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Question Number 184633 by ajfour last updated on 09/Jan/23

	Answered by mr W last updated on 09/Jan/23

	Commented by mr W last updated on 09/Jan/23

	${A D}^{2} = a^{2} + b^{2} + 2 a b \cos 30 °$ $\Rightarrow A D = \sqrt{a^{2} + b^{2} + \sqrt{3} a b}$ $\sin \frac{A}{2} = \frac{b}{A D} = \frac{b}{\sqrt{a^{2} + b^{2} + \sqrt{3} a b}}$ $\cos \frac{A}{2} = \frac{\sqrt{a^{2} + \sqrt{3} a b}}{\sqrt{a^{2} + b^{2} + \sqrt{3} a b}}$ $\Rightarrow \sin A = \frac{2 b \sqrt{a^{2} + \sqrt{3} a b}}{a^{2} + b^{2} + \sqrt{3} a b}$ $\frac{p}{\sin A} = 2 a$ $\Rightarrow p = \frac{4 a b \sqrt{a^{2} + \sqrt{3} a b}}{a^{2} + b^{2} + \sqrt{3} a b}$ $\frac{\sin α}{b} = \frac{\sin 30 °}{A D}$ $\Rightarrow \sin α = \frac{b}{2 \sqrt{a^{2} + b^{2} + \sqrt{3} a b}}$ $r = 2 a \cos (\frac{A}{2} - α)$ $q = 2 a \cos (\frac{A}{2} + α)$ $q + r = 4 a \cos \frac{A}{2} \cos α = 4 a \times \frac{\sqrt{a^{2} + \sqrt{3} a b}}{\sqrt{a^{2} + b^{2} + \sqrt{3} a b}} \times \frac{2 a + \sqrt{3} b}{2 \sqrt{a^{2} + b^{2} + \sqrt{3} a b}}$ $q + r = \frac{2 a (2 a + \sqrt{3} b) \sqrt{a^{2} + \sqrt{3} a b}}{a^{2} + b^{2} + \sqrt{3} a b}$ $q + r - 2 \sqrt{{A D}^{2} - b^{2}} = p$ $\frac{2 a (2 a + \sqrt{3} b) \sqrt{a^{2} + \sqrt{3} a b}}{a^{2} + b^{2} + \sqrt{3} a b} - 2 \sqrt{a^{2} + \sqrt{3} a b} = \frac{4 a b \sqrt{a^{2} + \sqrt{3} a b}}{a^{2} + b^{2} + \sqrt{3} a b}$ $\frac{a (2 a + \sqrt{3} b)}{a^{2} + b^{2} + \sqrt{3} a b} - 1 = \frac{2 a b}{a^{2} + b^{2} + \sqrt{3} a b}$ $a^{2} - 2 a b = b^{2}$ ${(\frac{b}{a})}^{2} + 2 (\frac{b}{a}) - 1 = 0$ $\Rightarrow \frac{b}{a} = \sqrt{2} - 1 ✓$
	Commented by mr W last updated on 10/Jan/23

	$t h e r e s u l t \frac{b}{a} = \sqrt{2} - 1 i s i n d e p e n d e n t$ $f r o m t h e a n g l e 60 ° . a n o t h e r e x a m p l e :$
	Commented by mr W last updated on 10/Jan/23

	Commented by mr W last updated on 10/Jan/23

	$\sin θ = \frac{b}{a + b}$ $\cos 2 θ = \frac{2 b}{a}$ $\frac{2 b}{a} = 1 - 2 {(\frac{b}{a + b})}^{2} = \frac{a^{2} - b^{2} + 2 a b}{a^{2} + b^{2} + 2 a b}$ $2 b^{3} + 5 {a b}^{2} = a^{3}$ $λ^{3} - 5 λ - 2 = 0$ $(λ + 2) (λ^{2} - 2 λ - 1) = 0$ $\Rightarrow λ = \frac{a}{b} = 1 + \sqrt{2}$ $\Rightarrow \frac{b}{a} = \sqrt{2} - 1 ✓$
	Commented by MJS_new last updated on 10/Jan/23

	$this means if the center of the circumcircle$ $lies on the incircle the ratio of \frac{r}{R} is constant .$ $same as the distance of the centers of both$ $circles equals the radius of the incircle .$ $\Rightarrow$ $if we let a ⩽ b ⩽ c and a = 1$ $\Rightarrow 1 ⩽ b ⩽ 1 + \frac{\sqrt{2}}{2} \land \sqrt{2} ⩽ c ⩽ 1 + \frac{\sqrt{2}}{2}$ $with the ‘ ‘ {borders}^{″} being the isosceles triangles$ $a = b = 1 \land c = \sqrt{2} and a = 1 \land b = c = 1 + \frac{\sqrt{2}}{2} .$ $can we prove there are no other triangles with$ $\frac{r}{R} = \sqrt{2} - 1; meaning with the center of the$ $circumcircle n o t on the incircle ?$
	Commented by mr W last updated on 10/Jan/23

	$y e s!$ $a c c . t o {E u l e r}^{'} s t h e o r e m t h e d i s t a n c e$ $f r o m i n c e n t e r t o c i r c u m c e n t e r o f a n$ $a r b i t r a r y t r i a n g l e i s$ $d = \sqrt{R^{2} - 2 R r}$ $w h e n d = r, w e a l w a y s h a v e$ $R^{2} - 2 R r = r^{2}$ $\Rightarrow \frac{r}{R} = - 1 + \sqrt{2}$ $v i c e v e r s a, i f \frac{r}{R} = \sqrt{2} - 1, w e h a v e$ $R^{2} - 2 R r = r^{2}, t h e n d = r, t h a t m e a n s$ $c i r c u m c e n t e r l i e s o n t h e i n c i r c l e .$
	Commented by mr W last updated on 10/Jan/23

	Commented by MJS_new last updated on 10/Jan/23

	$thank you, I {didn}^{'} t know this theorem$
	Commented by mr W last updated on 10/Jan/23

	$i o n c e k n e w t h i s t h e o r e m . b u t a s i$ $t r i e d t o s o l v e t h i s q u e s t i o n, i {d i d n}^{'} t$ $h a v e i t i n m i n d . o t h e r w i s e i h a v e h a d$ $t a k e n a s h o r t c u t .$
	Commented by mr W last updated on 10/Jan/23

	$s i r, h a v e y o u g o t a n i d e a h o w t o s o l v e$ $t h e e q u a t i o n s y s t e m i n Q 184607$ $g e n e r a l l y ?$
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