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Thread: problem

  1. #1
    Super Member dhiab's Avatar
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    problem

    problem-37246809_176464033220423_2765168011806507008_n.jpg
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  2. #2
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    Re: problem

    From left to right: 15, 12, 20.
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  3. #3
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    Re: problem

    Quote Originally Posted by SlipEternal View Post
    From left to right: 15, 12, 20.
    Because $\overline{CM}$ is a mean proportional between $9 ~\&~16$ so we have:
    $ \begin{align*}\dfrac{9}{CM}&=\dfrac{CM}{16}\\CM^2& =9\cdot 16\\CM&=12 \end{align*}$

    Thus $AC=\sqrt{(~?~)^2+(12^2)}~\&~BC=\sqrt{(~?~)^2+(12^ 2)}$

    problem-mp.gif
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    Re: problem

    Using Plato's diagram:
    you DO know that angleABC = angleACM and angleBAC = angleBCM ?
    In case your teacher wants you to solve using similar triangles......
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    Re: problem

    Quote Originally Posted by DenisB View Post
    Using Plato's diagram:
    you DO know that angleABC = angleACM and angleBAC = angleBCM ?
    In case your teacher wants you to solve using similar triangles......
    @Denis, two angles cannot be equal. If $\angle ACB=\angle AMC$ then that is only one angle.
    But $\angle ACB=\overrightarrow {CA} \cup \overrightarrow {CB}$ AND $\angle AMC=\overrightarrow {MA} \cup \overrightarrow {MC}$ which shows that those angles cannot be the same angle.

    Now it is true that $m(\angle ACB)=\dfrac{\pi}{2}=m(\angle AMC)$
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    Re: problem

    Quote Originally Posted by Plato View Post
    Now it is true that $m(\angle ACB)=\dfrac{\pi}{2}=m(\angle AMC)$
    mmmmm.....
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    Re: problem

    Click image for larger version. 

Name:	MP.gif 
Views:	2 
Size:	2.2 KB 
ID:	38854

    I'll refer to Plato's diagram. Because Triangle ACB is a right triangle, and Line segment CM is perpendicular to AB, and meets Triangle ACB at vertex C
    (which is across from the hypotenuse), then this forms three right triangles, all of which are similar to each other:

    Triangle AMC, Triangle MCB, Triangle ACB

    Let a = measure of Line segment AC
    Let b = measure of Line segment CM
    Let c = measure of Line segment CB

    I will look at the respective ratios of the shorter legs to the hypotenuses, in order from smaller to larger areas of triangles:

    $\displaystyle \dfrac{9}{a} \ = \ \dfrac{b}{c} \ = \ \dfrac{a}{25}$

    $\displaystyle \dfrac{9}{a} \ = \ \dfrac{a}{25} \ \implies \ a \ = \ 15$


    You might see the value of b right away as being a multiple from the common 3, 4, 5 Pythagorean Triple.

    Or, use a version of the Pythagorean Theorem to find it. (b = 12)


    Then you could use this to solve for c:

    $\displaystyle \dfrac{b}{c} \ = \ \dfrac{a}{25} $

    $\displaystyle \dfrac{12}{c} \ = \ \dfrac{15}{25} $

    $\displaystyle \dfrac{12}{c} \ = \ \dfrac{3}{5} $


    $\displaystyle \cdots$
    Last edited by greg1313; Jul 18th 2018 at 08:57 AM.
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  8. #8
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    Re: problem

    Quote Originally Posted by greg1313 View Post
    Click image for larger version. 

Name:	MP.gif 
Views:	2 
Size:	2.2 KB 
ID:	38854

    I'll refer to Plato's diagram. Because triangle ACB is a right triangle, and line segment CM is perpendicular to AB, and meets the triangle ACB at vertex C
    (which is across from the hypotenuse), then this forms three right triangles, all of which are similar to each other:

    triangle AMC, triangle MCB, triangle ACB

    Let a = measure of line segment AC
    Let b = measure of line segment CM
    Let c = measure of line segment CB

    I will look at the respective ratios of the shorter legs to the hypotenuses, in order from smaller to larger areas of triangles:

    $\displaystyle \dfrac{9}{a} \ = \ \dfrac{b}{c} \ = \ \dfrac{a}{25}$

    $\displaystyle \dfrac{9}{a} \ = \ \dfrac{a}{25} \ \implies \ a \ = \ 15$


    You might see b right away as being a multiple from the common 3, 4, 5 Pythagorean Triple.

    Or, use a version of the Pythagorean Theorem to find it. (b = 12)


    Then you could use this to solve for c:

    $\displaystyle \dfrac{b}{c} \ = \ \dfrac{a}{25} $

    $\displaystyle \dfrac{12}{c} \ = \ \dfrac{15}{25} $

    $\displaystyle \dfrac{12}{c} \ = \ \dfrac{3}{5} $


    $\displaystyle \cdots$
    I used the Pythagorean Theorem three times to get:

    $$\begin{align*}b^2+9^2 & = a^2 \\ b^2+16^2 & =c^2 \\ a^2+c^2 & = (9+16)^2\end{align*}$$

    Let $x=a^2, y=b^2, z=c^2$. This gives:

    $$\begin{align*}y+81 & = x \\ y+256 & = z \\ x+z & = 625\end{align*}$$

    And from here, we can easily solve the system of three equations:

    Plugging in for $x$ from the first equation into the last equation gives:

    $$y+z=544$$

    Added to the middle equation $z-y=256$ gives:

    $$2z=800 \Longrightarrow z=400$$

    $$400+y=544 \Longrightarrow y=144$$

    $$x=144+81 \Longrightarrow x = 225$$

    Taking square roots gives the lengths of the sides, as well.
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    Re: problem

    Triangle AMC is similar to triangle BMC; so:

    AM/CM = CM/BM
    9/CM = CM/16
    CM^2 = 144
    CM = 12

    Use Big Pete twice to get AC = 15 and BC = 20
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  10. #10
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    Re: problem

    @All, note that is question was on an 1869 admission test for MIT.
    I think that the question was designed to test the use of the mean proportional theorem.
    Recall that in 1869 these answers would require the showing of all work.
    The grader would have been look for certain methods.
    Thanks from topsquark
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  11. #11
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    Re: problem

    a=BC, b=AC, m=CM, u=AM, v=BM

    Givens: u and v

    General case:
    m = sqrt(u*v)
    a = sqrt(v^2 + m^2)
    b = sqrt(u^2 + m^2)

    ....and he rode into the sunset!!
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