Incircle Frustation

Geometry Level 3

Let Δ A B C \Delta ABC be right-angled, with A = 9 0 \angle A = 90 ^ \circ , & let its circumcircle be τ \tau . Let ω \omega be the circle which touches sides A B AB & A C AC & circle τ \tau (internally). Show that radius of ω \omega is given by

r = 2 b c a + b + c r = \frac{2bc}{a+b+c} ,

& find the ratio : r r c \frac {r}{r^c}

where r c r^c is the radius of the incircle of Δ A B C \Delta ABC .

Note : You can use the formula for r r directly, but prove it if you want an extra challenge!


The answer is 2.

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1 solution

If you hadn't told the formula it would have a harder question. Cause finding the radius of incirle is easy. Formula for incircle is a r e a / s e m i p e r i m e t e r area /semiperimeter

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