Circular Segment Party

Geometry Level 4

A circle centered at O O has a radius r r . The line A B ‾ \overline { AB } is the diameter of this circle and the line O E ‾ \overline { OE } is perpendicular with A B ‾ \overline { AB } . The line C D ‾ \overline { CD } is perpendicular with O E ‾ \overline { OE } and passes through point E E .

x x is the length of O E ‾ \overline { OE } that make the area of green area, pink area, and blue area equal.

Which of the following equation is correct?

x r 2 r 2 − x 2 + π 6 = arctan ⁡ r 2 − x 2 r \frac { x }{ { { r }^{ 2 } } } \sqrt { { r }^{ 2 }-{ x }^{ 2 } } +\frac { \pi }{ 6 } =\arctan ^{ }{ \frac { \sqrt { { r }^{ 2 }-{ x }^{ 2 } } }{ r } } x r 2 r 2 − x 2 + π 3 = arcsin ⁡ x r \frac { x }{ { { r }^{ 2 } } } \sqrt { { r }^{ 2 }-{ x }^{ 2 } } +\frac { \pi }{ 3 } =\arcsin { \frac { x }{ r } } x r 2 + x 2 = r 2 arccos ⁡ x r 2 + x 2 − π r 2 6 x\sqrt { { r }^{ 2 }+{ x }^{ 2 } } ={ r }^{ 2 }\arccos ^{ }{ \frac { x }{ \sqrt { { r }^{ 2 }+{ x }^{ 2 } } } } -\frac { \pi { r }^{ 2 } }{ 6 } x r 2 r 2 − x 2 = arcsin ⁡ r 2 − x 2 r − π 6 \frac { x }{ { { r }^{ 2 } } } \sqrt { { r }^{ 2 }-{ x }^{ 2 } } =\arcsin ^{ }{ \frac { \sqrt { { r }^{ 2 }-{ x }^{ 2 } } }{ r } } -\frac { \pi }{ 6 } x r = arccos ⁡ r r 2 + x 2 − π 3 \frac { x }{ { r } } =\arccos ^{ }{ \frac { r }{ \sqrt { { r }^{ 2 }+{ x }^{ 2 } } } } -\frac { \pi }{ 3 } x r 2 r 2 + x 2 + π 6 = arccos ⁡ x r \frac { x }{ { { r }^{ 2 } } } \sqrt { { r }^{ 2 }+{ x }^{ 2 } } +\frac { \pi }{ 6 } =\arccos ^{ }{ \frac { x }{ r } } r x 2 r 2 − x 2 = r 2 arccos ⁡ r 2 − x 2 r − π r 2 6 \frac { r }{ { { x }^{ 2 } } } \sqrt { { r }^{ 2 }-{ x }^{ 2 } } ={ r }^{ 2 }\arccos { \frac { \sqrt { { r }^{ 2 }-{ x }^{ 2 } } }{ r } } -\frac { \pi { r }^{ 2 } }{ 6 } x r 2 r 2 − x 2 = arccos ⁡ r x − π 6 \frac { x }{ { { r }^{ 2 } } } \sqrt { { r }^{ 2 }-{ x }^{ 2 } } =\arccos ^{ }{ \frac { r }{ x } } -\frac { \pi }{ 6 }

Sorawit Nick by Sorawit Nick , 18, Thailand

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1 solution

David Vreken
Mar 6, 2020

Draw C O CO and D O DO and let θ = ∠ C O E = ∠ D O E \theta = \angle COE = \angle DOE , let h = C E h = CE , and let A G A_G be the area of the green section.

Since the green, pink, and blue areas are equal, the green area is one third of half the circle, so A G = 1 6 π r 2 A_G = \frac{1}{6}\pi r^2 .

The green area is also the difference between the area of sector C O D COD and the area of △ C O D \triangle COD , so A G = 1 2 ⋅ 2 θ ⋅ r 2 − 1 2 ⋅ x ⋅ 2 h = θ r 2 − x h A_G = \frac{1}{2} \cdot 2\theta \cdot r^2 - \frac{1}{2} \cdot x \cdot 2h = \theta r^2 - xh .

Therefore, A G = 1 6 π r 2 = θ r 2 − x h A_G = \frac{1}{6}\pi r^2 = \theta r^2 - xh , which can be rearranged to x h r 2 = θ − π 6 \frac{xh}{r^2} = \theta - \frac{\pi}{6}

From right triangle △ C O E \triangle COE , since C O = r CO = r , O E = x OE = x , C E = h CE = h , and ∠ C O E = θ \angle COE = \theta , by trigonometry h = r 2 − x 2 h = \sqrt{r^2 - x^2} and θ = arcsin ⁡ r 2 − x 2 x \theta = \arcsin \frac{\sqrt{r^2 - x^2}}{x} .

Substituting these into x h r 2 = θ − π 6 \frac{xh}{r^2} = \theta - \frac{\pi}{6} gives x r 2 r 2 − x 2 = arcsin ⁡ r 2 − x 2 x − π 6 \boxed{\frac{x}{r^2}\sqrt{r^2 - x^2} = \arcsin \frac{\sqrt{r^2 - x^2}}{x} - \frac{\pi}{6}}

1 Helpful 1 Interesting 0 Brilliant 0 Confused

may be coe = arc sin h/r

trupal panchal - 1 year, 3 months ago

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