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# ​ Extremely Difficult German Problem

+2
206
8 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

#2
+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:

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

$$\mathbf{x = 24\ cm}$$ Jul 2, 2019

#1
+5

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
+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:

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

$$\mathbf{x = 24\ cm}$$ heureka Jul 2, 2019
#3
+3

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

-$$tommarvoloriddle$$

tommarvoloriddle  Jul 2, 2019
#4
+2

Hallo tommarvoloriddle,

thank you,

i did make the diagram with Microsoft Word ! heureka  Jul 2, 2019
#5
+3

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
+3

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

Jul 2, 2019
#8
+2

Thank you, CalculatorUser ! heureka  Jul 3, 2019
#7
+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