# Check below? (geometry involved)

## Given isosceles triangle $A \hat{B} C$ $\left(A B = A C\right)$ circumscribed in a circle with radius $r = 1$ . Consider $x$ the height of the triangle $A \hat{B} C$ from the vertex $A$. a) Prove that $B C = 2 \sqrt{\frac{x}{x - 2}}$ $\textcolor{w h i t e}{a a}$ , $x > 2$ b) Find the value of $x$ for which the area of the triangle $A \hat{B} C$ is minimum c) The side $B C$ of the triangle is changing with a rate of $\sqrt{3}$ $\frac{c m}{\sec}$. Find the rate of change for the angle $\hat{A}$ when the triangle becomes equilateral (Area is given as a function of $x$)

May 28, 2018

PART a):

#### Explanation:

Have a look:

I tried this:

May 28, 2018

PART b): (but check my maths anyway)

#### Explanation:

Have a look:

May 28, 2018

PART c) BUT I am not sure about it...I think it is wrong...

Have a look:

May 29, 2018

Part c

#### Explanation:

c)

Take into account that while the base $B C$ of the triangle increases, the height $A M$ decreases.

Based on the above,

Consider hatA=2φ, $\textcolor{w h i t e}{a a}$ φ$\in$(0,π/2)

We have

• ΔAEI: sinφ=1/(AI) $\iff$ AI=1/sinφ

• AM=AI+IM=1/sinφ+1=(1+sinφ)/sinφ

In ΔAMB: tanφ=(MB)/(MA) $\iff$ MB=MAtanφ

$\iff$ y=(1+sinφ)/sinφ*sinφ/cosφ $\iff$

y=(1+sinφ)/cosφ $\iff$ y=1/cosφ+tanφ

$\iff$ y(t)=1/cos(φ(t))+tan(φ(t))

Differentiating in respect to $t$ we get

y'(t)=(sin(φ(t))/cos^2(φ(t))+1/cos^2(φ(t)))φ'(t)

For $t = {t}_{0}$ ,

φ=30°

and $y ' \left({t}_{0}\right) = \frac{\sqrt{3}}{2}$

Thus, since cosφ=cos30°=sqrt3/2 and sinφ=sin30°=1/2

we have

sqrt3/2=((1/2)/(3/4)+(1/3)/(3/4))φ'(t_0) $\iff$

sqrt3/2=(2/3+4/3)φ'(t_0) $\iff$

sqrt3/2=2φ'(t_0) $\iff$

φ'(t_0)=sqrt3/4

But hatA=ω(t), ω(t)=2φ(t)

therefore, ω'(t_0)=2φ'(t_0)=2sqrt3/4=sqrt3/2 $\frac{r a d}{\sec}$

(Note: The moment when the triangle becomes equilateral $A I$ is also the center of mass and $A M = 3 A I = 3$, $x = 3$ and height= $\sqrt{3}$)