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Math Help - Geometry problems

  1. #1
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    Geometry problems

    Hello everyone! I'm getting stuck at this problem and I dont know how to solve it. I'm just curious about it. Any help will be greatly appreciated
    Geometry problems-triangle.png
    Last edited by chubakueno; November 23rd 2012 at 08:52 PM.
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  2. #2
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    Re: Geometry problems

    Could you resubmit, including a diagram, with labels for the various points so that we can refer to lines and angles ?
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  3. #3
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    Re: Geometry problems

    thanks for your attention, here is the diagram
    Geometry problems-triangle.png
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  4. #4
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    Re: Geometry problems

    Quote Originally Posted by chubakueno View Post
    thanks for your attention, here is the diagram
    Click image for larger version. 

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    1. You are dealing with 2 right triangles: \Delta(ABD) and \Delta(ACE).

    So the point A is located on a half circle (circle of Thales) over BD and point A is located on a half circle (circle of Thales) over CE.

    2. You certainely have made an exact drawing(?). See attachment.

    a) The point A is the intersection of the green circle and the circle of Thales over BD. The center of the green circle has the coordinates \left(\frac32 , \frac32 \right) with the radius \frac32 \sqrt{2} (why?)

    b) Draw the line AC. Construct a right angle in A on AC. The leg of this right angle intersects the prolonged line BD in E.

    3. I've done the construction in a coordinate system so you can read the length of x.

    4. The line segment BD is divided harmonically by C (inner point) and E (outer point). According to the harmonic partition you'll get the proportion:

    \frac{BC}{CD}=\frac{BE}{DE}

    With your values:

    \frac{3}{2}=\frac{5+x}{x}

    Solve for x.
    Attached Thumbnails Attached Thumbnails Geometry problems-zweirwdineinemkonstr.png  
    Thanks from MarkFL and chubakueno
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  5. #5
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    Re: Geometry problems

    No, it was just illustrative, but that's the way the problem was given to me. Thank you!
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  6. #6
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    Re: Geometry problems

    My solution is rather more workmanlike !
    Start with the right angled triangle ABD and let the side AB = a.
    Then
    a^{2}+AD^{2}=5^{2}, so AD = \sqrt{25-a^{2}}.
    Now let the angle ACD = \theta, and use the sine rule in each of the triangles ABC and ACD.
    From ACD,
    \frac{2}{\sin 45}=\frac{\sqrt{25-a^{2}}}{\sin \theta},
    so
    \sin \theta = \frac{\sin 45\sqrt{25-a^{2}}}{2}.
    From ABC,
    \frac{a}{\sin(180-\theta)}=\frac{3}{\sin 45},
    so
    \sin \theta=\frac{a\sin 45}{3}.
    Equate the two expessions for \sin \theta, simplify, and you find that a=\frac{15}{\sqrt{13}},
    and from which it follows that AD = 10/\sqrt{13}.
    Call the angle ABD \phi, then from the triangle ABD, \tan \phi=\frac{10\sqrt{13}}{15\sqrt{13}}=\frac{2}{3}, and from that you can deduce that \sin \phi=2/\sqrt{13} and \cos \phi=3/\sqrt{13}.
    Now, finally, turn your attention to the triangle at the other end, ADE.
    The angle AED will equal 45 - \phi, so making use of the sine rule again,
    \frac{AD}{\sin (45-\phi)}=\frac{x}{\sin 45},
    so
    x = \frac{AD\sin 45}{\sin (45 - \phi)}=\frac{10\sin 45}{\sqrt{13}(\sin 45 \cos \phi-\cos 45 \sin \phi)}.
    \sin 45 = \cos 45, so after cancelling and then substituting for \cos \phi and \sin \phi,
    we finish up with
    x = \frac{10}{\sqrt{13}(3/\sqrt{13}-2/\sqrt{13})}=10.
    Thanks from MarkFL and Petrus
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