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Math Help - A problem on triangles (area, ratios)

  1. #31
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    BC =6rad2. If y =BP then PC= 6-rad2-y. PT = TC = (6rad2-y/rad2).PTC is an isosceles rt triangle
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  2. #32
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    I was able to get the TC and PT now. But I was only able to make one x and y equation. Which is triangle APT. Which is very long and complicated btw. Where do I get my other equation from then?
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  3. #33
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    Triangle APC is a triangle on the base AC.The perpendicular PT on AC divides this triangle into APT andPCT. The pythagorean theorem for APT provides one equation in x and y.Triangle BHP using same theorem provides the other.




    bjh
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  4. #34
    MHF Contributor Unknown008's Avatar
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    Oh gosh, I didn't realise I'll come down to having something like 0 = 0 if I used TC in terms of y...

    Anyway, maybe a way avoiding the calculator using my previous method:

    Let \sin\theta_1 = \frac{1}{\sqrt5}

    ( \theta_1 is angle HAM after cancelling the x that I introduced.)

    Let \sin\theta_2 = \frac{2}{\sqrt5}

    ( \theta_2 is angle HAB)

    Angle HBP is then \frac{\pi}{4} - \theta_1

    Angle APB = \pi - \frac{\pi}{4} - \theta_2 = \frac{3\pi}{4} - \theta_2

    Angle APC = \frac{\pi}{4} + \theta_2

    then, using the sine rule:

    Okay, let AB = 1 now for the sake of simplicity and BP = x

    \frac{x}{\sin \left(\theta_2\right)} = \frac{1}{\sin \left(\frac{3\pi}{4} - \theta_2\right)}

    x = \frac{\frac{2}{\sqrt5}}{\sin \left(\frac{3\pi}{4}\right)\cos(\theta_2) - \cos\left(\frac{3\pi}{4}\right)\sin (\theta_2)}

    x = \frac{\frac{2}{\sqrt5}}{\left(\frac{1}{\sqrt2} \right)\left(\frac{1}{\sqrt5}\right) - \left(-\frac{1}{\sqrt2}\right)\left(\frac{2}{\sqrt5} \right)}

    x = \frac{\frac{2}{\sqrt5}}{\left(\frac{3}{\sqrt{10}} \right)}

    x = \frac{2\sqrt2}{3}

    Now, the same with PC, which I call y.

    \frac{y}{\sin \left(\theta_1\right)} = \frac{1}{\sin \left(\frac{\pi}{4} + \theta_2\right)}

    y = \frac{\frac{1}{\sqrt5}}{\sin \left(\frac{\pi}{4} \right)\cos\left(\theta_2\right) + \cos\left(\frac{\pi}{4} \right)\sin\left(\theta_2\right)}

    y = \frac{\frac{1}{\sqrt5}}{\left(\frac{1}{\sqrt2} \right)\left(\frac{1}{\sqrt5}\right) + \left(\frac{1}{\sqrt2}\right)\left(\frac{2}{\sqrt5  } \right)}

    y = \frac{\frac{1}{\sqrt5}}{\left(\frac{3}{\sqrt{10}} \right)}

    y = \frac{\sqrt2}{3}

    You immediately see that x:y = 2:1
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  5. #35
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    My posts related only to showing that there was a solution not requiring trig.And when it is used the answer to this problem is very simple




    bjh
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  6. #36
    MHF Contributor Unknown008's Avatar
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    Well, if you don't want to use trigonometry, maybe a sketch will do:



    When I complete the sketch some image of the initial triangle, I get a square, with two parallel lines running across the diagonal. These lines start at corners and then the next line start right where the previous line started (in this case, the black line parallel to the initial blue line starts at the left, directly opposite to where the blue line ended). This divides the diagonal into three pieces of the same length.

    This applies to rectangles too (second picture) and even a greater number of lines drawn across the diagonal. In the last diagram, the diagonal is equally divided into 4 parts by the three lines.
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  7. #37
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    Thanks a lot guys. But hold on. It's taking me a really long while to absorb these solutions... But am still trying!
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  8. #38
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    Thanks bjhopper,

    But I've been trying to solve this using your x and y method. And I come up with such messy solutions for triangle APT. Something like (12-6rad2-6/2)^2 + (6rad2-y/2)^2 = (rad(171/5) + x)^2 and when your try to simplify it, it becomes HELL. There are like 4 questions like these for a 60 min exam so I don't think it's practical. And this exam is meant for Graduating High school students :/

    Now to try Unknow008's solution!
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  9. #39
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    Thank you very much bjhopper and Unknown008! You two have been a great help! I finally completed my solution using both your approaches
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