Find the maximum value of ∠OMA

Geometry Level 3

Let O O be the origin of the Cartesian coordinate, given a circle C C : x 2 + y 2 = 8 x^2+y^2=8 , and a point A ( 2 , 0 ) A(2,0) . Let M M be an arbitrary point on circle C C . What is the maximum value of ∠ O M A ∠OMA (in degrees)?


The answer is 45.

Alice Smith by Alice Smith , 20, China

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2 solutions

Consider the △ O M A \triangle OMA . We note that O M = 2 2 OM=2 \sqrt 2 , the radius of circle C C , and O A = 2 OA = 2 , the x x -coordinate of A A . Let ∠ O M A = θ \angle OMA = \theta and ∠ M A O = α \angle MAO = \alpha . By sine rule , we have:

sin ⁡ ∠ O M A O A = sin ⁡ ∠ M A O O M sin ⁡ θ 2 = sin ⁡ α 2 2 ⟹ sin ⁡ θ = sin ⁡ α 2 \begin{aligned} \frac {\sin \angle OMA}{OA} & = \frac {\sin \angle MAO}{OM} \\ \frac {\sin \theta}2 & = \frac {\sin \alpha}{2\sqrt 2} \\ \implies \sin \theta & = \frac {\sin \alpha}{\sqrt 2} \end{aligned}

For 0 ∘ ≤ θ ≤ 9 0 ∘ 0^\circ \le \theta \le 90^\circ , the larger θ \theta , the larger sin ⁡ θ \sin \theta and θ \theta is maximum when sin ⁡ θ \sin \theta is maximum or when sin ⁡ α = 1 \sin \alpha = 1 . Then max ⁡ ( sin ⁡ θ ) = 1 2 ⟹ max ⁡ ( θ ) = 45 ∘ \max (\sin \theta) = \frac 1{\sqrt 2} \implies \max (\theta) = \boxed{45}^\circ .

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@Alice Smith , we use backslash "\" a lot in LaTex. For examples: \triangle △ \triangle , \angle ∠ \angle , \in ∈ \in , \frac \pi 2 π 2 \frac \pi 2 , \dfrac 3{23} 3 23 \dfrac 3{23} , \sin \alpha sin ⁡ α \sin \alpha , \cos \beta cos ⁡ β \cos \beta , \tan \gamma tan ⁡ γ \tan \gamma , \ln x ln ⁡ x \ln x , \int 0^\frac \pi 2, \sum {k=0}^\infty (there is an underscore after \int and \sum not shown on screen here), \frac 12 ∫ 0 π 2 , ∑ k = 0 ∞ , 1 2 \int_0^\frac \pi 2, \sum_{k=0}^\infty, \frac 12 and put them after \displaystyle \int 0^\frac \pi 2, \sum {k=0}^\infty, \frac 12 ∫ 0 π 2 , ∑ k = 0 ∞ , 1 2 \displaystyle \int_0^\frac \pi 2, \sum_{k=0}^\infty, \frac 12 . We need only one \displaystyle in front. If you are using \ [ ⋯ \ ] \backslash [ \cdots \backslash] instead of \ ( ⋯ \ ) \backslash( \cdots \backslash) , you don't need \displaystyle.

Chew-Seong Cheong - 1 year, 11 months ago
Zhengxi Gao
Jul 7, 2019

In the triangle OMA , we will get 2 s i n Θ \frac{2}{sinΘ} = 2 2 s i n α \frac{2\sqrt{2}}{sinα} ,because of the sine theorem.
Observing this equation, we will find that sinα gets maximum value when Θ get maximum value.
So when MA is perpendicular to x-axis,we get the maximum value of ∠OMA .
We can easily figure out that sinΘ = 2 2 \frac{\sqrt{2}}{2} .
hence, the maximum value of ∠OMA=45°



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