There are 3 tangent circumferences inscribed in an isosceles right triangle Two of these circumferences have radius R and are tangent to the hypotenuse and to the two cathetus. The smaller circumference has radius r and is tangent to the two cathetus. How can I find the radius of the smaller circumference as a function of R? (I want a tip on how to solve the problem).


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Question Number 71819 by Maclaurin Stickker last updated on 20/Oct/19

$T h e r e a r e 3 t a n g e n t c i r c u m f e r e n c e s$ $i n s c r i b e d i n a n i s o s c e l e s r i g h t t r i a n g l e$ $T w o o f t h e s e c i r c u m f e r e n c e s h a v e$ $r a d i u s R a n d a r e t a n g e n t t o t h e$ $h y p o t e n u s e a n d t o t h e t w o c a t h e t u s .$ $T h e s m a l l e r c i r c u m f e r e n c e h a s$ $r a d i u s r a n d i s t a n g e n t t o t h e t w o$ $c a t h e t u s . H o w c a n I f i n d t h e r a d i u s$ $o f t h e s m a l l e r c i r c u m f e r e n c e a s a$ $f u n c t i o n o f R ?$ $(I w a n t a t i p o n h o w t o s o l v e t h e$ $p r o b l e m) .$
Commented by mr W last updated on 20/Oct/19

Commented by mr W last updated on 20/Oct/19

$i s t h i s w h a t y o u m e a n ?$
Commented by Maclaurin Stickker last updated on 20/Oct/19

$Y e s! I^{'} d l i k e a t i p o n h o w t o s o l v e .$
	Answered by mr W last updated on 20/Oct/19

	Commented by mr W last updated on 20/Oct/19

	$A B = \sqrt{2} r + \sqrt{{(r + R)}^{2} - R^{2}} + R = \sqrt{2} r + \sqrt{r (2 R + r)} + R$ $\tan 22.5 ° = \frac{\sin 45 °}{1 + \cos 45 °} = \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \sqrt{2} - 1$ $D C = \frac{R}{\tan 22.5 °} = \frac{R}{\sqrt{2} - 1} = (\sqrt{2} + 1) R$ $B C = R + (\sqrt{2} + 1) R = (2 + \sqrt{2}) R$ $A B = B C$ $\sqrt{2} r + \sqrt{r (2 R + r)} + R = (2 + \sqrt{2}) R$ $\sqrt{r (2 R + r)} = (1 + \sqrt{2}) R - \sqrt{2} r$ $r (2 R + r) = {(1 + \sqrt{2})}^{2} R^{2} + 2 r^{2} - 2 (1 + \sqrt{2}) \sqrt{2} R r$ $\Rightarrow r^{2} - 2 (\sqrt{2} + 3) R r + {(1 + \sqrt{2})}^{2} R^{2} = 0$ $\Rightarrow r = (3 + \sqrt{2} - 2 \sqrt{2 + \sqrt{2}}) R$ $\Rightarrow r = {(\sqrt{2 + \sqrt{2}} - 1)}^{2} R \approx 0.7187 R$
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