We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
70
8
avatar+646 

 

This problem stumped me, especially annoying the fact the 2cm and 7cm look exactly the same length on this *not to scale* image!

 

 

This problem is a real brain smasher

 Jul 2, 2019

Best Answer 

 #2
avatar+22523 
+4

Extremely Difficult German Problem

 

1.

\(\begin{array}{|rcll|} \hline \mathbf{\sin(\alpha)} &=& \mathbf{\dfrac{1}{x-7}} \qquad (1) \\ \hline \end{array}\)

 

2.

\(\begin{array}{|rcll|} \hline \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\sin(180^\circ-(90^\circ-\alpha) ) }{x} \\\\ \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\sin(90^\circ+\alpha) ) }{x} \\\\ \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\cos(\alpha) }{x} \quad | \quad \sin(45^\circ) = \dfrac{\sqrt{2}}{2} \\\\ \dfrac{ \sqrt{2}}{2(x-7)} &=& \dfrac{\cos(\alpha) }{x} \\\\ \mathbf{\cos(\alpha)} &=& \mathbf{ \dfrac{ x\sqrt{2}}{2(x-7)} } \qquad (2) \\ \hline \end{array} \)

 

Phythagoras:

\(\begin{array}{|rcll|} \hline \sin^2(\alpha) + \cos^2(\alpha) &=& 1 \\\\ \left( \dfrac{1}{x-7}\right)^2 + \left( \dfrac{x\sqrt{2}}{2(x-7)}\right)^2 &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{2x^2}{4(x-7)^2} &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{ x^2}{2(x-7)^2} &=& 1 \quad | \quad \cdot 2(x-7)^2 \\\\ 2 + x^2 &=& 2(x-7)^2 \\ 2 + x^2 &=& 2(x^2-14x+49) \\ 2 + x^2 &=& 2x^2-28x+98 \\ x^2-28x+96 &=& 0 \\\\ x &=& \dfrac{28\pm\sqrt{28^2-4\cdot 96}} {2} \\ &=& \dfrac{28\pm\sqrt{400}} {2} \\ &=& \dfrac{28\pm 20} {2} \\\\ x_1 &=& \dfrac{28+ 20} {2} \\ x_1 &=& \dfrac{48} {2} \\ \mathbf{x_1} &=& \mathbf{24} \\\\ x_2 &=& \dfrac{28- 20} {2} \\ x_2 &=& \dfrac{8} {2} \\ \mathbf{x_2} &=& \mathbf{4} \quad | \quad \text{no solution, while } x > 7cm ! \\\\ \hline \end{array}\)

 

\(\mathbf{x = 24\ cm}\)

 

laugh

 Jul 2, 2019
 #1
avatar+625 
0

Really off topic- wish there was a comment thing- for me, on my screen, the 2 length is greater than the 7 length.

 Jul 2, 2019
 #2
avatar+22523 
+4
Best Answer

Extremely Difficult German Problem

 

1.

\(\begin{array}{|rcll|} \hline \mathbf{\sin(\alpha)} &=& \mathbf{\dfrac{1}{x-7}} \qquad (1) \\ \hline \end{array}\)

 

2.

\(\begin{array}{|rcll|} \hline \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\sin(180^\circ-(90^\circ-\alpha) ) }{x} \\\\ \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\sin(90^\circ+\alpha) ) }{x} \\\\ \dfrac{ \sin(45^\circ) }{x-7} &=& \dfrac{\cos(\alpha) }{x} \quad | \quad \sin(45^\circ) = \dfrac{\sqrt{2}}{2} \\\\ \dfrac{ \sqrt{2}}{2(x-7)} &=& \dfrac{\cos(\alpha) }{x} \\\\ \mathbf{\cos(\alpha)} &=& \mathbf{ \dfrac{ x\sqrt{2}}{2(x-7)} } \qquad (2) \\ \hline \end{array} \)

 

Phythagoras:

\(\begin{array}{|rcll|} \hline \sin^2(\alpha) + \cos^2(\alpha) &=& 1 \\\\ \left( \dfrac{1}{x-7}\right)^2 + \left( \dfrac{x\sqrt{2}}{2(x-7)}\right)^2 &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{2x^2}{4(x-7)^2} &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{ x^2}{2(x-7)^2} &=& 1 \quad | \quad \cdot 2(x-7)^2 \\\\ 2 + x^2 &=& 2(x-7)^2 \\ 2 + x^2 &=& 2(x^2-14x+49) \\ 2 + x^2 &=& 2x^2-28x+98 \\ x^2-28x+96 &=& 0 \\\\ x &=& \dfrac{28\pm\sqrt{28^2-4\cdot 96}} {2} \\ &=& \dfrac{28\pm\sqrt{400}} {2} \\ &=& \dfrac{28\pm 20} {2} \\\\ x_1 &=& \dfrac{28+ 20} {2} \\ x_1 &=& \dfrac{48} {2} \\ \mathbf{x_1} &=& \mathbf{24} \\\\ x_2 &=& \dfrac{28- 20} {2} \\ x_2 &=& \dfrac{8} {2} \\ \mathbf{x_2} &=& \mathbf{4} \quad | \quad \text{no solution, while } x > 7cm ! \\\\ \hline \end{array}\)

 

\(\mathbf{x = 24\ cm}\)

 

laugh

heureka Jul 2, 2019
 #3
avatar+625 
0

Beautiful- and a better diagram! Where did you make the diagram?

 

-\(tommarvoloriddle\)

tommarvoloriddle  Jul 2, 2019
 #4
avatar+22523 
+2

Hallo tommarvoloriddle,

 

thank you,

i did make the diagram with Microsoft Word !

 

laugh

heureka  Jul 2, 2019
 #5
avatar+625 
0

Hallo Heureka, your name is hard to spell, thank you a ton, how do you make a diagram in Microsoft word?🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️🤷‍♀️ and why does this feel like a mirror on the wall thing? It kinda feels like it rhymes.

 

-\(tommarvoloriddle\)

tommarvoloriddle  Jul 2, 2019
 #6
avatar+646 
+2

Thank you Heureka! That is a very elegant solution, this problem was so hard all my friends were stumped.

 Jul 2, 2019
 #8
avatar+22523 
+2

Thank you, CalculatorUser !

 

laugh

heureka  Jul 3, 2019
 #7
avatar
+3

alternatively, you can derive the following equation from heureka's diagram-

 

label top left corner A, center of square B, and the bottom (or top, it's completely symmetrical) point where the circle intersects the diagonal of the square C. the length of AB is \(\frac{x}{\sqrt{2}}\) , the length of BC is 1 and the length of AC is x-7. angle ABC is a right angle so we can use pythagoras and solve for x

 Jul 2, 2019
edited by Guest  Jul 2, 2019

5 Online Users