in-AB-C-cos-A-cos-B-cos-C-7-4-R-r-

Question Number 170856 by mnjuly1970 last updated on 01/Jun/22

i n A \overset{Δ}{B C} : c o s (A) + c o s (B) + c o s (C) = \frac{7}{4}

\frac{R}{r} = ?

Commented by mr W last updated on 01/Jun/22

\cos A + \cos B + \cos C = \frac{7}{4} n o t p o s s i b l e,

s i n c e \cos A + \cos B + \cos C ⩽ \frac{3}{2} .

Commented by mnjuly1970 last updated on 02/Jun/22

y e s y o u a r e r i g h t s i r \dots

i a p p o l o g i z e \dots i t h i n k \dots

s i n (A) + s i n (B) + s i n (B) \overset{?}{=} 1 + \frac{r}{R}

\overset{R ⩾ 2 r}{⩽} 1 + \frac{1}{2} = \frac{3}{2}

Answered by mr W last updated on 01/Jun/22

a s s u m e \cos A + \cos B + \cos C = \frac{5}{4} .

a = 2 R \sin A, b = 2 R \sin B, c = 2 R \sin C

Δ = \frac{a b \sin C}{2}

r = \frac{2 Δ}{a + b + c}

\frac{R}{r} = \frac{R (a + b + c)}{2 Δ}

\frac{R}{r} = \frac{R (a + b + c)}{a b \sin C}

\frac{R}{r} = \frac{2 R^{2} (\sin A + \sin B + \sin C)}{4 R^{2} \sin A \sin B \sin C}

\frac{R}{r} = \frac{\sin A + \sin B + \sin C}{2 \sin A \sin B \sin C}

\frac{R}{r} = \frac{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{16 \sin \frac{A}{2} \cos \frac{C}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}}

\frac{R}{r} = \frac{1}{4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}

\frac{R}{r} = \frac{1}{\cos A + \cos B + \cos C - 1}

\frac{R}{r} = \frac{1}{\frac{5}{4} - 1} = 4 ✓

Commented by MJS_new last updated on 01/Jun/22

I think {there}^{'} s no such triangle

Commented by mr W last updated on 01/Jun/22

y o u a r e r i g h t s i r!

\cos A + \cos B + \cos C m u s t b e ⩽ \frac{3}{2} .

Commented by mnjuly1970 last updated on 02/Jun/22

t h a n k s a l o t \dots