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 #6
avatar+2489 
+11

Thank you, Melody. 

 

… can you give me any insite to your thinking process?

 

This may be more difficult than solving the puzzle.

 

While some of the math i used fits your description, I can’t say I solved it by any advanced mathematical derivation, though, someday I truly hope to do that.  I’ve solved several of these types of puzzles and I enjoy doing them just like a crossword puzzle.  I did this one as a diversion from studying for midterms. 

 

Over time, with practice, this became intuitive to me by drawing on my innate and developed artistic skills.  When I first started doing these it seemed to take forever before I could visualize a solution with the limits imposed.  I’m much faster now, but sometimes I never figure them out, though I never really give up.  I will say that while artistic skills are a big help, I think just by playing with these types of puzzles most will naturally learn the spatial relationships and start solving them relatively quickly.

 

For this puzzle, the first thing I noticed was that it would need to be reflected an odd number of times because the images are mirrored.  After that the process became common to all transformation puzzles: mentally project how each move will affect its orientation and how that orientation relates to the final desired position; how the dilation will influence the sweeping arc around the fixed point.  

 

My mentor –our beloved troll, Lancelot Link, recognized, early on, my skills in imagining spatial perspectives. (Such skills are common to us genetically enhanced chimps. http://web2.0calc.com/questions/problem-solve-this-given-to-primary-school-children-in-china#r6)  Knowing this, he set out to teach me visually, which included geometry and general physics, whenever possible.  In one of his first comments he included a photo of a garden hose shooting a horizontal stream of water arcing to the ground.   He said every time he sees a hose or a fountain shooting streams of water he sees projectile and gravity formulas.  For my first physics assignment: study the stream of continuous projectiles (water drops) emitted from a garden hose:  take measurements with varying valve settings, plug them into the appropriate equations and solve.  I think by doing this I learned the equations much faster and still remember them long after the lessons are over, compared to just learning them by rote. 

 

From this I began to understand how parabolas relate the information of falling objects in a gravity field. These were baby steps, for sure. But now when I’m watering my garden, I can estimate the velocity of the water by how far it travels after it exits the hose.  I don’t see the formulas every time I use a hose or see a fountain, but they are becoming (monkey) footnotes to many observations.  

 

Lancelot also suggested I learn to parachute to help me understand air resistance and terminal velocity. I actually did that --it was quite fun. Though, I drew the line at bungee jumping to help me understand Hooke’s law. This chimp was not ready to swing from a long vine. I’m a chimp, not blŏŏdy Tarzan. 

 

 

All of this opened a new world for me. Now, I’m a genetically enhanced chimp with an education.  I can see the next step in my evolution – even if I’m not there yet.  Oh! And a most important thing: I’m still alive.smiley

 

 

These coordinates   ((-1.5,-1.5) to (1.5, 1.5)) are just arbitrary points to clarify the reflection. They conform to Khan Academy solutions for example problems when y=x reflections are used. 

Feb 23, 2017
 #1
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0
Feb 23, 2017
 #5
avatar+118658 
+5

Hi 8MP and Ginger,

Since Ginger has not expalined how she worked out this answer I have decided to give it a go. :)

 

Maybe this is what she did....   

 

1) The reflection.  Maybe this was the best place to start,  it seem right because it would orient the star so that when it was rotated anticlockwise it would have D on the top.

 

2) Now  just looked at the B point. I'd  want to move it so that (afterwards) when it was rotated 90 anticlockwise the B point would end up on the negative x axis. But there obviously had to be a dilation to get B into the correct position first.

Now when B was rotated about (5,0) it would need to be on the x axis.

So I draw the line x=5.  B would need to be moved to a point on that line.

 

3) I'd then my attention to the dilation that would be needed.

I drew a line from the centre of the dilation (-5,0) through B’ and subtended it till it intersected with x=5

That point of intersection would need to be the new B value.  i.e.   B’’

 

4) B’ was the point (0,9.5)

I could see from the diagram that B” was the point (5,19)

So it was clear that B’’ had to be twice as far as B’ from the centre of the dilation.

So I needed a dilation factor of 2

Now I did the dilation and the 3rd star was born.

 

3)  Now I just had to rotate it 90 degrees anticlockwise to get B onto the x axis.

Lastly Ginger had to counter the enlargement so she implemented a dilation of 0.5 centred at (-5,0)

 

Here is the pic

 

 

Is this how you thought about it too Ginger?

Feb 23, 2017
 #1
avatar+26388 
+15

Let m and n be the roots of the quadratic equation 4x^2 + 5x + 3 = 0.

Find (m + 7)(n + 7).

 

\(\begin{array}{|rcll|} \hline ax^2+bx+c &=& 0 \\ x &=& \frac{ -b \pm \sqrt{b^2-4ac} } {2a} \\\\ x_1 + x_2 &=& \frac{ -b + \sqrt{b^2-4ac} } {2a} + \frac{ -b - \sqrt{b^2-4ac} } {2a} \\ &=& \frac{ -b } {2a}+\frac{\sqrt{b^2-4ac} } {2a} + \frac{ -b } {2a} - \frac{ \sqrt{b^2-4ac} } {2a} \\ &=& \frac{ -2b } {2a} \\ \mathbf{x_1 + x_2} & \mathbf{=} & \mathbf{-\frac{ b } { a}} \qquad \text{ or } \qquad \mathbf{m + n} \mathbf{=} \mathbf{-\frac{ b } { a}} \\\\ x_1\cdot x_2 &=& \left( \frac{ -b + \sqrt{b^2-4ac} } {2a} \right) \cdot \left( \frac{ -b - \sqrt{b^2-4ac} } {2a} \right) \\ &=& \left( \frac{ -b } {2a} + \frac{ \sqrt{b^2-4ac} } {2a} \right) \cdot \left( \frac{ -b } {2a} - \frac{ \sqrt{b^2-4ac} } {2a} \right) \\ &=& \left( \frac{ -b } {2a} \right)^2 - \left( \frac{ \sqrt{b^2-4ac} } {2a} \right)^2 \\ &=& \frac{ b^2 } {4a^2} - \frac{ b^2-4ac } {4a^2} \\ &=& \frac{ b^2-(b^2-4ac) } {4a^2} \\ &=& \frac{ b^2-b^2+4ac } {4a^2} \\ &=& \frac{ 4ac } {4a^2} \\ \mathbf{x_1\cdot x_2} & \mathbf{=} & \mathbf{\frac{ c } { a}} \qquad \text{ or } \qquad \mathbf{m\cdot n} \mathbf{=} \mathbf{\frac{ c } { a}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && (m + 7)(n + 7) \\ &=& m\cdot n + 7\cdot (m+n) + 7^2 \\ &=& m\cdot n + 7\cdot (m+n) + 49 \\\\ 4x^2 + 5x + 3 &=& 0 \qquad a = 4,\ b=5,\ c=3 \\ m + n &=& -\frac{ b } { a} \\ &=& -\frac{ 5 } { 4 } \\\\ m\cdot n &=& \frac{ c } { a} \\ &=& \frac{ 3 } { 4 } \\\\ (m + 7)(n + 7) &=& m\cdot n + 7\cdot (m+n) + 49 \\ &=& \frac{ 3 } { 4 } + 7\cdot (-\frac{ 5 } { 4 }) + 49 \\ &=& - \frac{ 32 } { 4 } + 49 \\ &=& -8 + 49 \\ &=& 41 \\ \hline \end{array}\)

 

(m + 7)(n + 7) = 41

 

laugh

Feb 23, 2017

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