+2864 
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
+26393 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:
\(\begin{array}{|rcll|} \hline \sin^2(\alpha) + \cos^2(\alpha) &=& 1 \\\\ \left( \dfrac{1}{x-7}\right)^2 + \left( \dfrac{x\sqrt{2}}{2(x-7)}\right)^2 &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{2x^2}{4(x-7)^2} &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{ x^2}{2(x-7)^2} &=& 1 \quad | \quad \cdot 2(x-7)^2 \\\\ 2 + x^2 &=& 2(x-7)^2 \\ 2 + x^2 &=& 2(x^2-14x+49) \\ 2 + x^2 &=& 2x^2-28x+98 \\ x^2-28x+96 &=& 0 \\\\ x &=& \dfrac{28\pm\sqrt{28^2-4\cdot 96}} {2} \\ &=& \dfrac{28\pm\sqrt{400}} {2} \\ &=& \dfrac{28\pm 20} {2} \\\\ x_1 &=& \dfrac{28+ 20} {2} \\ x_1 &=& \dfrac{48} {2} \\ \mathbf{x_1} &=& \mathbf{24} \\\\ x_2 &=& \dfrac{28- 20} {2} \\ x_2 &=& \dfrac{8} {2} \\ \mathbf{x_2} &=& \mathbf{4} \quad | \quad \text{no solution, while } x > 7cm ! \\\\ \hline \end{array}\)
\(\mathbf{x = 24\ cm}\)
+1713 Really off topic- wish there was a comment thing- for me, on my screen, the 2 length is greater than the 7 length.
+26393 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:
\(\begin{array}{|rcll|} \hline \sin^2(\alpha) + \cos^2(\alpha) &=& 1 \\\\ \left( \dfrac{1}{x-7}\right)^2 + \left( \dfrac{x\sqrt{2}}{2(x-7)}\right)^2 &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{2x^2}{4(x-7)^2} &=& 1 \\\\ \dfrac{1}{(x-7)^2} + \dfrac{ x^2}{2(x-7)^2} &=& 1 \quad | \quad \cdot 2(x-7)^2 \\\\ 2 + x^2 &=& 2(x-7)^2 \\ 2 + x^2 &=& 2(x^2-14x+49) \\ 2 + x^2 &=& 2x^2-28x+98 \\ x^2-28x+96 &=& 0 \\\\ x &=& \dfrac{28\pm\sqrt{28^2-4\cdot 96}} {2} \\ &=& \dfrac{28\pm\sqrt{400}} {2} \\ &=& \dfrac{28\pm 20} {2} \\\\ x_1 &=& \dfrac{28+ 20} {2} \\ x_1 &=& \dfrac{48} {2} \\ \mathbf{x_1} &=& \mathbf{24} \\\\ x_2 &=& \dfrac{28- 20} {2} \\ x_2 &=& \dfrac{8} {2} \\ \mathbf{x_2} &=& \mathbf{4} \quad | \quad \text{no solution, while } x > 7cm ! \\\\ \hline \end{array}\)
\(\mathbf{x = 24\ cm}\)
+1713 Beautiful- and a better diagram! Where did you make the diagram?
-\(tommarvoloriddle\)
+26393 Hallo tommarvoloriddle,
thank you,
i did make the diagram with Microsoft Word !
+1713 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.
-\(tommarvoloriddle\)
+2864 Thank you Heureka! That is a very elegant solution, this problem was so hard all my friends were stumped.
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
