Solveeit Logo

Question

Question: Let \(ABC\) be an equilateral triangle inscribed in a circle \(O\) . \(M\) is a point on the arc \(B...

Let ABCABC be an equilateral triangle inscribed in a circle OO . MM is a point on the arc BCBC . Lines AM,BMAM,BM and CMCM are drawn. Then \mathop {AM}\limits^{\\_\\_\\_\\_\\_} is:
\left( a \right){\text{ equal to }}\mathop {BM}\limits^{\\_\\_\\_\\_\\_} + \mathop {CM}\limits^{\\_\\_\\_\\_\\_}
\left( b \right){\text{ less than }}\mathop {BM}\limits^{\\_\\_\\_\\_\\_} + \mathop {CM}\limits^{\\_\\_\\_\\_\\_}
\left( c \right){\text{ greater than }}\mathop {BM}\limits^{\\_\\_\\_\\_\\_} + \mathop {CM}\limits^{\\_\\_\\_\\_\\_}
\left( d \right){\text{ equal, less than, or greater than }}\mathop {BM}\limits^{\\_\\_\\_\\_\\_} + \mathop {CM}\limits^{\\_\\_\\_\\_\\_} ,{\text{ depending upon position upon M}}

Explanation

Solution

For proving this type of question, all we can do is proceed with the information and figure provided. So we can see that since the OO is the center which lies on AMAM hence, AMAM will be equal to 2R2R. And also AMAM is perpendicular to BCBC and also ADAD is perpendicular to the same. So by using this we will equate both of them and get the value. And on solving it we will be able to find the required equation.

Complete answer:
First of all, we will assume this is an ideal case. And since, from the information provided, we can say that OO is the center which lies on AMAM .
Therefore, AM=2RAM = 2R and here RR will be the radius of the circle.
Also, we can see that AMBCAM \bot BC , so from this OO will be the point which lies on AMAM and BCBC .
Here, ADBCAD \bot BC , so from the figure we can write it in fraction as
BDOB=32\Rightarrow \dfrac{{BD}}{{OB}} = \dfrac{{\sqrt 3 }}{2}
Now let us assume AB=BC=CA=a , OA = OB = OC = RAB = BC = CA = a{\text{ , OA = OB = OC = R}}
Therefore, on substituting the values, we will get the relation as
a2R=32\Rightarrow \dfrac{a}{{2R}} = \dfrac{{\sqrt 3 }}{2}
And by doing the cross multiplication and solving the above equation, we get the equation as
a=3R\Rightarrow a = \sqrt 3 R
Since, we have AM=2RAM = 2R ,
So by using the trigonometric properties, we have
BM=CM=BDcos30\Rightarrow BM = CM = \dfrac{{BD}}{{\cos {{30}^ \circ }}}
And on solving the above line, we get the equation as
a3\Rightarrow \dfrac{a}{{\sqrt 3 }}
So from this now we can have the sum of BM&CMBM\& CM
On adding the sum, we will get 2a3\dfrac{{2a}}{{\sqrt 3 }}
Now on substituting the value of aa , we get the equation as
2×3R3\Rightarrow \dfrac{{2 \times \sqrt 3 R}}{{\sqrt 3 }}
And on solving the above line we get
2R=AM\Rightarrow 2R = AM
Therefore, from this, we can say \mathop {AM}\limits^{\\_\\_\\_\\_\\_} = \mathop {BM}\limits^{\\_\\_\\_\\_\\_} + \mathop {CM}\limits^{\\_\\_\\_\\_\\_} .

Hence, the option (a)\left( a \right) is correct.

Note: For solving this type of question we need to know the geometry and its properties ideas so that we can equate the equation. Also, the use of trigonometry should be known to us as they are used sometimes. So by using the relation we will expand the equation and solve it then we can easily get to the solution.