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Can you solve this?

 Dec 4, 2018
 #1
avatar+4471 
+2

well...

 

the answer is t=12

 

The algebra involved is formidable w/o software so there is probably a very clever geometric solution.

 

Stay tuned.

 Dec 5, 2018
 #2
avatar+4471 
+2

 

 

we can reduce the the problem to the following

 

\((\text{length of the blue line})^2 =2^2 + (\text{length of the red line }-1)^2\)

 

\((\text{length of blue line})^2 = (t-2)^2 + (t-2)^2 = 2(t-2)^2\\ (\text{length of red line}-1)^2 =(\sqrt{ (t-3)^2+t^2}-1)^2 \\ \text{a modest amount of algebra gets us }\\ 6+2t = 2\sqrt{2 t^2-6 t+9}\\ 3+t=\sqrt{2 t^2-6 t+9}\\ 9+6t+t^2 = 2t^2 - 6t+9\\ 12t=t^2\\ t=12 \)

Rom  Dec 5, 2018

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