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Question Number 201037 by mr W last updated on 28/Nov/23

Commented by mr W last updated on 28/Nov/23
$a triangle has sides a, b, c. find the fraction of its area covered by all (infinite) inscribed circles as shown.$
$a t r i a n g l e h a s s i d e s a, b, c . f i n d t h e$ $f r a c t i o n o f i t s a r e a c o v e r e d b y a l l$ $(i n f i n i t e) i n s c r i b e d c i r c l e s a s s h o w n .$
	Answered by mr W last updated on 29/Nov/23

	Commented by mr W last updated on 29/Nov/23

	$\frac{r_{n} - r_{n + 1}}{r_{n} + r_{n + 1}} = \sin \frac{θ}{2} = λ$ $\Rightarrow \frac{r_{n + 1}}{r_{n}} = \frac{1 - λ}{1 + λ} = c o n s t a n t \Rightarrow G . P .$
	Commented by mr W last updated on 29/Nov/23

	Commented by mr W last updated on 29/Nov/23

	$l e t λ_{A} = \sin \frac{A}{2}$ $\frac{r_{A, n + 1}}{r_{A, n}} = . . . = \frac{r_{A, 1}}{r_{0}} = \frac{1 - λ_{A}}{1 + λ_{A}} = k_{A}$ $a r e a o f a l l i n c r i b e d c i r c l e s :$ $A_{I C} = π (\underset{n = 0}{\sum^{\infty}} r_{A, n}^{2} + \underset{n = 0}{\sum^{\infty}} r_{B, n}^{2} + \underset{n = 0}{\sum^{\infty}} r_{C, n}^{2}) - 2 π r_{0}^{2}$ $= π r_{0}^{2} (\frac{1}{1 - k_{A}^{2}} + \frac{1}{1 - k_{B}^{2}} + \frac{1}{1 - k_{C}^{2}} - 2)$ $= π r_{0}^{2} [\frac{1}{1 - {(\frac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}})}^{2}} + \frac{1}{1 - {(\frac{1 - \sin \frac{B}{2}}{1 + \sin \frac{B}{2}})}^{2}} + \frac{1}{1 - {(\frac{1 - \sin \frac{C}{2}}{1 + \sin \frac{C}{2}})}^{2}} - 2]$ $= \frac{4 π Δ^{2}}{{(a + b + c)}^{2}} [\frac{1}{1 - {(\frac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}})}^{2}} + \frac{1}{1 - {(\frac{1 - \sin \frac{B}{2}}{1 + \sin \frac{B}{2}})}^{2}} + \frac{1}{1 - {(\frac{1 - \sin \frac{C}{2}}{1 + \sin \frac{C}{2}})}^{2}} - 2]$ $\frac{A_{I C}}{Δ} = \frac{4 π Δ}{{(a + b + c)}^{2}} [\frac{1}{1 - {(\frac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}})}^{2}} + \frac{1}{1 - {(\frac{1 - \sin \frac{B}{2}}{1 + \sin \frac{B}{2}})}^{2}} + \frac{1}{1 - {(\frac{1 - \sin \frac{C}{2}}{1 + \sin \frac{C}{2}})}^{2}} - 2]$ $= \frac{8 π \sin A \sin B \sin C}{{(\sin A + \sin B + \sin C)}^{2}} [\frac{1}{1 - {(\frac{1 - \sin \frac{A}{2}}{1 + \sin \frac{A}{2}})}^{2}} + \frac{1}{1 - {(\frac{1 - \sin \frac{B}{2}}{1 + \sin \frac{B}{2}})}^{2}} + \frac{1}{1 - {(\frac{1 - \sin \frac{C}{2}}{1 + \sin \frac{C}{2}})}^{2}} - 2]$
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