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Thread: inequality last

  1. #1
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    inequality last

    ABC is a triangle. BC is perpendicular to CD. |AD|=|DB|=6 units.
    |CA|=x units. What is the sum of the biggest and smallest values that x can assume?inequality last-ads-z.png

    I drew paralel AK to DC. |DC|=a and |BC|=s |AK|=2a and |CK|=s

    let s=5.9 and a=0.8. Then smallest value of x is 6.
    But if we sum 5.9+6 =11.9. Then biggest integer value for x is 11. Then the answer must be 17. But the book says 18.
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    Re: inequality last

    $s^2+a^2 = 6^2$
    $s^2+(2a)^2 = x^2$

    Subtracting gives $3a^2 = x^2-6^2$ so $a = \sqrt{\dfrac{x^2-36}{3}}$

    Since $a$ is a positive real, it must be that $x^2>36$ so $x>6$.

    Next, $s^2 = 6^2-a^2 = 36-\dfrac{x^2-36}{3}$.

    $3s^2 = 36\cdot 3 - x^2-36 = 72-x^2$
    $s^2 = \dfrac{72-x^2}{3}$
    $s = \sqrt{\dfrac{72-x^2}{3}}$

    Again, $s$ is a positive real number, so $72>x^2$ implies $6\sqrt{2}>x$.

    Now, you have $6<x<6\sqrt{2}$. So, if $x$ is an integer, then $x=7$ or $x=8$.
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    Re: inequality last

    Isn't it -x^2+36/3? There is - in front of the x^2-36/3.
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    Re: inequality last

    IF we multiply x^2+s^2=36 with (-),it becomes a^2=x^2-36, Am I wrong?
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    Re: inequality last

    Quote Originally Posted by kastamonu View Post
    Isn't it (-x^2+36)/3? There is - in front of the (x^2-36)/3.
    Yes, and when you distribute you get exactly what I showed. I just skipped steps. I also added parentheses because what you wrote was not correct without them.
    Last edited by SlipEternal; Oct 17th 2017 at 04:07 PM.
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    Re: inequality last

    Quote Originally Posted by kastamonu View Post
    IF we multiply x^2+s^2=36 with (-),it becomes a^2=x^2-36, Am I wrong?
    Check the math again
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    Re: inequality last

    Quote Originally Posted by SlipEternal View Post

    Since $a$ is a positive real, it must be that $x^2>36$ so $x>6$.

    Next, $s^2 = 6^2-a^2 = 36-\dfrac{x^2-36}{3}$.

    $3s^2 = 36\cdot 3 - x^2-36 = 72-x^2$

    $s^2 = \dfrac{72-x^2}{3}$

    $s = \sqrt{\dfrac{72-x^2}{3}}$

    Again, $s$ is a positive real number, so $72>x^2$ implies $6\sqrt{2}>x$.

    Now, you have $6<x<6\sqrt{2}$. So, if $x$ is an integer, then $x=7$ or $x=8$.
    SlipEternal, I was checking some of your math:


    3s^2 = 36\cdot3 - (x^2 - 36) \implies

    3s^2 = 108 - x^2 + 36 \implies

    3s^2 = 144 - x^2 \implies

    s^2 = \dfrac{144 - x^2}{3}

    For s a positive number,

    s = \sqrt{\dfrac{144 - x^2}{3}}

    144 > x^2 \implies

     12 > x
    Last edited by greg1313; Oct 17th 2017 at 04:25 PM.
    Thanks from SlipEternal
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  8. #8
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    Re: inequality last

    Quote Originally Posted by greg1313 View Post
    SlipEternal, I was checking some of your math:


    3s^2 = 36\cdot3 - (x^2 - 36) \implies

    3s^2 = 108 - x^2 + 36 \implies

    3s^2 = 144 - x^2 \implies

    s^2 = \dfrac{144 - x^2}{3}

    For s a positive number,

    s = \sqrt{\dfrac{144 - x^2}{3}}

    144 > x^2 \implies

     12 > x
    Thanks, that makes sense. I completely misunderstood what the OP was saying. A full description of what he meant along with correct grouping symbols and it is obvious that I made a mistake.
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    Re: inequality last

    Your solution seems to be very long and time consuming. Follow my steps:
    Angle cda is an obtuse angel so c<6 ad+dc<12 then ac<12 7 and 11 is the answer.
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    Re: inequality last

    Quote Originally Posted by kastamonu View Post
    ABC is a triangle. BC is perpendicular to CD. |AD|=|DB|=6 units.
    |CA|=x units. What is the sum of the biggest and smallest values that x can assume?Click image for larger version. 

Name:	Adsız.png 
Views:	10 
Size:	6.4 KB 
ID:	38211

    I drew paralel AK to DC. |DC|=a and |BC|=s |AK|=2a and |CK|=s
    Suggestion:
    make right triangle ABK : BK=6, AK=8, AB=10
    so triangle BCD = 3-4-5 and AD=BD=5
    AC = sqrt(3^2 + 8^2)

    It'll be easier to work with.
    Then apply same procedure to your triangle.
    Last edited by DenisB; Oct 18th 2017 at 05:21 AM.
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