angle{DEF} = 180^\circ - 2(68^\circ) = 44^\circ \\ \large \space \space \space \angle {FEG} = 180^\circ - 51^\circ - 44^\circ = 85^\circ \\ \large \space \space \space \angle{FGE} = 85^\circ \\ \large \space \space \space \angle{EFG} = 180^\circ - 2(85^\circ) = 10^\circ \text{( } \triangle{EFG} \text{(Triangle 5))}$

$\large \text{ So there is}\space \color{#D61F06}{\boxed{5}}\space \text {isosceles triangle that can fit in}$

$\large \text{ It's the only solution i can think of :)}$

We let T(n) = 2T (n-1) - T (n-2) with T (1)=17 and T (0)=0. n=2, T (2)=34,...,n=5, T (5)=85. n=6 and T (6)=102 is impossible as 2T (6)>180. Hence, n=5.