A Part of Circle

Geometry Level 5

A quadrant A 0 A 90 A_0 A_{90} of a unit circle is divided into 90 equal parts by inserting the points A 1 , A 2 , … , A 90 A_1,A_2, \ldots, A_{90} on the arc of the circle .

Let k = ∑ i = 1 90 A r ( Δ A 0 O A i ) \displaystyle k = \sum_{i=1}^{90} Ar(\Delta A_0 O A_i) .

Report your answer as ⌊ k ⌋ \lfloor k \rfloor .

Notation : ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 28.

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A Former Brilliant Member by A Former Brilliant Member

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1 solution

Each of the 90 triangles is an isosceles triangles with equal sides =1.
The vertex angles are sin(1^o), sin(2^o), . . .,sin(90^o).
So the total area= 1 2 ∗ ∑ n = 1 90 sin ⁡ ( π 180 ∗ n ) = 28.897. \displaystyle \frac 1 2 * \sum_{n=1}^{90} \sin(\dfrac \pi {180} *n)=\large \color{#D61F06}{ 28.897}.

I obtained the sum of sines from calculator as follows with degree mode.
f(n)=x+sin(10 n+1)+sin(10 n+2)+sin(10 n+3)+sin(10 n+4)+sin(10 n+5)+sin(10 n+6)+sin(10 n+7)+sin(10 n+8)+sin(10 n+9)+sin(10 n+10) . 1st x=0, n=0.
2nd x=answer I get from previous calculation. n=previous n +1.
Till n=8.
I got 57.7943. Divided by 2.
If the calculator takes only radians use π 180 ∗ ( 10 ∗ n + 1 t o 10 ) . \dfrac \pi {180}*(10*n+1~ to~ 10).

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