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Geometry Level 4

Diameter A B AB of a circle with center O O has length 2. From the midpoint Q Q of O A OA , a perpendicular is drawn intersecting the circle at P P . Find the radius of the circle which can be inscribed in triangle A P B . APB.

Round your answer to 3 decimal places.


The answer is 0.366.

Sathvik Acharya by Sathvik Acharya , 17, India

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1 solution

c o s A = 0.5 1 cos~A=\dfrac{0.5}{1} ⟹ \implies A = 6 0 ∘ A=60^\circ

It follows that, B = 90 − 60 = 3 0 ∘ B=90-60=30^\circ .

Isolating the triangle with the inscribed cirle,

t a n 30 = r 2 − x tan~30=\dfrac{r}{2-x} ⟹ \implies r = 2 t a n 30 − x t a n 30 r=2tan~30-xtan~30 ( 1 ) \color{#D61F06}(1)

t a n 15 = r x tan~15=\dfrac{r}{x} ⟹ \implies r = x t a n 15 r=xtan~15 ( 2 ) \color{#D61F06}(2)

Equating r = r r=r , we get

2 t a n 30 − x t a n 30 = x t a n 15 2tan~30-xtan~30=xtan~15

x = 1.366025404 x=1.366025404

Finally,

r = 1.366025404 ( t a n 15 ) = r=1.366025404(tan~15)= 0.366 \boxed{0.366}

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we can also use in-radius formula: r = 2 Δ a + b + c r=\dfrac{2\Delta}{a+b+c} where a,b,c are the sides of the triangle.

Ayush G Rai - 4 years, 1 month ago

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