+0

# 2015 SS 28

0
284
5 Answer: 3sqrt(5)

Mar 18, 2019

#1
+5

2015 SS 28  1.

cos-theorem:

$$\begin{array}{|rcll|} \hline BM^2&=&x^2+\left(\dfrac{x}{2}\right)^2-2\cdot x \cdot \dfrac{x}{2}\cos(108^\circ) \\ BM^2 &=& x^2\left(1+\dfrac{1}{4}-\cos(108^\circ)\right) \quad |\quad \cos(108^\circ)=\cos(180^\circ-72^\circ)=-\cos(72^\circ) \\ BM^2 &=& x^2\left( \dfrac{5}{4}+\cos(72^\circ)\right) \\ \mathbf{BM} &\mathbf{=}& \mathbf{x \sqrt{\dfrac{5}{4}+\cos(72^\circ)}} \\ \hline \end{array}$$

2.
sin-theorem:

$$\begin{array}{|rcll|} \hline \dfrac{\sin(z)}{x} &=& \dfrac{\sin(36^\circ)}{ZB} \\\\ \mathbf{ZB} &\mathbf{=}& \mathbf{\dfrac{\sin(36^\circ)}{\sin(z)}x }\\ \hline \end{array}$$

3.

sin-theorem:

$$\begin{array}{|rcll|} \hline \dfrac{\sin(180^\circ-z)}{\frac{x}{2}} &=& \dfrac{\sin(72^\circ)}{ZM} \\\\ \dfrac{2\sin(z)}{x} &=& \dfrac{\sin(72^\circ)}{ZM} \\\\ \mathbf{ZM} &\mathbf{=}& \mathbf{\dfrac{\sin(72^\circ)}{2\sin(z)}x }\\ \hline \end{array}$$

$$\mathbf{z=\ ?}$$

$$\begin{array}{|rcll|} \hline BM &=& ZB+ZM \\ BM&=& \dfrac{\sin(36^\circ)}{\sin(z)}x+\dfrac{\sin(72^\circ)}{2\sin(z)}x \\\\ BM&=& \dfrac{x}{\sin(z)} \left( \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2}\right) \quad | \quad BM=x \sqrt{\dfrac{5}{4}+\cos(72^\circ)} \\\\ x \sqrt{\dfrac{5}{4}+\cos(72^\circ)}&=& \dfrac{x}{\sin(z)} \left( \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2}\right) \\\\ \sqrt{\dfrac{5}{4}+\cos(72^\circ)}&=& \dfrac{1}{\sin(z)} \left( \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2}\right) \\\\ \mathbf{\sin(z)} & \mathbf{=} & \mathbf{ \dfrac{ \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2} }{\sqrt{\dfrac{5}{4}+\cos(72^\circ)}} } \\\\ z &=& 180^\circ- \arcsin\left( \dfrac{ \sin(36^\circ) + \dfrac{\sin(72^\circ)}{2} }{\sqrt{\dfrac{5}{4}+\cos(72^\circ)}}\right) \\ &=& 180^\circ- \arcsin\left( \dfrac{ 1.06331351044 }{1.24860602048} \right) \\ &=& 180^\circ- \arcsin\left( 0.85160049928 \right) \\ &=& 180^\circ- 58.3861775592^\circ \\ \mathbf{z} &\mathbf{=}& \mathbf{121.613822441^\circ} \\ \hline \end{array}$$

$$\text{Let \angle ABM =180^\circ-(36^\circ+z) } \\ \text{Let  36^\circ+z = 36^\circ+ 121.613822441^\circ=157.613822441^\circ}$$

4.
sin-theorem:

$$\begin{array}{|rcll|} \hline \dfrac{\sin(z)}{x} &=& \dfrac{\sin(ABM)}{3} \quad | \quad \angle ABM =180^\circ-(36^\circ+z) \\\\ \dfrac{\sin(z)}{x} &=& \dfrac{\sin\Big(180^\circ-(36^\circ+z)\Big)}{3} \\\\ \dfrac{\sin(z)}{x} &=& \dfrac{\sin( 36^\circ+z )}{3} \\\\ \mathbf{x} &\mathbf{=}& \mathbf{3\cdot \dfrac{\sin(z)}{\sin( 36^\circ+z )} } \\\\ x & = & 3\cdot \dfrac{0.85160049928}{\sin( 157.613822441^\circ )} \\\\ x & = & 3\cdot \dfrac{0.85160049928}{0.38084732121} \\\\ x & = & 3\cdot 2.23606797751 \quad | \quad 2.23606797751 = \sqrt{5} \\ \mathbf{x} &\mathbf{=}& \mathbf{3\cdot \sqrt{5}} \\ \hline \end{array}$$ Mar 20, 2019
#2
+1

Thanks you. This is a great solution! But is there a way to do it without calc, because in this competition, calculators are not allowed.

dgfgrafgdfge111  Mar 21, 2019
edited by dgfgrafgdfge111  Mar 21, 2019
#3
0

dgf.....

Is that how you thank someone when they do hours ( quite possibly)  of work for you at no cost to you ?

I wonder how you would feel if it was the other way around?

I notice that you gave yourself a point for your questions but never even gave Heureka one for his beautifully laid out answer.

No point, no thankyou.

It seems that my first impression of you was correct after all.

Melody  Mar 21, 2019
edited by Melody  Mar 21, 2019
#4
+1

Okay, okay. I'm sorry. I just forgot to give him a point. I was just trying to ask if there was a way to do it without a calc. That's all.

dgfgrafgdfge111  Mar 21, 2019
#5
0

You never forget to give yourself points!

And you still have not said thank you.

You have not left any comment or questions about the answer either.

Did you even bother to look at it?

Melody  Mar 21, 2019